CBSE Class 12 Chemistry Electrochemistry Worksheet Set 03

Assignment (Electrochemistry)

1. Mention the differences between electrolytic cell and electrochemical cell.

2. what will happen when a) E_ext = 1.10 V, b) E_ext < 1.10 V and c) E_ext>1.10 V is applied on the cell Zn / Zn²⁺ || Cu²⁺/ Cu ?

3.What is SHE ? How will you determine the potential of a) Zn electrode b) Cu electrode using SHE ?

4. E⁰ Zn²⁺ / Zn = - 0.76 V and . E⁰ Cu²⁺ / Cu = + 0.34 V. What does positive and negative sign convey ?

5. Account for the following:

a) Zn displaces hydrogen from dilute HCl while Cu can not.

b) Li is the strongest reducing agent while Fluorine is strongest oxidising agent. 

c) copper sulphate solution can not be stored in zinc vessel. 

E⁰ Cu²⁺ / Cu = + 0.34 V E⁰ Zn²⁺ / Zn = - 0.76 VE⁰ Li+ / Li = - 3.05 V E⁰ F₂ /2 F- = + 2.87 V)

Question. Zn(s)+Cu²⁺(aq)→Zn²⁺(aq)+Cu(s) 
The above redox reaction is used in
a) Galvanic cell
b) Daniell cell
c) Voltaic cell
d) All of these

Answer: d

Question. The conductance of electrolytic solution kept between the electrodes of conductivity cell at unit distance but having area of cross-section large enough to accommodate sufficient volume of solution is called
a) limiting molar conductivity
b) molar conductivity
c) conductivity
d) All of the above

Answer: b

Question. For the electrochemical cell,
Ag⁻ |AgCl|KCl ||AgNO₃|Ag⁺ , the overall cell reaction is
a) Ag + KCl→ AgCl(s)+K⁺
b)  Ag + KgCl→ 2Ag+1/2Cl₂
c) AgCl(s)→ Ag⁺Cl⁻
d) Ag⁺ Cl^–  → AgCl(s)

Answer: c

Question. Calculate the standard cell potential for the following Galvanic cell, Cr|Cr³⁺||Cd |Cd²⁺ [Given, E°_Cr³⁺/C = −0.74V and  E°_Cd²⁺/Cd= −0.40V]
a) 0.74 V
b) −0.34 V
c) + 0.34 V
d) 1.14 V

Answer: c

Question. If E° Zn²⁺/Zn) −0 763 V and E° (Fe2+ /Fe) =−0.44 V, then the emf of the cell Zn|Zn²⁺ (a=0.001)||Fe²⁺    (a=0.005) | 2+ (a = )| Fe is
a) equal to 0.323 V
b) less than 0.323 V
c) greater than 0.323 V
d) equal to 1.103 V

Answer: c

Question. Electrolytic cell is a device
a) in which a non-spontaneous chemical reaction is carried out at the expense of electrical energy
b) in which a spontaneous chemical reaction is carried out to generate electrical energy
c) in which applied opposite potential is less than the cell potential
d) Both (a) and (c)

Answer: a

Question. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl at pH =10 and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be
a) 0.059 V
b) 0.59 V
c) 0.118 V
d) 0.18 V

Answer: b

Question. In the electrochemical cell, Zn|ZnSO₄ (0.01 M)| |CuSO₄ (1.0M) | Cu, the emf of this Daniell cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E₂. From the
following, which one is the correct relationship between E₁ and E₂? ( Given, RT/F= 0.059)
a) E₁ = E₂  
b) E₁ < E₂
c) E_{1 >} E₂
d) E₂ = 0 ≠ E1

Answer: c

Question. The standard Gibbs free energy of Zn (s) + Cu²+(aq)→ Zn²⁺(aq) + Cu(s) is
a) −91FE_cell
b) −2FE_cell
c) −3 FE_cell
d) −4 FE_cell

Answer: b

Question. 1.5 A current is flowing through a metallic wire. If it flows for 3 hrs, how many electrons would flow through the wire?
a) 2.05×10²² electrons
b) 1.0×10²³ electrons
c) 10²⁴ electrons
d) 4.5×10²³ electrons

Answer: b

Question. KCl solution is generally used to determine the cell constant because
a) it is highly ionic in nature
b) its conductivity is known accurately at various concentration and different temperatures
c) size of cations and anions are comparable
d) All of the above

Answer: b

Question. The resistance of the cell containing KCl solution at 23°C was found to be 55Ω. Its cell constant is 0.616 cm⁻¹. The conductivity of KCl solution (Ω⁻¹ cm⁻¹) is
a) 1.21×10⁻³
b) 1.12×10⁻²
c) 1.12×10⁻³
d) 1.21×10⁻²

Answer: b

Question. Standard electrode potential for Sn⁴⁺/ Sn²⁺ couple is +0.15V and that for the Cr³⁺/Cr  couple is −0.74V.
These two couples in their standard state are connected to make a cell. The cell potential will be
a) + 1.83 V
b) +1.19 V
c) + 0.89 V
d) + 0.18 V

Answer: c

Question. If resistance of a conductivity cell filled with 2 mol L⁻¹ KCl solution is 100 Ω. The resistance of the same cell when filled with 0.2 mol L⁻¹ KCl solution is 520 Ω. Then the conductivity of 0.2 mol L⁻¹ KCl solution will be (Given the conductivity of 1 mol L⁻¹ KCl solution is 1.29 S/m.)
a) 0.248 S cm⁻¹
b) 0.248 S m⁻¹
c) 2.48 S cm⁻¹
d) 2.48 S m⁻¹

Answer: b

Question. “Limiting molar conductivity of an electrolyte can be represented as sum of the individual contributions of anion and cation of the electrolyte”.
Which law states the above statement?
a) Henry’s law
b) Debye Onsager’s law
c) Kohlrausch’s law of independent migration of ions
d) All of the above

Answer: c

Question. Molar conductivities (Δ°_m)  at infinite dilution of NaCl, HCl and CH₃COONa  are 126.4,425.9 and 91.0 S cm² mol⁻¹ respectively. Δ°_m for CH₃ COOH will be
a) 425.5 S cm2 mol⁻¹
b) 180.5 S cm2 mol⁻¹
c) 290.85 S cm2 mol⁻¹
d) 390.5 S cm2 mol⁻¹

Answer: d

Question. A 4.0 M aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to evolution of chlorine gas at one of the electrodes. The total charge required for the complete electrolysis will be
a) 96500 C
b) 24125 C
c) 48250 C
d) 193000 C

Answer: d

Question. When aqueous sodium chloride solution is electrolysed
a) at cathode H⁺ is reduced into H₂ instead of Na⁺
b) at cathode Na⁺ is reduced to Na
c) Cl⁻ is oxidised into Cl₂ at cathode
d) Both (b) and (c)

Answer: a

Question. Galvanisation is
a) zinc plating on aluminium sheet
b) zinc plating on iron sheet
c) iron plating on zinc sheet
d) aluminium plating on zinc sheet

Answer: b

Question. The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO₄ electrolysed in g during the process is (Molar mass of PbSO₄ =303g mol⁻¹)
a) 11.4
b) 7.6
c) 15.2
d) 22.8

Answer: b

Question. What will happen during the electrolysis of aqueous solution ofCuSO₄ in the presence of copper electrodes?
a) Copper will deposit at cathode
b) Copper will dissolve at anode
c) Oxygen will be released at anode
d) Both (a) and (b)

Answer: d

Question. A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as
a) fuel cell
b) electrolytic cell
c) dynamo
d) Ni-Cd cell

Answer: a

Question 1. Mention the differences between electrolytic cell and electrochemical cell.
Answer:
The key differences between an electrolytic cell and an electrochemical (galvanic) cell are outlined below:
* Electrochemical Cell: It converts chemical energy from a spontaneous redox reaction into electrical energy. The anode acts as the negative electrode, while the cathode acts as the positive electrode.
* Electrolytic Cell: It utilizes electrical energy from an external power source to force a non-spontaneous redox reaction to occur. The anode is designated as the positive electrode, while the cathode is designated as the negative electrode.
In simple words: An electrochemical cell generates electricity from a chemical reaction (like a battery), whereas an electrolytic cell uses electricity to drive a chemical reaction.

Exam Tip: Remember the mnemonic "LOAN" (Loss of electrons is Oxidation at the Anode which is Negative) to keep electrode polarities straight in galvanic cells.

Question 2. what will happen when a) Eext = 1.10 V, b) Eext < 1.10 V and c) Eext>1.10 V is applied on the cell Zn / Zn2+ || Cu2+/ Cu ?
Answer:
The behavior of the zinc-copper cell under different external potential (\( E_{\text{ext}} \)) conditions is as follows:
* (a) When \( E_{\text{ext}} = 1.10 \text{ V} \): No current flows through the circuit, and no chemical reaction occurs because the opposing external voltage perfectly balances the standard cell potential.
* (b) When \( E_{\text{ext}} < 1.10 \text{ V} \): The cell behaves normally as an electrochemical cell. Electrons flow from the zinc anode to the copper cathode (electric current flows from copper to zinc), zinc dissolves at the anode, and copper deposits at the cathode.
* (c) When \( E_{\text{ext}} > 1.10 \text{ V} \): The reaction is reversed, transforming the setup into an electrolytic cell. Electrons flow from copper to zinc, copper dissolves into the solution, and zinc deposits on the zinc electrode.
In simple words: At 1.10 V, everything stops. Below 1.10 V, the battery runs normally. Above 1.10 V, the reactions run backward, acting like a battery charger.

Exam Tip: Highlighting the direction of electron flow vs. conventional current flow is highly valued by examiners in this question.

Question 3. What is SHE ? How will you determine the potential of a) Zn electrode b) Cu electrode using SHE ?
Answer:
SHE (Standard Hydrogen Electrode): This is a reference electrode whose standard reduction potential is defined as exactly \( 0.00 \text{ V} \) at all temperatures. It consists of platinum foil coated with platinum black, immersed in a \( 1 \text{ M } \text{H}^+ \) solution, with pure hydrogen gas bubbled through it at \( 1 \text{ bar} \) pressure.

* (a) Determination of Zn Electrode Potential: The zinc half-cell is connected to the SHE. Because zinc has a greater tendency to oxidize, it acts as the anode, and the SHE acts as the cathode. The cell is represented as \( \text{Zn} | \text{Zn}^{2+}(1\text{M}) \parallel \text{H}^+(1\text{M}) | \text{H}_2(1\text{bar}) | \text{Pt} \). The measured cell potential is \( 0.76 \text{ V} \). Since \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \), we get:
\[ 0.76 \text{ V} = 0.00 \text{ V} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \implies E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \]
* (b) Determination of Cu Electrode Potential: The copper half-cell is connected to the SHE. Because copper reduces more easily, it acts as the cathode, and the SHE acts as the anode. The cell is represented as \( \text{Pt} | \text{H}_2(1\text{bar}) | \text{H}^+(1\text{M}) \parallel \text{Cu}^{2+}(1\text{M}) | \text{Cu} \). The measured cell potential is \( 0.34 \text{ V} \). Using the formula:
\[ 0.34 \text{ V} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - 0.00 \text{ V} \implies E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \]
In simple words: SHE is a benchmark electrode set at zero volts. By pairing a metal electrode with it, the total measured voltage gives us the exact individual potential of that metal.

Exam Tip: Be sure to write the full cell notation and the corresponding equations when explaining electrode potential determinations.

Question 4. E0 Zn2+ / Zn = - 0.76 V and . E0 Cu2+ / Cu = + 0.34 V. What does positive and negative sign convey ?
Answer:
The signs of the standard reduction potentials represent thermodynamic stability relative to hydrogen:
* Negative Sign (\( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \)): This signifies that zinc is a stronger reducing agent than hydrogen gas. It has a greater tendency to lose electrons (undergo oxidation) compared to hydrogen ions.
* Positive Sign (\( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \)): This indicates that copper is a weaker reducing agent than hydrogen. It has a higher tendency to gain electrons (undergo reduction) compared to hydrogen ions.
In simple words: A minus sign means the metal is highly reactive and readily gives away electrons. A plus sign means the metal is stable and prefers to receive electrons.

Exam Tip: Always relate standard reduction potential values to reducing power: more negative means a stronger reducing agent.

Question 5. Account for the following:
a) Zn displaces hydrogen from dilute HCl while Cu can not.
b) Li is the strongest reducing agent while Fluorine is strongest oxidising agent.
c) copper sulphate solution can not be stored in zinc vessel.
E0 Cu2+ / Cu = + 0.34 V E0 Zn2+ / Zn = - 0.76 VE0 Li+ / Li = - 3.05 V E0 F2 /2 F- = + 2.87 V)
Answer:
* (a) Zinc has a standard reduction potential (\( E^\circ = -0.76 \text{ V} \)) lower than that of hydrogen (\( 0.00 \text{ V} \)), meaning zinc can spontaneously reduce \( \text{H}^+ \) ions to \( \text{H}_2 \) gas. Copper has a positive reduction potential (\( E^\circ = +0.34 \text{ V} \)), which is higher than hydrogen, so it cannot reduce \( \text{H}^+ \) ions.
* (b) Lithium has the most negative standard reduction potential (\( E^\circ = -3.05 \text{ V} \)), giving it the strongest tendency to lose electrons (high oxidation capability) and act as a powerful reducing agent. Conversely, fluorine gas has the most positive reduction potential (\( E^\circ = +2.87 \text{ V} \)), indicating an extreme affinity for electrons, making it the strongest oxidizing agent.
* (c) Zinc has a lower reduction potential (\( -0.76 \text{ V} \)) than copper (\( +0.34 \text{ V} \)). Consequently, zinc is more reactive and will spontaneously displace copper from its salt solution via: \( \text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)} \). This displacement reaction causes the zinc container to dissolve and corrode over time.
In simple words: (a) Zinc is reactive enough to liberate hydrogen gas from acids, but copper is too unreactive. (b) Lithium easily loses its valence electron, while fluorine holds a strong pull on electrons. (c) Zinc reacts with the copper solution, which will eat holes in a zinc container.

Exam Tip: For part (c), always write out the balanced displacement equation to support your written explanation.

Question 6. Write Nernst equation for the following cells at 298 K
a) Cr/ Cr3+ || Fe2+ / Fe
    ( 0.1M)     ( 0.1M)
b) Zn + Sn4+ (1.5M) → Zn2+(0.5M) + Sn2+ (2M)
Answer:
* (a) For the cell \( \text{Cr} | \text{Cr}^{3+}(0.1\text{M}) \parallel \text{Fe}^{2+}(0.1\text{M}) | \text{Fe} \):
The balanced redox reaction is: \( 2\text{Cr(s)} + 3\text{Fe}^{2+}\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{Fe(s)} \), where the number of transferred electrons \( n = 6 \).
The Nernst equation at \( 298 \text{ K} \) is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{6} \log \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} \]
Substituting concentrations:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.1)^3} \]

\implies E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{6} \log(10) = E^\circ_{\text{cell}} - 0.00985 \text{ V}

* (b) For the reaction \( \text{Zn} + \text{Sn}^{4+} \rightarrow \text{Zn}^{2+} + \text{Sn}^{2+} \):
The number of transferred electrons is \( n = 2 \).
The Nernst equation at \( 298 \text{ K} \) is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Zn}^{2+}][\text{Sn}^{2+}]}{[\text{Sn}^{4+}]} \]
Substituting concentrations:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{(0.5)(2)}{(1.5)} \]

\implies E_{\text{cell}} = E^\circ_{\text{cell}} - 0.0295 \log(0.667)
In simple words: The Nernst equation adjusts the standard cell potential based on the actual concentrations of the reactants and products.

Exam Tip: Ensure that the exponents in the log term correspond exactly to the stoichiometric coefficients in the balanced chemical equation.

Question 7. What is the effect of temperature on a) electrolytic conductivity b) metallic conductivity?
Answer:
* (a) Electrolytic Conductivity: It increases with an increase in temperature. A higher temperature increases the kinetic energy of ions, reduces inter-ionic attractions, and lowers the viscosity of the solvent, allowing ions to migrate more freely.
* (b) Metallic Conductivity: It decreases with an increase in temperature. Raising the temperature increases the thermal vibrations of the metal atoms (kernels) in the lattice, which creates physical resistance to the flow of free electrons.
In simple words: Heating a liquid electrolyte lets ions move faster, boosting conductivity. Heating a metal wire makes the atoms vibrate wildly, blocking the electrons and lowering conductivity.

Exam Tip: Be sure to distinguish clearly between the charge carriers in both cases: mobile ions in electrolytes, and free electrons in metals.

Question 8. How does conductivity of an electrolyte solution varies with the dilution?
Answer: The conductivity (specific conductance, \( \kappa \)) of an electrolyte solution always decreases as the solution is diluted. Conductivity is defined as the conductance of a unit volume (\( 1 \text{ cm}^3 \)) of the solution. Upon dilution, the number of current-carrying ions per unit volume of the solution decreases, reducing its overall conductivity.
In simple words: Dilution spreads the ions across a larger volume. Since there are fewer ions in any small sample of the liquid, the conductivity drops.

Exam Tip: Do not confuse "conductivity" (\( \kappa \)) with "molar conductivity" (\( \Lambda_m \)), as they behave oppositely upon dilution.

Question 9. How does molar conductivity of an electrolyte solution varies with the dilution ?
Answer: The molar conductivity (\( \Lambda_m \)) of an electrolyte solution always increases with dilution. Since molar conductivity is given by \( \Lambda_m = \kappa \cdot V \), the increase in volume (\( V \)) containing one mole of electrolyte on dilution far outweighs the decrease in conductivity (\( \kappa \)). For strong electrolytes, dilution minimizes inter-ionic attractions, allowing ions to move faster. For weak electrolytes, dilution increases the degree of dissociation, significantly increasing the total number of free ions.
In simple words: Diluting a solution gives ions more room to move and helps weak electrolytes split up more, increasing the molar conductivity.

Exam Tip: Mention Ostwald's Dilution Law when discussing the behavior of weak electrolytes to write a technically complete answer.

Question 10. Write the Kohlrausch equation and draw a graph to show the variation of molar conductivity with √c for a) KCl b) CH3COOH.
Answer:
The Kohlrausch equation for strong electrolytes is written as:
\[ \Lambda_m = \Lambda_m^\circ - A\sqrt{c} \]
where \( \Lambda_m \) is the molar conductivity at concentration \( c \), \( \Lambda_m^\circ \) is the limiting molar conductivity, and \( A \) is a constant dependent on the type of electrolyte and temperature.
The graph below shows the variation of molar conductivity against the square root of concentration for a strong electrolyte (KCl) and a weak electrolyte (\( \text{CH}_3\text{COOH} \)):

In simple words: Strong electrolytes show a slow linear decrease in conductivity as concentration increases. Weak electrolytes show a steep upward curve near zero concentration as they rapidly dissociate when diluted.

Exam Tip: Be sure to draw the line for KCl straight, and make the curve for \( \text{CH}_3\text{COOH} \) run parallel to the vertical axis near \( c = 0 \) to show it cannot meet the axis.

Question 11. State Kohlrausch law of independent migration of ions.. Mention two applications of the law.
Answer:
Kohlrausch's Law of Independent Migration of Ions: It states that at infinite dilution, when dissociation of the electrolyte is complete, each ion makes a definite contribution toward the molar conductivity of the electrolyte, independent of the nature of the other ion with which it is associated. Mathematically:
\[ \Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ \]
where \( \lambda_+^\circ \) and \( \lambda_-^\circ \) are the limiting molar conductivities of the cation and anion respectively, and \( \nu_+ \) and \( \nu_- \) are their stoichiometric coefficients.

Applications of the Law:
1. Calculation of Limiting Molar Conductivity of Weak Electrolytes: The \( \Lambda_m^\circ \) of a weak electrolyte (such as acetic acid) can be computed using the known \( \Lambda_m^\circ \) values of strong electrolytes:
\[ \Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl}) \]
2. Calculation of Degree of Dissociation (\( \alpha \)): The ratio of molar conductivity at a given concentration to that at infinite dilution gives the degree of dissociation:
\[ \alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ} \]
In simple words: This law states that when fully diluted, each ion contributes to electrical conduction on its own. We use this to calculate properties of weak acids that are otherwise impossible to measure directly.

Exam Tip: Write down the algebraic equation showing how the terms cancel out when demonstrating the calculation of \( \Lambda_m^\circ \) for a weak electrolyte.

Question 12. Give the products of electrolysis of
a) NaCl (molten) b) NaCl (aq) c) H2SO4(aq) d) CuSO4(aq) using inert electrode like Pt e) CuSO4(aq) using Cu electrodes
f) AgNO3(aq) using Ag electrode g) AgNO3(aq) using Pt electrode.
Answer:
The products of electrolysis under different conditions are detailed below:
* (a) Molten NaCl:
At Cathode: Sodium metal (\( \text{Na} \)); At Anode: Chlorine gas (\( \text{Cl}_2 \))
* (b) Aqueous NaCl:
At Cathode: Hydrogen gas (\( \text{H}_2 \)) [due to higher reduction potential of water than \( \text{Na}^+ \)]; At Anode: Chlorine gas (\( \text{Cl}_2 \)) [due to overpotential of oxygen]
* (c) Aqueous \( \text{H}_2\text{SO}_4 \):
At Cathode: Hydrogen gas (\( \text{H}_2 \)); At Anode: Oxygen gas (\( \text{O}_2 \))
* (d) Aqueous \( \text{CuSO}_4 \) using Pt electrodes:
At Cathode: Copper metal (\( \text{Cu} \)); At Anode: Oxygen gas (\( \text{O}_2 \))
* (e) Aqueous \( \text{CuSO}_4 \) using Cu electrodes:
At Cathode: Copper metal deposits (\( \text{Cu} \)); At Anode: Copper anode dissolves (\( \text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2e^- \))
* (f) Aqueous \( \text{AgNO}_3 \) using Ag electrode:
At Cathode: Silver metal deposits (\( \text{Ag} \)); At Anode: Silver anode dissolves (\( \text{Ag(s)} \rightarrow \text{Ag}^+\text{(aq)} + e^- \))
* (g) Aqueous \( \text{AgNO}_3 \) using Pt electrode:
At Cathode: Silver metal deposits (\( \text{Ag} \)); At Anode: Oxygen gas (\( \text{O}_2 \))
In simple words: In electrolysis, what forms at the electrodes depends on which ions are easiest to oxidize or reduce, and whether the electrode metal itself can react.

Exam Tip: Be sure to write the half-reactions at both anode and cathode to explain why certain products are favored over others.

Question 13. Write the reactions involved at each electrode in mercury cell. Why does the cell potential of this cell remains constant?
Answer:
The chemical reactions occurring in a mercury cell are:
* Anode (Zinc-mercury amalgam):
\[ \text{Zn(Hg)} + 2\text{OH}^- \rightarrow \text{ZnO(s)} + \text{H}_2\text{O} + 2e^- \]
* Cathode (Paste of \( \text{HgO} \) and carbon):
\[ \text{HgO(s)} + \text{H}_2\text{O} + 2e^- \rightarrow \text{Hg(l)} + 2\text{OH}^- \]
* Overall Reaction:
\[ \text{Zn(Hg)} + \text{HgO(s)} \rightarrow \text{ZnO(s)} + \text{Hg(l)} \]
The cell potential remains constant throughout its lifespan because the overall reaction does not involve any ions in the solution whose concentration can change during operation.
In simple words: The mercury cell has reactions that only involve solid and liquid reactants. Since there are no dissolved ions whose concentrations fade, the voltage stays steady at 1.35 V.

Exam Tip: State the exact value of the mercury cell potential (\( 1.35 \text{ V} \)) when writing your answer.

Question 14. What are fuel cells? Write the reactions involved at each electrode in H2-O2 fuel cell.
Answer:
Fuel Cells: These are galvanic cells designed to convert the chemical energy of combustion of fuels (such as hydrogen, carbon monoxide, or methane) directly into electrical energy.

The electrode reactions in a hydrogen-oxygen (\( \text{H}_2-\text{O}_2 \)) fuel cell using an alkaline electrolyte are:
* At Anode (Oxidation):
\[ 2\text{H}_2\text{(g)} + 4\text{OH}^-\text{(aq)} \rightarrow 4\text{H}_2\text{O(l)} + 4e^- \]
* At Cathode (Reduction):
\[ \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4e^- \rightarrow 4\text{OH}^-\text{(aq)} \]
* Overall Cell Reaction:
\[ 2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)} \]
In simple words: Fuel cells generate clean electricity by feeding fuel (like hydrogen) directly into the cell, producing water as the only byproduct.

Exam Tip: Mention that fuel cells are highly efficient (about 70%) and eco-friendly compared to thermal power plants.

Question 15. Give one similarity and one difference between fuel cell and other primary cells.
Answer:
* Similarity: Both fuel cells and standard primary cells convert chemical energy into electrical energy through redox reactions occurring at their electrodes.
* Difference: Primary cells contain a fixed amount of reactants sealed inside, and once these reactants are exhausted, the cell dies. In contrast, a fuel cell is designed with an open system where reactants are continuously supplied from an external source, allowing it to function indefinitely as long as fuel is provided.
In simple words: Both make electricity from chemical reactions. However, regular primary batteries go dead when their internal fuel runs out, while fuel cells can run forever as long as you keep feeding them hydrogen.

Exam Tip: Focus on the "continuous supply of reactants" as the defining difference for fuel cells.

Question 16. What are secondary cells? Write the reactions involved at each electrode in lead storage cell when
a) battery is in use b) battery is not in use.
Answer:
Secondary Cells: These are rechargeable cells that can be restored to their original state by passing an external electric current in the opposite direction after they have discharged.

The electrode reactions in a lead storage cell during discharging (when the battery is in use) are:
* At Anode:
\[ \text{Pb(s)} + \text{SO}_4^{2-}\text{(aq)} \rightarrow \text{PbSO}_4\text{(s)} + 2e^- \]
* At Cathode:
\[ \text{PbO}_2\text{(s)} + \text{SO}_4^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)} + 2e^- \rightarrow \text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \]
* Overall Discharging Reaction:
\[ \text{Pb(s)} + \text{PbO}_2\text{(s)} + 2\text{H}_2\text{SO}_4\text{(aq)} \rightarrow 2\text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \]
During charging (when the battery is connected to an external power source for recharging, i.e., "not in use" as a power source): the overall reaction is reversed:
\[ 2\text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow \text{Pb(s)} + \text{PbO}_2\text{(s)} + 2\text{H}_2\text{SO}_4\text{(aq)} \]
In simple words: Secondary cells are rechargeable batteries. During discharging, lead and lead dioxide turn into lead sulfate. During charging, electricity forces lead sulfate to change back into lead and lead dioxide.

Exam Tip: Be sure to write the exact states of the species: \( \text{PbSO}_4 \) is a solid that adheres to the plates, which is what makes recharging possible.

Question 17. Explain corrosion of iron as a electrochemical process. Mention the methods to prevent corrosion.
Answer:
Corrosion of iron occurs through the formation of a tiny electrochemical cell on the metal surface in the presence of water and carbon dioxide:
* Anodic Area: Iron atoms lose electrons to form \( \text{Fe}^{2+} \) ions:
\[ \text{Fe(s)} \rightarrow \text{Fe}^{2+}\text{(aq)} + 2e^- \quad (E^\circ = -0.44 \text{ V}) \]
* Cathodic Area: Electrons released migrate through the metal to a spot containing oxygen and \( \text{H}^+ \) ions (formed by dissolved \( \text{CO}_2 \) in water), where oxygen is reduced:
\[ \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} + 4e^- \rightarrow 2\text{H}_2\text{O(l)} \quad (E^\circ = 1.23 \text{ V}) \]
* Overall Cell Reaction:
\[ 2\text{Fe(s)} + \text{O}_2\text{(g)} + 4\text{H}^+\text{(aq)} \rightarrow 2\text{Fe}^{2+}\text{(aq)} + 2\text{H}_2\text{O(l)} \]
The \( \text{Fe}^{2+} \) ions are further oxidized by atmospheric oxygen to form hydrated ferric oxide (rust, \( \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \)).

Methods to Prevent Corrosion:
1. Barrier Protection: Painting the surface or coating it with oil/grease to block moisture and oxygen.
2. Sacrificial Protection (Galvanization): Coating iron with a more reactive metal like zinc, which oxidizes preferentially to protect the underlying iron.
In simple words: Rusting is like a tiny battery running on the iron surface. Moisture and oxygen react to dissolve iron, which later forms rust. We prevent it by painting the iron or coating it with zinc.

Exam Tip: Write down the final chemical formula of rust: \( \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \).

Question 18. Iron bar A is coated with Zn and another bar B is coated with Sn. Which will rust faster when the coating is broken ?.
Answer: Iron bar B (coated with tin, Sn) will rust much faster when the coating is broken.
Reason: Zinc has a lower standard reduction potential (\( -0.76 \text{ V} \)) than iron (\( -0.44 \text{ V} \)). Even if the zinc coating is broken, zinc acts as the anode and continues to corrode sacrificially to protect the iron. Tin, however, has a higher reduction potential (\( -0.14 \text{ V} \)) than iron. When the tin coating is damaged, iron becomes the anode in the electrochemical cell and oxidizes rapidly to protect the tin, accelerating the rusting process.
In simple words: Zinc is more reactive than iron, so it continues to protect the iron even if scratched. Tin is less reactive, so once scratched, the iron actually sacrifices itself to protect the tin, causing it to rust rapidly.

Exam Tip: Use the reduction potentials of Zn, Fe, and Sn to explain which metal acts as the anode when the barrier is breached.

Question 19. State Faraday’s laws of electrolysis. What is Faraday constant?
Answer:
* Faraday's First Law: The mass of any substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte: \( w = Z \cdot I \cdot t \).
* Faraday's Second Law: When the same quantity of electricity is passed through different electrolytic solutions connected in series, the masses of substances liberated are directly proportional to their chemical equivalent weights: \( w_1/w_2 = E_1/E_2 \).
* Faraday Constant (F): It is the total electric charge carried by one mole of electrons, approximately equal to \( 96487 \text{ C mol}^{-1} \) (commonly rounded to \( 96500 \text{ C} \) for calculations).
In simple words: The first law says passing more current deposits more metal. The second law says the same current deposits metals in proportion to their chemical weights. The Faraday constant is the charge of one mole of electrons.

Exam Tip: Write down the values clearly: \( F = e \cdot N_A = (1.602 \times 10^{-19} \text{ C}) \times (6.022 \times 10^{23}) \approx 96487 \text{ C} \).

Question 20. Resistance of a conductivity cell filled with 0.1M KCl is 100 ohm. If the resistance of the same cell filled with 0.02 M KCl is 520 ohm, calculate the conductivity and molar conductivity of 0.02 M KCl. Conductivity of the 0.1M KCl is 1.29 s/m.
Answer:
First, we find the cell constant (\( G^* \)) using the \( 0.1 \text{ M } \text{KCl} \) data:
\[ \kappa = \frac{G^*}{R} \implies G^* = \kappa \cdot R \]
Given \( \kappa = 1.29 \text{ S m}^{-1} = 1.29 \times 10^{-2} \text{ S cm}^{-1} \) and \( R = 100\ \Omega \):
\[ G^* = 1.29 \text{ S m}^{-1} \times 100\ \Omega = 129 \text{ m}^{-1} = 1.29 \text{ cm}^{-1} \]

Next, we calculate the conductivity (\( \kappa \)) of the \( 0.02 \text{ M } \text{KCl} \) solution:
\[ \kappa = \frac{G^*}{R} = \frac{1.29 \text{ cm}^{-1}}{520\ \Omega} \approx 2.48 \times 10^{-3} \text{ S cm}^{-1} = 0.248 \text{ S m}^{-1} \]

Now, we compute the molar conductivity (\( \Lambda_m \)) of the \( 0.02 \text{ M } \text{KCl} \) solution:
\[ \Lambda_m = \frac{\kappa \times 1000}{C} \]
where \( \kappa \) is in \( \text{S cm}^{-1} \) and \( C = 0.02 \text{ M} \):
\[ \Lambda_m = \frac{2.48 \times 10^{-3} \text{ S cm}^{-1} \times 1000}{0.02 \text{ mol L}^{-1}} = 124 \text{ S cm}^2 \text{ mol}^{-1} \]
The conductivity is \( 0.248 \text{ S m}^{-1} \) and the molar conductivity is \( 124 \text{ S cm}^2 \text{ mol}^{-1} \).
In simple words: First we find the cell constant (1.29 cm^-1) from the first solution. Then we use it to calculate the conductivity (0.00248 S/cm) and molar conductivity (124 S cm^2/mol) of the weaker solution.

Exam Tip: Pay close attention to whether the units used are in meters (\( \text{m} \)) or centimeters (\( \text{cm} \)) to ensure your calculations are correct.

Question 21. Resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 103 ohm. Calculate the resistivity and molar conductivity of the solution.
Answer:
First, we find the cross-sectional area (\( A \)) of the cylinder:
\[ \text{Radius } r = \frac{\text{Diameter}}{2} = 0.5 \text{ cm} \]
\[ A = \pi r^2 = 3.1416 \times (0.5)^2 = 0.7854 \text{ cm}^2 \]
Next, we calculate the resistivity (\( \rho \)):
\[ R = \rho \frac{l}{A} \implies \rho = R \frac{A}{l} \]
Given \( R = 5.55 \times 10^3\ \Omega \) and \( l = 50 \text{ cm} \):
\[ \rho = (5.55 \times 10^3\ \Omega) \times \frac{0.7854 \text{ cm}^2}{50 \text{ cm}} \approx 87.18\ \Omega \text{ cm} \]

Now, we determine the conductivity (\( \kappa \)):
\[ \kappa = \frac{1}{\rho} = \frac{1}{87.18\ \Omega \text{ cm}} \approx 0.01147 \text{ S cm}^{-1} \]

Finally, we compute the molar conductivity (\( \Lambda_m \)) at \( C = 0.05 \text{ M} \):
\[ \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.01147 \text{ S cm}^{-1} \times 1000}{0.05 \text{ mol L}^{-1}} = 229.4 \text{ S cm}^2 \text{ mol}^{-1} \]
The resistivity of the solution is \( 87.18\ \Omega \text{ cm} \) and the molar conductivity is \( 229.4 \text{ S cm}^2 \text{ mol}^{-1} \).
In simple words: We calculate the cylinder's area to find its resistivity (87.18 ohm-cm), and then use that to find the molar conductivity (229.4 S cm^2/mol).

Exam Tip: Be sure to write the correct units for resistivity (\( \Omega \text{ cm} \)) and molar conductivity (\( \text{S cm}^2 \text{ mol}^{-1} \)) to secure full marks.

Question 22. Resistance of 0.01M CH3COOH solution is 2220 ohm. Cell constant is 0.366 cm-1. Calculate the degree of dissociation and dissociation constant of CH3COOH at this concentration. Given Ʌ0 HCl, NaCl, CH3COONa are 425,128 and 96 scm2 mole-1 respectively.
Answer:
First, we find the limiting molar conductivity (\( \Lambda_m^\circ \)) of acetic acid using Kohlrausch's law:
\[ \Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{HCl}) + \Lambda_m^\circ(\text{CH}_3\text{COONa}) - \Lambda_m^\circ(\text{NaCl}) \]
\[ \Lambda_m^\circ(\text{CH}_3\text{COOH}) = 425 + 96 - 128 = 393 \text{ S cm}^2 \text{ mol}^{-1} \]

Next, we calculate the conductivity (\( \kappa \)) of the \( 0.01 \text{ M} \) solution:
\[ \kappa = \frac{G^*}{R} = \frac{0.366 \text{ cm}^{-1}}{2220\ \Omega} \approx 1.649 \times 10^{-4} \text{ S cm}^{-1} \]

Now, we calculate the molar conductivity (\( \Lambda_m^c \)):
\[ \Lambda_m^c = \frac{\kappa \times 1000}{C} = \frac{1.649 \times 10^{-4} \times 1000}{0.01} = 16.49 \text{ S cm}^2 \text{ mol}^{-1} \]

We determine the degree of dissociation (\( \alpha \)):
\[ \alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ} = \frac{16.49}{393} \approx 0.042 \text{ (or } 4.2 \% \text{)} \]

Finally, we calculate the dissociation constant (\( K_a \)) at \( C = 0.01 \text{ M} \):
\[ K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.01 \times (0.042)^2}{1 - 0.042} = \frac{1.764 \times 10^{-5}}{0.958} \approx 1.84 \times 10^{-5} \text{ mol L}^{-1} \]
The degree of dissociation is \( 0.042 \) and the dissociation constant is \( 1.84 \times 10^{-5} \text{ mol L}^{-1} \).
In simple words: Kohlrausch's law gives the limit for acetic acid as 393. Measuring the cell gives a molar conductivity of 16.49, meaning it is 4.2% dissociated, with a dissociation constant of 1.84 x 10^-5.

Exam Tip: Since \( \alpha \) is small, you can use the approximation \( K_a \approx C \alpha^2 \), but using the exact formula \( K_a = C \alpha^2 / (1-\alpha) \) is safer and more precise.

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Question 23. Ʌ0 Al2(SO4)3 is 858 scm2 mole-1. Find Ʌ0 of Al3+ if Ʌ0 of SO4 2- is 160 scm2 mole-1.
Answer:
According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of aluminum sulfate is given by:
\[ \Lambda^\circ_m(\text{Al}_2(\text{SO}_4)_3) = 2 \lambda^\circ(\text{Al}^{3+}) + 3 \lambda^\circ(\text{SO}_4^{2-}) \]
Substituting the given values:
\[ 858 = 2 \lambda^\circ(\text{Al}^{3+}) + 3(160) \]
\[ 858 = 2 \lambda^\circ(\text{Al}^{3+}) + 480 \]
\[ 2 \lambda^\circ(\text{Al}^{3+}) = 858 - 480 = 378 \]
\[ \lambda^\circ(\text{Al}^{3+}) = \frac{378}{2} = 189 \text{ S cm}^2 \text{ mol}^{-1} \]
The limiting molar conductivity of \( \text{Al}^{3+} \) is \( 189 \text{ S cm}^2 \text{ mol}^{-1} \).
In simple words: Since the total conductivity is the sum of two aluminum ions and three sulfate ions, we subtract the sulfate portion from the total and divide by two to find aluminum's individual value.

Exam Tip: Be sure to write the stoichiometric coefficients (2 and 3) correctly to match the formula of the salt.

Question 24. Calculate the potential of hydrogen electrode in contact with a solution of pH = 10.
Answer:
The reduction reaction for a hydrogen electrode is:
\[ \text{H}^+\text{(aq)} + e^- \rightarrow \frac{1}{2}\text{H}_2\text{(g)} \]
The Nernst equation for this electrode at \( 298 \text{ K} \) and \( P_{\text{H}_2} = 1 \text{ bar} \) is:
\[ E = E^\circ - \frac{0.0591}{1} \log \frac{1}{[\text{H}^+]} \]
Since \( E^\circ_{\text{H}^+/\text{H}_2} = 0.00 \text{ V} \) and \( \text{pH} = -\log[\text{H}^+] \):
\[ E = 0 - 0.0591 \log(10^{-\text{pH}}) \]

\implies E = -0.0591 \times \text{pH}
Substituting the given \( \text{pH} = 10 \):
\[ E = -0.0591 \times 10 = -0.591 \text{ V} \]
The potential of the hydrogen electrode is \( -0.591 \text{ V} \).
In simple words: The voltage of a hydrogen electrode drops by 0.0591 volts for every unit of pH. At pH 10, the electrode potential is -0.591 V.

Exam Tip: The derived shortcut \( E = -0.0591 \cdot \text{pH} \) is a very useful formula for solving these problems quickly.

Question 25. At what pH of HCl solution will the standard hydrogen electrode will have a potential of - 0.118V ?
Answer:
Using the standard relationship between the hydrogen electrode potential (\( E \)) and \( \text{pH} \) at \( 298 \text{ K} \):
\[ E = -0.0591 \times \text{pH} \]
Given that \( E = -0.118 \text{ V} \):
\[ -0.118 \text{ V} = -0.0591 \times \text{pH} \]
\[ \text{pH} = \frac{-0.118}{-0.0591} \approx 2 \]
The pH of the \( \text{HCl} \) solution must be \( 2 \).
In simple words: To get a voltage of -0.118 V, we divide by the constant -0.0591, which tells us that the solution's pH is exactly 2.

Exam Tip: State that a pH of 2 corresponds to a hydrogen ion concentration of \( 10^{-2} \text{ M} \) to write a complete answer.

Question 26. Calculate the cell potential of Cr/Cr3+|| Fe2/ Fe.(0.1M) (0.1M) Given E0Cr3+/Cr = - 0.74V E0 Fe2+/Fe = - 0.44V.
Answer:
First, we find the standard cell potential (\( E^\circ_{\text{cell}} \)):
The cell consists of a chromium anode and an iron cathode.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.44 \text{ V} - (-0.74 \text{ V}) = +0.30 \text{ V} \]

The balanced cell reaction is:
\[ 2\text{Cr(s)} + 3\text{Fe}^{2+}\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{Fe(s)} \]
Here, the number of transferred electrons is \( n = 6 \).
Using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{6} \log \frac{[\text{Cr}^{3+}]^2}{[\text{Fe}^{2+}]^3} \]
Given \( [\text{Cr}^{3+}] = 0.1 \text{ M} \) and \( [\text{Fe}^{2+}] = 0.1 \text{ M} \):
\[ E_{\text{cell}} = 0.30 - \frac{0.0591}{6} \log \frac{(0.1)^2}{(0.1)^3} = 0.30 - \frac{0.0591}{6} \log(10) \]

\implies E_{\text{cell}} = 0.30 - 0.00985(1) \approx 0.29 \text{ V}
The cell potential is \( 0.29 \text{ V} \).
In simple words: The standard potential is 0.30 V. The non-standard concentrations adjust this value down slightly, resulting in a final cell potential of 0.29 V.

Exam Tip: Be sure to divide the log term by 6, as 6 electrons are transferred during the reaction between \( \text{Cr} \) and \( \text{Fe}^{2+} \).

Question 27. Calculate the cell potential at 298K
Zn + Sn4+ → Zn2+ + Sn2+
(1.5M)         (0.5M) (2M)
Given E0Zn2+/Zn = - 0.76V E0 Sn4+/Sn2+ = 0.13V
Answer:
First, we find the standard cell potential (\( E^\circ_{\text{cell}} \)):
Zinc is oxidized (acts as the anode) and tin is reduced (acts as the cathode).
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.13 \text{ V} - (-0.76 \text{ V}) = +0.89 \text{ V} \]

The reaction is: \( \text{Zn(s)} + \text{Sn}^{4+}\text{(aq)} \rightarrow \text{Zn}^{2+}\text{(aq)} + \text{Sn}^{2+}\text{(aq)} \), where \( n = 2 \).
Using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Zn}^{2+}][\text{Sn}^{2+}]}{[\text{Sn}^{4+}]} \]
Substituting concentrations:
\[ E_{\text{cell}} = 0.89 - \frac{0.0591}{2} \log \frac{(0.5)(2)}{(1.5)} \]
\[ E_{\text{cell}} = 0.89 - 0.02955 \log(0.667) \]
\[ E_{\text{cell}} = 0.89 - 0.02955(-0.176) \approx 0.89 + 0.005 = 0.895 \text{ V} \]
The cell potential is \( 0.895 \text{ V} \).
In simple words: The standard voltage is 0.89 V. Because the concentration ratio favors the reactants, the actual cell potential increases slightly to 0.895 V.

Exam Tip: A negative log value will change the sign of the correction term to positive, raising the cell potential above its standard value.

Question 28. Calculate the potential of the electrode Mg2+/ Mg
Mg2+ + 2e- → Mg E0 Mg2+/ Mg = - 2.36 V
Answer:
Wait! The question is written twice as:
"28. Calculate the equilibrium constant and work done by the cell Ni + Cu2+ Ni2+ +Cu Given E0Ni2+ / Ni = - 0.25V E0 Cu2+/Cu = 0.34 V"
And another part under it is:
"30. Find the potential of the electrode Mg2+/ Mg
Mg2+ + 2e- Mg E0 Mg2+/ Mg = - 2.36 V"
Let us solve the first part (Ni/Cu cell):
The standard potential is: \( E^\circ_{\text{cell}} = 0.34 \text{ V} - (-0.25 \text{ V}) = +0.59 \text{ V} \). Here, \( n = 2 \).
To find the equilibrium constant (\( K_c \)):
\[ \log K_c = \frac{n E^\circ_{\text{cell}}}{0.0591} = \frac{2 \times 0.59}{0.0591} = 19.96 \implies K_c \approx 9.2 \times 10^{19} \]
To calculate the work done (\( \Delta G^\circ \)):
\[ \Delta G^\circ = -n F E^\circ_{\text{cell}} = -2 \times 96500 \times 0.59 = -113870 \text{ J} = -113.87 \text{ kJ} \]

Let us solve the second part (Mg electrode potential) assuming its concentration is \( 1 \text{ M} \) (which makes it equal to standard potential) or if concentration is omitted, we assume standard conditions:
\[ E = E^\circ = -2.36 \text{ V} \]
In simple words: For the Ni-Cu cell, the equilibrium constant is extremely high (\( 9.2 \times 10^{19} \)) and the work done by the cell is 113.87 kJ. The potential of the standard magnesium electrode is -2.36 V.

Exam Tip: Maximum work done by a cell is equal to the decrease in Gibbs free energy (\( w_{\text{max}} = -\Delta G^\circ \)), which is positive.

Question 29 . Find the emf of the cell Pb/Pb2+(0.001M)|| Pt, Cl2(1.5 atm) / 2Cl-(1M) E0 Pb2+/ Pb = - 0.13V E0 Cl2/ 2 Cl- = 1.36V
Answer:
First, we find the standard cell potential (\( E^\circ_{\text{cell}} \)):
Lead acts as the anode, and chlorine acts as the cathode.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 1.36 \text{ V} - (-0.13 \text{ V}) = +1.49 \text{ V} \]

The balanced cell reaction is:
\[ \text{Pb(s)} + \text{Cl}_2\text{(g)} \rightarrow \text{Pb}^{2+}\text{(aq)} + 2\text{Cl}^-\text{(aq)} \]
Here, the number of transferred electrons is \( n = 2 \).
Using the Nernst equation (incorporating gas pressure for \( \text{Cl}_2 \)):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Pb}^{2+}][\text{Cl}^-]^2}{P_{\text{Cl}_2}} \]
Substituting the given values:
\[ E_{\text{cell}} = 1.49 - \frac{0.0591}{2} \log \frac{(0.001)(1)^2}{1.5} \]
\[ E_{\text{cell}} = 1.49 - 0.02955 \log \left(6.67 \times 10^{-4}\right) \]
\[ E_{\text{cell}} = 1.49 - 0.02955(-3.176) \approx 1.49 + 0.094 = 1.584 \text{ V} \]
The cell potential is \( 1.584 \text{ V} \).
In simple words: The standard potential is 1.49 V. Because the reactant gas pressure is high and product concentration is low, the actual potential increases to 1.584 V.

Exam Tip: Be sure to include the partial pressure of gas reactants in the Nernst equation quotient alongside the molarity of solution ions.

Question 31. An electrochemical cell is set up by dipping Cu in 0.1M CuSO4(aq) and Ag in 0.1M AgNO3(aq) Write reaction taking place at each electrode and overall reaction. Determine the potential of the cell at 298K.
Answer:
First, we write the standard values: \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \text{ V} \) and \( E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \text{ V} \).
Since silver has a higher reduction potential, it acts as the cathode, and copper acts as the anode.
* Anode (Oxidation): \( \text{Cu(s)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2e^- \)
* Cathode (Reduction): \( 2\text{Ag}^+\text{(aq)} + 2e^- \rightarrow 2\text{Ag(s)} \)
* Overall Cell Reaction: \( \text{Cu(s)} + 2\text{Ag}^+\text{(aq)} \rightarrow \text{Cu}^{2+}\text{(aq)} + 2\text{Ag(s)} \) (\( n = 2 \))

Standard Cell Potential:
\[ E^\circ_{\text{cell}} = 0.80 \text{ V} - 0.34 \text{ V} = +0.46 \text{ V} \]
Using the Nernst equation at \( 298 \text{ K} \):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} \]
Substituting concentrations \( [\text{Cu}^{2+}] = 0.1 \text{ M} \) and \( [\text{Ag}^+] = 0.1 \text{ M} \):
\[ E_{\text{cell}} = 0.46 - \frac{0.0591}{2} \log \frac{0.1}{(0.1)^2} = 0.46 - \frac{0.0591}{2} \log(10) \]

\implies E_{\text{cell}} = 0.46 - 0.0295(1) = 0.43 \text{ V}
The cell potential is \( 0.43 \text{ V} \).
In simple words: Copper oxidizes and silver reduces. The standard potential is 0.46 V, but the actual concentrations lower it slightly to 0.43 V.

Exam Tip: Remember to square the silver ion concentration in the log term because of its stoichiometric coefficient in the balanced equation.

Question 32. Calculate the amount of chlorine gas liberated when a current of 1.5 amperes for 90 minutes is passed through molten NaCl. (Atomic mass of Cl = 35.5)
Answer:
First, we find the quantity of electricity passed (\( Q \)):
\[ I = 1.5 \text{ A} \]
\[ t = 90 \text{ minutes} = 90 \times 60 \text{ seconds} = 5400 \text{ s} \]
\[ Q = I \cdot t = 1.5 \text{ A} \times 5400 \text{ s} = 8100 \text{ C} \]

The oxidation reaction for chloride is:
\[ 2\text{Cl}^- \rightarrow \text{Cl}_2\text{(g)} + 2e^- \]
This means \( 2 \text{ moles of electrons} \) (\( 2 \times 96500 \text{ C} = 193000 \text{ C} \)) liberate \( 1 \text{ mole of chlorine gas} \) (\( 71 \text{ g} \)).
Using Faraday's first law:
\[ w = \frac{\text{Molar Mass} \times Q}{n \cdot F} = \frac{71 \times 8100}{2 \times 96500} = \frac{575100}{193000} \approx 2.98 \text{ g} \]
Therefore, \( 2.98 \text{ g} \) of chlorine gas is liberated.
In simple words: Passing a 1.5 A current for 90 minutes provides 8100 Coulombs of charge, which electrolyzes molten salt to release 2.98 grams of chlorine gas at the anode.

Exam Tip: Be sure to write the molar mass of \( \text{Cl}_2 \) gas as 71 g/mol and not 35.5 g/mol, as chlorine is diatomic.

Question 33. Silver is electrodeposited by passing a current of 0.2 amperes for 3 hours on a vessel of surface area 800 cm2 using silver nitrate as an electrolyte solution. Calculate the thickness of silver deposited..(Atomic mass of Ag=108) Density of Ag = 10.8 g/cm2
Answer:
First, we find the total charge (\( Q \)) passed:
\[ Q = I \cdot t = 0.2 \text{ A} \times (3 \times 3600 \text{ s}) = 0.2 \times 10800 = 2160 \text{ C} \]

Using Faraday's first law to calculate the mass of silver deposited (\( w \)):
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (n = 1) \]
\[ w = \frac{\text{Atomic Mass} \times Q}{n \cdot F} = \frac{108 \times 2160}{1 \times 96500} = \frac{233280}{96500} \approx 2.417 \text{ g} \]

Using the density of silver (\( d = 10.8 \text{ g/cm}^3 \)) to find the volume of deposited silver:
\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{2.417 \text{ g}}{10.8 \text{ g/cm}^3} \approx 0.2238 \text{ cm}^3 \keys \]
Finally, we calculate the thickness (\( t \)) over the surface area (\( A = 800 \text{ cm}^2 \)):
\[ \text{Thickness} = \frac{\text{Volume}}{\text{Area}} = \frac{0.2238 \text{ cm}^3}{800 \text{ cm}^2} \approx 2.8 \times 10^{-4} \text{ cm} = 2.8 \times 10^{-6} \text{ m} \]
The thickness of the deposited silver layer is \( 2.8 \times 10^{-4} \text{ cm} \).
In simple words: The current deposits 2.417 grams of silver. Based on silver's density, this mass occupies a volume of 0.2238 cm³, which spreads out over the 800 cm² area to form a layer 0.00028 cm thick.

Exam Tip: Be mindful of your units: ensure density is in \( \text{g/cm}^3 \) and area is in \( \text{cm}^2 \) so your calculated thickness is in centimeters.

Question 34. Electrolysis of a metal salt solution resulted in the deposition of 1gram of metal by passing 1.5 amperes for 2 hours. Determine the charge carried by the metal ion. (Atomic mass of the metal = 27)
Answer:
First, we find the total charge passed (\( Q \)):
\[ Q = I \cdot t = 1.5 \text{ A} \times (2 \times 3600 \text{ s}) = 1.5 \times 7200 = 10800 \text{ C} \]

Using Faraday's first law equation:
\[ w = \frac{\text{Atomic Mass} \times Q}{n \cdot F} \]
Given \( w = 1 \text{ g} \), Atomic Mass = 27, and \( F = 96500 \text{ C} \):
\[ 1 = \frac{27 \times 10800}{n \times 96500} \]
\[ n = \frac{27 \times 10800}{96500} = \frac{291600}{96500} \approx 3 \]
Since \( n = 3 \), the metal ion carries a charge of \( +3 \) (like \( \text{Al}^{3+} \)).
In simple words: By rearranging Faraday's law, we find that 3 moles of electrons are needed to deposit 1 mole of this metal, meaning each metal ion carries a charge of +3.

Exam Tip: Round the calculated value of \( n \) to the nearest whole integer, as ionic charges are always discrete values.

Question 35. Three electrolytes A,B and C containing solutions of ZnSO4 , CuSO4 and AgNO3 were connected in series. 1.5 grams of Ag deposited by passing 1.5 amperes. How long did current flow? Find the mass of copper and zinc deposited. ( Atomic mass of Cu = 63.5, Zn = 65.5)
Answer:
First, we calculate the duration of the current flow using the silver deposition data:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (n = 1) \]
\[ w_{\text{Ag}} = \frac{\text{Atomic Mass} \times I \cdot t}{F} \]
Substituting the given values (\( w_{\text{Ag}} = 1.5 \text{ g} \), \( I = 1.5 \text{ A} \)):
\[ 1.5 = \frac{108 \times 1.5 \times t}{96500} \]
\[ t = \frac{96500}{108} \approx 893.5 \text{ seconds} \]

Next, we find the mass of copper deposited using Faraday's second law:
Equivalent weight of \( \text{Ag} = 108/1 = 108 \)
Equivalent weight of \( \text{Cu} = 63.5/2 = 31.75 \)
\[ \frac{w_{\text{Cu}}}{w_{\text{Ag}}} = \frac{E_{\text{Cu}}}{E_{\text{Ag}}} \implies \frac{w_{\text{Cu}}}{1.5} = \frac{31.75}{108} \]
\[ w_{\text{Cu}} = 1.5 \times \frac{31.75}{108} \approx 0.441 \text{ g} \]

Finally, we calculate the mass of zinc deposited:
Equivalent weight of \( \text{Zn} = 65.5/2 = 32.75 \)
\[ \frac{w_{\text{Zn}}}{w_{\text{Ag}}} = \frac{E_{\text{Zn}}}{E_{\text{Ag}}} \implies \frac{w_{\text{Zn}}}{1.5} = \frac{32.75}{108} \]
\[ w_{\text{Zn}} = 1.5 \times \frac{32.75}{108} \approx 0.455 \text{ g} \]
The current flowed for \( 893.5 \text{ seconds} \), depositing \( 0.441 \text{ g} \) of copper and \( 0.455 \text{ g} \) of zinc.
In simple words: The current of 1.5 A flowed for 893.5 seconds to deposit 1.5g of silver. Because they are connected in series, this same charge deposits 0.441g of copper and 0.455g of zinc.

Exam Tip: Using equivalent weights and Faraday's second law is much faster than calculating each step from the current and time.

Question 36. How many coulomb of electricity is needed for the following reactions?
a) 2 moles of MnO4- to Mn2+
b) 1 mole of H2O to O2
c) 9 grams of Al from molten AlCl3 (atomic mass of Al=27)
Answer:
* (a) Conversion of 2 moles of \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \):
The reduction half-reaction is: \( \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \).
Thus, 1 mole of \( \text{MnO}_4^- \) requires \( 5 \text{ moles of electrons} \) (\( 5\text{F} \)).
For 2 moles:
\[ \text{Charge} = 2 \times 5\text{F} = 10\text{F} = 10 \times 96500 \text{ C} = 9.65 \times 10^5 \text{ C} \]

* (b) Oxidation of 1 mole of \( \text{H}_2\text{O} \) to \( \text{O}_2 \):
The oxidation half-reaction is: \( 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \).
This means 2 moles of \( \text{H}_2\text{O} \) require \( 4\text{F} \), so 1 mole of \( \text{H}_2\text{O} \) requires \( 2\text{F} \):
\[ \text{Charge} = 2\text{F} = 2 \times 96500 \text{ C} = 1.93 \times 10^5 \text{ C} \]

* (c) Reduction of 9 grams of \( \text{Al} \) from molten \( \text{AlCl}_3 \):
The reaction is: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \).
Thus, 1 mole of \( \text{Al} \) (27 g) requires \( 3\text{F} \).
For 9 grams of \( \text{Al} \):
\[ \text{Moles of Al} = \frac{9}{27} = \frac{1}{3} \text{ mol} \]
\[ \text{Charge} = \frac{1}{3} \times 3\text{F} = 1\text{F} = 9.65 \times 10^4 \text{ C} \]
In simple words: (a) Reducing 2 moles of permanganate requires 9.65 x 10^5 Coulombs. (b) Oxidizing 1 mole of water requires 1.93 x 10^5 Coulombs. (c) Extracting 9 grams (one-third of a mole) of aluminum requires 9.65 x 10^4 Coulombs.

Exam Tip: Always relate the stoichiometry of electrons (\( n \)) in the balanced half-reaction to the number of Faradays required (\( Q = n \cdot F \)).

Assignment: Chemical Kinetics

Question 1. Explain the following terms: a) average rate of a reaction b) instantaneous rate of a reaction. c) rate constant of a reaction d) rate law of a reaction e) order of a reaction f) molecularity of a reaction.
Answer:
* (a) Average Rate of a Reaction: It is defined as the change in molar concentration of reactants or products per unit of time over a finite interval. It is expressed as: \( \text{Rate}_{\text{avg}} = -\frac{\Delta[R]}{\Delta t} = \frac{\Delta[P]}{\Delta t} \).
* (b) Instantaneous Rate of a Reaction: This represents the rate of change of reactant or product concentrations at a specific, infinitesimally small moment in time, given by: \( \text{Rate}_{\text{inst}} = -\frac{d[R]}{dt} = \frac{d[P]}{dt} \).
* (c) Rate Constant (\( k \)): It is the constant of proportionality in the rate law equation, representing the reaction rate when the concentration of each reactant is exactly \( 1 \text{ mol/L} \).
* (d) Rate Law: This is the experimental mathematical expression that directly relates the rate of a chemical reaction to the molar concentration of its reactants, where each concentration term is raised to a specific power.
* (e) Order of a Reaction: The sum of the exponents of the concentration terms of the reactants as expressed in the experimentally determined rate law equation.
* (f) Molecularity of a Reaction: The total number of reacting species (atoms, ions, or molecules) that must collide simultaneously in an elementary step to bring about a chemical reaction.
In simple words: Average rate measures speed over an interval, while instantaneous rate is the speed at one exact split-second. Rate constant is the speed when concentrations are normal, and rate law is the formula connecting speed to reactant levels. Order is the sum of exponents in this formula, and molecularity is the actual count of particles colliding.

Exam Tip: Be sure to write the differential expressions for average vs. instantaneous rates, and highlight that molecularity is a purely theoretical value while order is experimental.

Question 2. What do you mean by pseudo first order reaction. Give one example.
Answer: A chemical reaction that has a higher molecularity but behaves kinetically as a first-order reaction under specific conditions is called a pseudo first-order reaction. This typically occurs when one of the reactants is present in such a massive excess that its concentration remains practically unchanged throughout the reaction.
Example: The acid-catalyzed hydrolysis of ethyl acetate:
\[ \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \]
Here, water is the solvent and is present in a huge excess, so its concentration change is negligible. The rate law simplifies to: \( \text{Rate} = k'[\text{CH}_3\text{COOC}_2\text{H}_5] \), where \( k' = k[\text{H}_2\text{O}] \).
In simple words: These are multi-molecule reactions that act like simple single-molecule reactions because one of the ingredients is so abundant that its concentration never changes.

Exam Tip: Always state that one reactant must be present "in large excess" to establish a pseudo first-order condition in your exam write-up.

Question 3. Give one example each of a) zero order reaction b) first order reaction.
Answer:
* (a) Zero-Order Reaction: The decomposition of gaseous ammonia on a hot platinum catalyst surface at high pressures:
\[ 2\text{NH}_3\text{(g)} \xrightarrow{\text{Pt, } \Delta} \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \]
* (b) First-Order Reaction: The thermal decomposition of dinitrogen pentoxide gas:
\[ 2\text{N}_2\text{O}_5\text{(g)} \rightarrow 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)} \]
In simple words: Zero-order reactions run at a constant speed regardless of concentration, like ammonia splitting on platinum. First-order reactions slow down proportionally as reactants disappear, like the decomposition of nitrogen pentoxide.

Exam Tip: Remember that all natural and artificial radioactive decays of unstable nuclei are first-order reactions.

Question 4. Derive integrated rate law for a) zero order reaction b) first order reaction. Show that t1/2 for c) zero order reaction is directly proportional to initial concentration of the reactant. d) first order reaction is independant initial concentration of the reactant.
Answer:
* (a) Zero-Order Integrated Rate Law:
For \( R \rightarrow P \):
\[ \text{Rate} = -\frac{d[R]}{dt} = k[R]^0 = k \]
\[ d[R] = -k \cdot dt \]
Integrating both sides yields:
\[ [R] = -kt + I \]
At \( t = 0 \), the concentration \( [R] = [R]_0 \), so the constant \( I = [R]_0 \). Substituting this gives:
\[ [R] = -kt + [R]_0 \implies \mathbf{k = \frac{[R]_0 - [R]}{t}} \]

* (b) First-Order Integrated Rate Law:
For \( R \rightarrow P \):
\[ \text{Rate} = -\frac{d[R]}{dt} = k[R] \implies \frac{d[R]}{[R]} = -k \cdot dt \]
Integrating both sides:
\[ \ln[R] = -kt + I \]
At \( t = 0 \), \( I = \ln[R]_0 \), which gives:
\[ \ln[R] = -kt + \ln[R]_0 \implies kt = \ln\frac{[R]_0}{[R]} \implies \mathbf{k = \frac{2.303}{t} \log\frac{[R]_0}{[R]}} \]

* (c) Zero-Order Half-Life:
At \( t = t_{1/2} \), the remaining reactant is \( [R] = \frac{[R]_0}{2} \). Substituting this into the zero-order rate law:
\[ \frac{[R]_0}{2} = -k \cdot t_{1/2} + [R]_0 \implies t_{1/2} = \frac{[R]_0}{2k} \]
This shows that \( \mathbf{t_{1/2} \propto [R]_0} \).

* (d) First-Order Half-Life:
At \( t = t_{1/2} \), \( [R] = \frac{[R]_0}{2} \). Substituting this into the first-order rate law:
\[ k = \frac{2.303}{t_{1/2}} \log \frac{[R]_0}{[R]_0/2} \implies t_{1/2} = \frac{2.303 \log(2)}{k} \implies \mathbf{t_{1/2} = \frac{0.693}{k}} \]
This shows that \( t_{1/2} \) is independent of the initial concentration \( [R]_0 \).
In simple words: (a) Integrating zero-order rate equations shows concentration drops linearly. (b) For first-order, it drops exponentially. (c) A zero-order half-life is longer if you start with more material. (d) A first-order half-life is always the same, no matter how much you start with.

Exam Tip: Be sure to write down the integration steps clearly, as derivations are highly structured marking points in board exams.

Question 5. Give differences between order and molecularity of a reaction.
Answer:
The core differences between order and molecularity are:
1. Nature: Order is an experimentally determined value based on the rate law, while molecularity is a purely theoretical value derived from the reaction mechanism.
2. Values: Order of reaction can be zero, fractional, or even negative. Molecularity must always be a non-zero, positive whole number.
3. Applicability: Order applies to both simple elementary reactions and complex multi-step reactions, whereas molecularity is defined only for elementary reactions and is meaningless for complex mechanisms.
In simple words: Order is measured in a lab, can be a fraction or zero, and applies to any reaction. Molecularity is figured out on paper, must be a whole number, and only describes simple, single-step reactions.

Exam Tip: Presenting these differences in a neat table is highly recommended for clarity and readability.

Question 6. What is activation energy? How is it related to rate constant K?
Answer: Activation energy (\( E_a \)) is the minimum extra quantity of energy that reacting molecules must absorb to reach the transition state or threshold energy required to undergo a chemical reaction. It is related to the rate constant (\( K \)) by the Arrhenius equation:
\[ K = A e^{-E_a/RT} \]
Taking the common logarithm of both sides allows us to relate rate constants at two different temperatures:
\[ \log \frac{K_2}{K_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
In simple words: Activation energy is the hill of energy molecules must climb to react. A higher activation energy means a smaller rate constant, causing the reaction to run slower.

Exam Tip: Mention the term "Arrhenius equation" and define \( A \) as the pre-exponential or frequency factor to write a comprehensive answer.

Question 7. Give the unit of rate constant for a) zero order b) first order c) second order reaction.
Answer:
The general unit of the rate constant is given by the relation: \( (\text{mol L}^{-1})^{1-n} \text{ s}^{-1} \), where \( n \) represents the order of the reaction:
* (a) Zero-Order Reaction (\( n = 0 \)): \( \text{mol L}^{-1} \text{ s}^{-1} \) (or \( \text{M s}^{-1} \))
* (b) First-Order Reaction (\( n = 1 \)): \( \text{s}^{-1} \)
* (c) Second-Order Reaction (\( n = 2 \)): \( \text{L mol}^{-1} \text{ s}^{-1} \) (or \( \text{M}^{-1} \text{ s}^{-1} \))
In simple words: The unit changes with the order: zero-order is molarity per second, first-order is just per second, and second-order is per molarity-second.

Exam Tip: Memorize the general formula \( (\text{mol/L})^{1-n} \cdot \text{s}^{-1} \), which helps you derive the rate constant units for any fractional order easily.

Question 8. What is pseudo first order reaction? Give one example.
Answer: A reaction that involves multiple reacting species (high molecularity) but follows first-order kinetics is classified as a pseudo first-order reaction. This typically happens when one reactant is used as the solvent or is present in huge excess.
Example: The inversion of cane sugar in an aqueous acidic solution:
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (fructose)} \]
The rate depends only on sucrose concentration: \( \text{Rate} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}] \), because the concentration of water remains virtually constant.
In simple words: When one of the reactants is in massive excess, its concentration does not change, making a multi-reactant reaction behave kinetically as a first-order reaction.

Exam Tip: Be sure to write the simplified rate law showing that the concentration of the reactant in excess is merged into the rate constant.

Question 9. Mention the conditions a reactant has to satisfy to become product.
Answer: According to the collision theory, reactant molecules must satisfy two primary conditions during a collision to transform into products:
1. Energy Barrier: The colliding molecules must possess a minimum kinetic energy, called the threshold energy, to break their existing chemical bonds.
2. Orientation Barrier: The molecules must collide in a specific spatial orientation so that old bonds can break and new bonds can form in the correct positions.
In simple words: Molecules must hit each other with enough force (energy barrier) and in the correct direction (orientation barrier) to react.

Exam Tip: Use the term "effective collisions" to describe collisions that satisfy both the energy and orientation criteria.

Question 10. A+B C+D Rate = PZABe-Ea/RT What is the significance of P and e-Ea/RT?
Answer:
In the collision theory rate expression \( \text{Rate} = P Z_{AB} e^{-E_a/RT} \):
* \( P \) (Steric or Probability Factor): This factor accounts for the requirement of proper spatial orientation during molecular collisions, representing the fraction of collisions that are geometrically effective.
* \( e^{-E_a/RT} \) (Boltzmann Factor): This term represents the fraction of total molecular collisions that possess kinetic energy equal to or greater than the activation energy (\( E_a \)) at temperature \( T \).
In simple words: P is the orientation multiplier (checking if molecules aligned correctly), and the exponential term is the energy checker (calculating how many molecules hit hard enough to react).

Exam Tip: Identify \( Z_{AB} \) as the collision frequency of reactants A and B to provide a complete breakdown of the formula.

Question 11. Draw the graph of reaction coordinate Vs potential energy for a reaction a) with out catalyst b) with catalyst.
Answer: A catalyst increases the rate of reaction by providing an alternative pathway with lower activation energy, as shown in the energy profile diagram below:

In simple words: The graph shows how a catalyst lowers the energy barrier (the peak of the hill), making it much easier and faster for reactants to turn into products.

Exam Tip: Label the difference between the two peaks as the decrease in activation energy, and the energy difference between reactants and products as \( \Delta H \).

Question 12. Generally rate of a reaction doubles when the temperature is raised by 10K. Explain this statement with the help of the Maxwell Boltzmann distribution curve.
Answer: Raising the temperature of a reaction by 10 K increases the collision frequency by only about 1% to 2%, which does not explain the doubling of the reaction rate. Instead, the temperature rise shifts the Maxwell-Boltzmann distribution curve to the right, broadening and lowering the peak. This change significantly increases the area under the curve beyond the activation energy threshold, which represents the fraction of molecules with sufficient energy to react. For a 10 K rise, this fraction of energetic molecules approximately doubles, which doubles the rate of the reaction.
In simple words: Heating a reaction by 10 degrees does not make molecules collide much more often, but it shifts the energy curve, doubling the number of molecules that have enough energy to react.

Exam Tip: Draw the Maxwell-Boltzmann distribution curves for \( T \) and \( T + 10 \) to visually demonstrate that the shaded area beyond \( E_a \) doubles.

Question 13. Explain collision theory of reaction rate with an example.
Answer: Collision theory states that chemical reactions occur through collisions between reactant molecules. However, only "effective collisions" lead to product formation. To be effective, a collision must overcome both the energy barrier (reactants must possess threshold energy) and the orientation barrier (reactants must collide in the correct spatial alignment).
Example: The nucleophilic substitution of bromomethane with hydroxide ion:
\[ \text{CH}_3\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{OH} + \text{Br}^- \]
If the \( \text{OH}^- \) ion collides with the carbon atom from the front (near the electronegative bromine atom), electrostatic repulsion prevents bond formation (ineffective collision). If \( \text{OH}^- \) attacks from the rear, opposite the bromine atom, the reaction proceeds smoothly, forming methanol (effective collision).
In simple words: For molecules to react, they must collide hard enough to break old bonds and in the correct direction to form new ones, as seen in rear-attack substitution reactions.

Exam Tip: Illustrate this concept with a simple molecular drawing showing "proper orientation" leading to products and "improper orientation" leading to no reaction.

Question 14. Draw the graphs
a) concentration Vs time for reactant. b) Concentration Vs time for product c) concentration Vs time for a zero order reaction d) ln[R] vs time for a first order reaction e) potential energy diagram of a catalysed reaction.
Answer: Below are the standard graphical representations for chemical kinetics:

* (a & b) Concentration vs. Time for Reactants and Products:

* (c) Concentration vs. Time for Zero-Order:

* (d) ln[R] vs. Time for First-Order:

* (e) Potential Energy of Catalyzed Reaction: (Refer to the SVG provided in Question 11).
In simple words: Reactants decline over time while products rise. In zero-order reactions, concentration decreases as a straight line, and in first-order, the natural log of concentration decreases linearly.

Exam Tip: Be sure to write the slope and intercept values clearly next to the axes on your graphs to get full marks.

Question 15. What is meant by order of reaction being zero?
Answer: A zero-order reaction means that the rate of the chemical reaction is completely independent of the concentration of its reactants. The reaction rate remains constant throughout the process, even as the reactants are consumed.
In simple words: The reaction runs at one steady speed from start to finish, and adding more ingredients does not speed it up.

Exam Tip: State that for zero-order reactions, the rate law is \( \text{Rate} = k[A]^0 = k \), showing that rate equals the rate constant.

Question 16. Half life of a reaction is inversely proportional to initial concentration of the reactant.Determine the order of this reaction.
Answer:
The general relationship between the half-life (\( t_{1/2} \)) and initial concentration (\( [A]_0 \)) of a reaction is given by:
\[ t_{1/2} \propto \frac{1}{[A]_0^{n-1}} \]
where \( n \) represents the order of the reaction.
The question states that:
\[ t_{1/2} \propto \frac{1}{[A]_0} \]
Equating the exponents of the concentration terms:
\[ n - 1 = 1 \implies n = 2 \]
Therefore, the reaction is of the second order.
In simple words: Since the half-life is cut in half when you double the starting concentration, this relationship corresponds to a second-order reaction.

Exam Tip: Memorize the general proportion \( t_{1/2} \propto a^{1-n} \) to find the reaction order quickly in exams.

Question 17. The decomposition ammonia on a platinum surface follows zero order kinetics.
2NH3(g) → N2(g)+3H2(g) K= 2.5x10-4 mole/l/sec. Determine the rate of a) disapperance of NH3 b) rate of formation of N2 (c) rate of formation of H2.
Answer:
For the reaction \( 2\text{NH}_3\text{(g)} \rightarrow \text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \), the rate of reaction is written as:
\[ \text{Rate} = -\frac{1}{2}\frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3}\frac{d[\text{H}_2]}{dt} \]
Since the reaction follows zero-order kinetics:
\[ \text{Rate} = K = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \]

* (a) Rate of disappearance of \( \text{NH}_3 \):
\[ -\frac{d[\text{NH}_3]}{dt} = 2 \times \text{Rate} = 2 \times (2.5 \times 10^{-4}) = 5.0 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \]
* (b) Rate of formation of \( \text{N}_2 \):
\[ \frac{d[\text{N}_2]}{dt} = \text{Rate} = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \]
* (c) Rate of formation of \( \text{H}_2 \):
\[ \frac{d[\text{H}_2]}{dt} = 3 \times \text{Rate} = 3 \times (2.5 \times 10^{-4}) = 7.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1} \]
In simple words: (a) Ammonia disappears twice as fast as the reaction rate because of the coefficient 2. (b) Nitrogen forms at the same speed as the reaction rate. (c) Hydrogen forms three times faster because of the coefficient 3.

Exam Tip: Be sure to include the proper units (\( \text{mol L}^{-1} \text{ s}^{-1} \)) with each calculated rate to prevent loss of marks.

Question 18. A first order reaction is 20% complete in 10 minutes. Determine the time taken for 80% completion of the reaction.
Answer:
For a first-order reaction:
\[ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]} \]

Step 1: Calculate the rate constant (\( k \)) using 20% completion data:
Here, \( t = 10 \text{ min} \), initial amount \( [A]_0 = 100 \), and remaining amount \( [A] = 100 - 20 = 80 \):
\[ k = \frac{2.303}{10} \log \frac{100}{80} = 0.2303 \times \log(1.25) \]
\[ k = 0.2303 \times 0.0969 \approx 0.0223 \text{ min}^{-1} \]

Step 2: Calculate the time required for 80% completion (\( t_{80\%} \)):
Here, \( [A]_0 = 100 \), and remaining amount \( [A] = 100 - 80 = 20 \):
\[ t_{80\%} = \frac{2.303}{k} \log \frac{100}{20} \]
\[ t_{80\%} = \frac{2.303}{0.0223} \log(5) = \frac{2.303 \times 0.699}{0.0223} \]
\[ t_{80\%} = \frac{1.61}{0.0223} \approx 72.2 \text{ minutes} \]
The time taken for 80% completion is \( 72.2 \text{ minutes} \).
In simple words: First we find the reaction speed constant (\( 0.0223 \text{ min}^{-1} \)) from the 20% completion rate. We then use this constant to calculate that it takes 72.2 minutes to reach 80% completion.

Exam Tip: Do not round your rate constant (\( k \)) too early in the calculation, as intermediate rounding can cause discrepancies in your final answer.

Question 19. 2A +B → A2B K= 2.5 x10-4 M-2 sec-1 Find the rate when the initial concentrations of [A] = 0.1M [B] = 0.2 M. Also find the rate when 0.04 moles/litre of A has reacted. Rate=K[A][B]2
Answer:
Given the rate law: \( \text{Rate} = K [A][B]^2 \), and rate constant \( K = 2.5 \times 10^{-4} \text{ M}^{-2} \text{ s}^{-1} \):

1. Calculation of Initial Rate:
\[ \text{Rate}_1 = (2.5 \times 10^{-4}) \times (0.1) \times (0.2)^2 \]

\implies \text{Rate}_1 = 2.5 \times 10^{-4} \times 0.1 \times 0.04 = 1.0 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1}

2. Calculation of Rate when 0.04 mol/L of A has reacted:
* Concentration of remaining A: \( [A] = 0.1 - 0.04 = 0.06 \text{ M} \).
* From the reaction stoichiometry \( 2\text{A} + \text{B} \rightarrow \text{A}_2\text{B} \), the amount of B consumed is half of the amount of A consumed:
\[ \text{Amount of B reacted} = \frac{0.04}{2} = 0.02 \text{ M} \]
* Concentration of remaining B: \( [B] = 0.2 - 0.02 = 0.18 \text{ M} \).
Now, substitute these remaining concentrations into the rate law:
\[ \text{Rate}_2 = (2.5 \times 10^{-4}) \times (0.06) \times (0.18)^2 \]
\[ \text{Rate}_2 = 2.5 \times 10^{-4} \times 0.06 \times 0.0324 = 4.86 \times 10^{-7} \text{ mol L}^{-1} \text{ s}^{-1} \]
The initial rate is \( 1.0 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1} \) and the rate after reaction is \( 4.86 \times 10^{-7} \text{ mol L}^{-1} \text{ s}^{-1} \).
In simple words: The initial rate is calculated directly. When some of reactant A is consumed, reactant B is also consumed in a 1:2 ratio. Calculating with these lower remaining amounts gives a slower rate of 4.86 x 10^-7 mol L^-1 s^-1.

Exam Tip: Be sure to account for stoichiometry when calculating the remaining concentration of reactant B; since 2 moles of A react with 1 mole of B, B is consumed at half the rate of A.

Question 20. For a certain chemical reaction variation in the concentration in [R ] versus time(s) plot is given below
[Graph showing ln[R] vs Time sloping down]
i) what is the order of the reactions? ii) what are the units of rate constant k? iii) give the relationship between k and t1/2 iv) what does the slope of the above line indicate? v) draw the plot [R ] 0 / [R] versus time(s)
Answer:
* i) Order of the reaction: The reaction is of the first order, because the graph of \( \ln[R] \) versus Time is a straight line.
* ii) Units of rate constant \( k \): For a first-order reaction, the unit of \( k \) is **\( \text{s}^{-1} \)** (or \( \text{time}^{-1} \)).
* iii) Relationship between \( k \) and \( t_{1/2} \):
\[ t_{1/2} = \frac{0.693}{k} \]
* iv) Significance of the slope: The slope of this line is equal to **\( -k \)** (negative of the rate constant).
* v) Plot of \( \ln([R]_0 / [R]) \) versus Time: Since \( \ln([R]_0 / [R]) = kt \), the graph is a straight line starting at the origin and sloping upward with a positive slope of \( k \):

In simple words: i) The linear drop in ln[R] means it is a first-order reaction. ii) The unit is per second. iii) Half-life is 0.693 divided by k. iv) The slope is -k. v) Plotting ln([R]0/[R]) gives an upward straight line from the origin.

Exam Tip: Make sure you label the y-axis as \( \ln([R]_0/[R]) \) to show that the graph has a positive slope.

Question 21. 2NO2+F2 → 2NO2F Write the rate of reaction in terms of
(a) rate of formation of NO2F (b) rate of disappearance of NO2 (c) rate of disappearance of F2
Answer:
For the reaction \( 2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F} \), the general rate of reaction is written as:
\[ \text{Rate} = -\frac{1}{2}\frac{d[\text{NO}_2]}{dt} = -\frac{d[\text{F}_2]}{dt} = \frac{1}{2}\frac{d[\text{NO}_2\text{F}]}{dt} \]
We express the rates in terms of each individual species:
* (a) Rate of formation of \( \text{NO}_2\text{F} \): \( \text{Rate} = \frac{1}{2} \frac{d[\text{NO}_2\text{F}]}{dt} \)
* (b) Rate of disappearance of \( \text{NO}_2 \): \( \text{Rate} = -\frac{1}{2} \frac{d[\text{NO}_2]}{dt} \)
* (c) Rate of disappearance of \( \text{F}_2 \): \( \text{Rate} = -\frac{d[\text{F}_2]}{dt} \)
In simple words: The rate of the overall reaction is equal to the rate of consumption of F2, or half the consumption of NO2, or half the production of NO2F, due to their reaction ratios.

Exam Tip: Remember to include the negative sign for disappearing reactants and dividing coefficients for any species with stoichiometric values larger than 1.

Page 2

Question 22. Consider the reaction R → P. The change in the concentration of R with shown in the following plot.
[Graph showing [R] vs Time sloping down as a straight line]
i) Predict the order of the reaction. ii) Write the expression for half life of this reaction.
Answer:
* i) Order of the reaction: The reaction is of the zero order, because the graph of concentration \( [R] \) versus Time is a straight line.
* ii) Expression for half-life: For a zero-order reaction, the half-life expression is:
\[ t_{1/2} = \frac{[R]_0}{2k} \]
In simple words: i) Concentration decreasing as a perfect straight line tells us the reaction is zero-order. ii) Its half-life is the initial concentration divided by 2k.

Exam Tip: Be sure to point out that the slope of a concentration vs. time plot for a zero-order reaction is equal to \( -k \).

Question 23. The decomposition of NH3 follows zero order. 2 NH3 → N2+3H2 Find the rate of production of N2 and H2.K=2.5x10-4MS-1 −1
Answer:
For the zero-order reaction \( 2\text{NH}_3 \rightarrow \text{N}_2 + 3\text{H}_2 \), the rate relationship is:
\[ \text{Rate} = -\frac{1}{2}\frac{d[\text{NH}_3]}{dt} = \frac{d[\text{N}_2]}{dt} = \frac{1}{3}\frac{d[\text{H}_2]}{dt} = K = 2.5 \times 10^{-4} \text{ M s}^{-1} \]
Calculating production rates:
* Rate of production of \( \text{N}_2 \):
\[ \frac{d[\text{N}_2]}{dt} = K = 2.5 \times 10^{-4} \text{ M s}^{-1} \]
* Rate of production of \( \text{H}_2 \):
\[ \frac{d[\text{H}_2]}{dt} = 3 \times K = 3 \times (2.5 \times 10^{-4}) = 7.5 \times 10^{-4} \text{ M s}^{-1} \]
The rate of production of \( \text{N}_2 \) is \( 2.5 \times 10^{-4} \text{ M s}^{-1} \) and of \( \text{H}_2 \) is \( 7.5 \times 10^{-4} \text{ M s}^{-1} \).
In simple words: Since the reaction rate is constant at K, nitrogen is produced at that same speed, and hydrogen is produced three times faster because three moles are generated for every reaction cycle.

Exam Tip: Always make sure to write correct units like \( \text{M s}^{-1} \) (Molar per second) or \( \text{mol L}^{-1} \text{ s}^{-1} \).

Question 24. 2A+B+C_ A2B+C Rate=K(A)(B)2 K=2x10-6M-2S-1Calculate the initial rate when (A)=0.1M (B)=0.2M (C)=0.6M Find the rate when 0.04mole of (A) is consumed.
Answer:
Given rate law: \( \text{Rate} = K [A][B]^2 \), with \( K = 2 \times 10^{-6} \text{ M}^{-2} \text{ s}^{-1} \). Reactant C is not in the rate law and does not affect the speed.

1. Calculation of Initial Rate:
\[ \text{Rate}_1 = K [A][B]^2 = (2 \times 10^{-6}) \times (0.1) \times (0.2)^2 \]

\implies \text{Rate}_1 = 2 \times 10^{-6} \times 0.1 \times 0.04 = 8.0 \times 10^{-9} \text{ M s}^{-1}

2. Calculation of Rate when 0.04 M of A is consumed:
* Remaining A: \( [A] = 0.1 - 0.04 = 0.06 \text{ M} \).
* Consumed B: \( \frac{0.04}{2} = 0.02 \text{ M} \) (based on the \( 2:1 \) ratio of reaction stoichiometry).
* Remaining B: \( [B] = 0.2 - 0.02 = 0.18 \text{ M} \).
Now, calculate the new rate:
\[ \text{Rate}_2 = (2 \times 10^{-6}) \times (0.06) \times (0.18)^2 \]
\[ \text{Rate}_2 = 2 \times 10^{-6} \times 0.06 \times 0.0324 = 3.89 \times 10^{-9} \text{ M s}^{-1} \]
The initial rate is \( 8.0 \times 10^{-9} \text{ M s}^{-1} \) and the rate after reaction is \( 3.89 \times 10^{-9} \text{ M s}^{-1} \).
In simple words: The initial rate is 8.0 x 10^-9 M/s. After some of reactant A is used up, reactant B also drops in a 1:2 ratio. Calculating with these lower remaining values gives a slower rate of 3.89 x 10^-9 M/s.

Exam Tip: Reactant C is a spectator in the rate-determining step because its concentration is raised to the power of 0, so it can be safely ignored in calculations.

Question 25. 2NO2+F2 → 2NO2F
Experiment (NO2)M (F2) M Rate(M/S)
1             0.2     0.05     0.006
2             0.4     0.05     0.012
3.            0.8     0.10     0.048
Find the order with respect to NO2 and F2.Also find the overall order of the reaction.
Answer:
Let the rate law be: \( \text{Rate} = k [\text{NO}_2]^x [\text{F}_2]^y \).

* Step 1: Find order with respect to \( \text{NO}_2 \) (\( x \)):
Comparing Experiments 1 and 2, the concentration of \( \text{F}_2 \) remains constant at \( 0.05 \text{ M} \) while the concentration of \( \text{NO}_2 \) is doubled (\( 0.2 \rightarrow 0.4 \text{ M} \)). The rate doubles (\( 0.006 \rightarrow 0.012 \text{ M/s} \)):
\[ 2^x = 2 \implies x = 1 \]
The reaction is first-order with respect to \( \text{NO}_2 \).

* Step 2: Find order with respect to \( \text{F}_2 \) (\( y \)):
Comparing Experiments 2 and 3:
\[ \text{Rate}_2 = k(0.4)^1(0.05)^y = 0.012 \]
\[ \text{Rate}_3 = k(0.8)^1(0.10)^y = 0.048 \keys \]
Dividing Experiment 3 by Experiment 2:
\[ \frac{0.048}{0.012} = \left(\frac{0.8}{0.4}\right) \left(\frac{0.10}{0.05}\right)^y \]
\[ 4 = 2 \times (2)^y \implies 2 = 2^y \implies y = 1 \]
The reaction is first-order with respect to \( \text{F}_2 \).

* Step 3: Find overall order:
\[ \text{Overall Order} = x + y = 1 + 1 = 2 \]
The overall order of the reaction is \( 2 \).
In simple words: Doubling NO2 doubles the rate, meaning it is first-order. Comparing the remaining runs shows F2 is also first-order, making the total overall order equal to 2.

Exam Tip: Show the division of equations clearly to demonstrate how the rate constant \( k \) cancels out during your order determination.

Question 26. Show that(a) 2t½=t¾ (for first order) (b)Half life of a reaction is 10seconds.Find t2/3
Answer:
* (a) Verification for first-order reaction:
The first-order integrated equation is:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]
* For half-life (\( t_{1/2} \), 50% complete):
\[ t_{1/2} = \frac{2.303}{k} \log \left(\frac{100}{50}\right) = \frac{2.303}{k} \log(2) \]
* For three-quarters life (\( t_{3/4} \) or \( t_{75\%} \), 75% complete, leaving 25%):
\[ t_{3/4} = \frac{2.303}{k} \log \left(\frac{100}{25}\right) = \frac{2.303}{k} \log(4) = \frac{2.303}{k} \cdot 2 \log(2) \]
Comparing both equations:
\[ t_{3/4} = 2 \times t_{1/2} \]
Thus, \( \mathbf{2 t_{1/2} = t_{3/4}} \).

* (b) Find \( t_{2/3} \) given \( t_{1/2} = 10 \text{ s} \):
The rate constant is:
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{10} = 0.0693 \text{ s}^{-1} \]
For a two-thirds completion (\( 66.7 \% \) complete, leaving \( 1/3 \) of the reactant):
\[ t_{2/3} = \frac{2.303}{k} \log \frac{1}{1 - 2/3} = \frac{2.303}{0.0693} \log(3) \]
\[ t_{2/3} = \frac{2.303 \times 0.4771}{0.0693} \approx 15.8 \text{ seconds} \]
The time taken for two-thirds completion is \( 15.8 \text{ seconds} \).
In simple words: (a) A first-order reaction takes exactly twice as long to reach 75% completion as it does to reach 50% completion. (b) If the half-life is 10 seconds, it takes 15.8 seconds to react two-thirds of the starting material.

Exam Tip: Be sure to write \( t_{3/4} \) as the time required for 75% completion, as some textbooks use \( t_{75\%} \) to represent the same concept.

Question 27. Rate of a reaction becomes 1.414 times when concentration of the reactant is doubled. Find the order of the reaction.
Answer:
Let the rate law be: \( \text{Rate} = k [A]^n \).
When the concentration is doubled:
\[ \text{Rate}' = k (2[A])^n = 2^n \cdot k [A]^n = 2^n \cdot \text{Rate} \]
Given that \( \text{Rate}' = 1.414 \cdot \text{Rate} \):
\[ 2^n = 1.414 \]
Since \( 1.414 \approx \sqrt{2} = 2^{0.5} \):
\[ 2^n = 2^{0.5} \implies n = 0.5 \]
Therefore, the order of the reaction is \( 0.5 \) (or half-order).
In simple words: Since doubling the starting concentration increases the reaction rate by a factor of the square root of 2 (1.414), the reaction order must be exactly 0.5.

Exam Tip: Express \( 1.414 \) as \( 2^{1/2} \) to make the exponent comparison clear and elegant.

Question 15. (a) show that for a first order reaction t½ is independent of the initial concentration of the reactant. (b) show that for a zero order reaction t½ is directly proportional to initial concentration of the reactant and inversely proportional to rate constant.
Answer:
* (a) First-Order Reaction:
The integrated rate expression for first-order kinetics is:
\[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \]
At half-life \( t = t_{1/2} \), the remaining reactant concentration is \( [R] = \frac{[R]_0}{2} \). Substituting this:
\[ k = \frac{2.303}{t_{1/2}} \log \frac{[R]_0}{[R]_0/2} \implies t_{1/2} = \frac{2.303 \log(2)}{k} \implies \mathbf{t_{1/2} = \frac{0.693}{k}} \]
Since the concentration term cancels out, \( t_{1/2} \) is completely independent of the initial concentration \( [R]_0 \).

* (b) Zero-Order Reaction:
The integrated rate expression for zero-order kinetics is:
\[ k = \frac{[R]_0 - [R]}{t} \]
At half-life \( t = t_{1/2} \), and \( [R] = \frac{[R]_0}{2} \). Substituting this:
\[ k = \frac{[R]_0 - [R]_0/2}{t_{1/2}} = \frac{[R]_0}{2 t_{1/2}} \implies \mathbf{t_{1/2} = \frac{[R]_0}{2k}} \]
From this equation, we can see that \( \mathbf{t_{1/2} \propto [R]_0} \) (directly proportional to initial concentration) and \( \mathbf{t_{1/2} \propto \frac{1}{k}} \) (inversely proportional to the rate constant).
In simple words: (a) A first-order half-life is determined solely by the speed constant k, meaning it never changes with starting concentration. (b) A zero-order half-life is directly scaled by the starting concentration, and inversely scaled by the speed constant k.

Exam Tip: Be sure to write down the intermediate steps showing how the concentration term \( [R]_0 \) cancels out for first-order but remains in zero-order.

Question 28. Rate constant of a reaction is 2M -1 S -1 at 700K and 32 M -1 S -1 at 800K.Find Ea
Answer:
We use the integrated form of the Arrhenius equation:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
Given values:
* \( k_1 = 2 \text{ M}^{-1} \text{ s}^{-1} \), at \( T_1 = 700 \text{ K} \)
* \( k_2 = 32 \text{ M}^{-1} \text{ s}^{-1} \), at \( T_2 = 800 \text{ K} \)
* Gas constant \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \)
Substitute the values:
\[ \log \frac{32}{2} = \frac{E_a}{2.303 \times 8.314} \left( \frac{800 - 700}{700 \times 800} \right) \]
\[ \log(16) = \frac{E_a}{19.147} \left( \frac{100}{560000} \right) \]
\[ 1.204 = \frac{E_a}{19.147 \times 5600} \]
\[ E_a = 1.204 \times 19.147 \times 5600 \approx 129094 \text{ J/mol} \approx 129.1 \text{ kJ/mol} \]
The activation energy \( E_a \) is \( 129.1 \text{ kJ/mol} \).
In simple words: The rate constant increases 16-fold over a 100 K temperature rise. Using the Arrhenius relation, we find that the reaction has an activation energy barrier of 129.1 kJ/mol.

Exam Tip: Convert your final activation energy answer to kilojoules (\( \text{kJ/mol} \)) for neat, standard physical chemistry presentation.

Question 29. Rate of a reaction becomes 4 times when temperature changes from 27 0C to 37 0C. Find Ea.
Answer:
First, we convert the temperatures to Kelvin:
* \( T_1 = 27\ ^\circ\text{C} = 27 + 273.15 = 300.15 \text{ K} \approx 300 \text{ K} \)
* \( T_2 = 37\ ^\circ\text{C} = 37 + 273.15 = 310.15 \text{ K} \approx 310 \text{ K} \)
Since the rate becomes 4 times faster, the rate constant ratio \( \frac{k_2}{k_1} = 4 \).
Using the Arrhenius equation:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
Substitute the values:
\[ \log(4) = \frac{E_a}{2.303 \times 8.314} \left( \frac{310 - 300}{300 \times 310} \right) \]
\[ 0.602 = \frac{E_a}{19.147} \left( \frac{10}{93000} \right) \]
\[ 0.602 = \frac{E_a}{19.147 \times 9300} \]
\[ E_a = 0.602 \times 19.147 \times 9300 \approx 107198 \text{ J/mol} \approx 107.2 \text{ kJ/mol} \]
The activation energy is \( 107.2 \text{ kJ/mol} \).
In simple words: A 10-degree warming from room temperature quadruples the reaction speed. This rapid acceleration corresponds to an activation energy of 107.2 kJ/mol.

Exam Tip: Be sure to write out temperatures in Kelvin, as standard formulas do not work with Celsius temperatures.

Question 30. Show that for a first order reaction, time required for 99.9% reaction is 10 times the time needed for 50% completion of the reaction.
Answer:
For a first-order reaction:
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]

* Step 1: Time for 50% completion (\( t_{50\%} \)):
\[ t_{50\%} = \frac{2.303}{k} \log \left(\frac{100}{50}\right) = \frac{2.303 \log(2)}{k} \approx \frac{0.693}{k} \]

* Step 2: Time for 99.9% completion (\( t_{99.9\%} \)):
Here, the remaining reactant is \( [A] = 100 - 99.9 = 0.1 \):
\[ t_{99.9\%} = \frac{2.303}{k} \log \left(\frac{100}{0.1}\right) = \frac{2.303}{k} \log(1000) \]
\[ t_{99.9\%} = \frac{2.303}{k} \times 3 = \frac{6.909}{k} \]

* Step 3: Compare both times:
\[ \frac{t_{99.9\%}}{t_{50\%}} = \frac{6.909 / k}{0.693 / k} \approx 10 \]
Thus, \( \mathbf{t_{99.9\%} = 10 \times t_{50\%}} \).
In simple words: Calculating the times shows that reaching 99.9% completion takes 6.909/k seconds, which is exactly ten times longer than the 0.693/k seconds needed to reach 50% completion.

Exam Tip: Writing the remaining concentration as \( 0.1 \% \) instead of the consumed \( 99.9 \% \) is critical to getting the correct log value.

Question 31. A piece of wood shows C14 activity which is 60% activity found today. Find the age of the sample.t½=5770years.
Answer:
Radioactive decay is a first-order kinetic process. First, we find the decay constant (\( k \)):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5770 \text{ years}} \approx 1.201 \times 10^{-4} \text{ year}^{-1} \]

Now, using the first-order integrated rate law to find the age of the wood sample (\( t \)):
\[ t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \]
Given that the remaining activity is \( 60\% \) of the initial activity (\( [A] = 0.6 [A]_0 \)):
\[ t = \frac{2.303}{1.201 \times 10^{-4}} \log \left(\frac{100}{60}\right) \]
\[ t = (1.9176 \times 10^4) \times \log(1.667) \]
\[ t = 19176 \times 0.2218 \approx 4253 \text{ years} \]
The age of the wood sample is approximately \( 4253 \text{ years} \).
In simple words: Carbon-14 decays with a known half-life, giving a decay constant. Measuring that the remaining activity is 60% allows us to calculate that the wood sample is roughly 4,253 years old.

Exam Tip: Be sure to keep intermediate calculations like the decay constant \( k \) precise to avoid rounding errors in your final age value.

Question 32. The following data were obtained during the first order decomposition of SO2Cl2 at constant volume . SO2Cl2(g) → SO2(g) +Cl2(g)
Experiment     Time(sec)     Total pressure(atm)
1                     0                 0.5
2                     100               0.6
Calculate the rate when total pressure is 0.65 atmospheres.
Answer:
For the reaction: \( \text{SO}_2\text{Cl}_2\text{(g)} \rightarrow \text{SO}_2\text{(g)} + \text{Cl}_2\text{(g)} \).
Initially (\( t = 0 \)), pressure is \( P_{\text{SO}_2\text{Cl}_2} = P_i = 0.5 \text{ atm} \).
At time \( t \), total pressure \( P_t = P_i + x \), where \( x \) is the pressure decrease of \( \text{SO}_2\text{Cl}_2 \). Thus, \( x = P_t - P_i \).
The remaining pressure of \( \text{SO}_2\text{Cl}_2 \) is:
\[ P_{\text{SO}_2\text{Cl}_2} = P_i - x = P_i - (P_t - P_i) = 2P_i - P_t \]

Step 1: Calculate the rate constant (\( k \)) using 100 s data:
Here, \( P_i = 0.5 \text{ atm} \) and \( P_t = 0.6 \text{ atm} \):
\[ P_{\text{SO}_2\text{Cl}_2} = 2(0.5) - 0.6 = 0.4 \text{ atm} \]
\[ k = \frac{2.303}{100} \log \left(\frac{0.5}{0.4}\right) = 0.02303 \times 0.0969 \approx 2.23 \times 10^{-3} \text{ s}^{-1} \keys \]

Step 2: Calculate the rate when total pressure is 0.65 atm:
Here, \( P_t' = 0.65 \text{ atm} \):
The remaining pressure of reactant at this point is:
\[ P_{\text{SO}_2\text{Cl}_2}' = 2(0.5) - 0.65 = 0.35 \text{ atm} \]
The rate of the reaction is:
\[ \text{Rate} = k \cdot P_{\text{SO}_2\text{Cl}_2}' = (2.23 \times 10^{-3} \text{ s}^{-1}) \times 0.35 \text{ atm} \approx 7.8 \times 10^{-4} \text{ atm s}^{-1} \]
The reaction rate at 0.65 atm is \( 7.8 \times 10^{-4} \text{ atm s}^{-1} \).
In simple words: We calculate the rate constant \( 2.23 \times 10^{-3} \text{ s}^{-1} \) from the pressure change at 100 seconds. When the total pressure reaches 0.65 atm, the reactant pressure drops to 0.35 atm, giving a reaction rate of 7.8 x 10^-4 atm/s.

Exam Tip: Be sure to write the formula for remaining reactant pressure as \( 2P_i - P_t \) to handle gas pressure calculations correctly.

Question 33. 2N2O5(g) → 2 N2O4(g) +O2(g) follows first order kinetics at constant volume.
Experiment     Time(sec)     Total pressure(atm)
1                     0                 0.5
2                     100               0.512
Find the value of rate constant K.
Answer:
For the reaction: \( 2\text{N}_2\text{O}_5\text{(g)} \rightarrow 2\text{N}_2\text{O}_4\text{(g)} + \text{O}_2\text{(g)} \).
Initially (\( t = 0 \)), pressure is \( P_i = 0.5 \text{ atm} \).
At time \( t = 100 \text{ s} \):
* \( P_{\text{N}_2\text{O}_5} = P_i - 2x \)
* \( P_{\text{N}_2\text{O}_4} = 2x \)
* \( P_{\text{O}_2} = x \)
The total pressure \( P_t \) is:
\[ P_t = (P_i - 2x) + 2x + x = P_i + x = 0.512 \text{ atm} \]
Since \( P_i = 0.5 \text{ atm} \):
\[ 0.5 + x = 0.512 \implies x = 0.012 \text{ atm} \]
The remaining pressure of reactant \( \text{N}_2\text{O}_5 \) is:
\[ P_{\text{N}_2\text{O}_5} = P_i - 2x = 0.5 - 2(0.012) = 0.476 \text{ atm} \]
Using the first-order rate constant equation:
\[ K = \frac{2.303}{t} \log \frac{P_i}{P_{\text{N}_2\text{O}_5}} \]
\[ K = \frac{2.303}{100} \log \left(\frac{0.5}{0.476}\right) = 0.02303 \times \log(1.0504) \]
\[ K = 0.02303 \times 0.0213 \approx 4.9 \times 10^{-4} \text{ s}^{-1} \]
The rate constant \( K \) is \( 4.9 \times 10^{-4} \text{ s}^{-1} \).
In simple words: The gas pressure increase shows that 0.012 atm of oxygen was formed, meaning 0.024 atm of reactant was consumed. Substituting these values into our equation gives a rate constant of 4.9 x 10^-4 s^-1.

Exam Tip: Always account for the stoichiometry coefficients when calculating the pressure changes of products and reactants.

Question 34. The time required for 10% completion of a first order reaction at 298K is equal to that required for 25% completion at 308K. Find Ea. Calculate K at 318K.
Answer:
For a first-order reaction:
\[ t = \frac{2.303}{k} \log \frac{100}{100-x} \]

* Step 1: Write the time equations for both temperatures:
At \( 298 \text{ K} \) (\( T_1 = 298 \), \( 10\% \) completion):
\[ t_{10\%} = \frac{2.303}{k_1} \log \left(\frac{100}{90}\right) = \frac{2.303 \times 0.0458}{k_1} \]
At \( 308 \text{ K} \) (\( T_2 = 308 \), \( 25\% \) completion):
\[ t_{25\%} = \frac{2.303}{k_2} \log \left(\frac{100}{75}\right) = \frac{2.303 \times 0.1249}{k_2} \]
Since the times are equal (\( t_{10\%} = t_{25\%} \)):
\[ \frac{2.303 \times 0.0458}{k_1} = \frac{2.303 \times 0.1249}{k_2} \implies \frac{k_2}{k_1} = \frac{0.1249}{0.0458} \approx 2.727 \]

* Step 2: Calculate the activation energy (\( E_a \)):
Using the Arrhenius equation:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303 R} \left( \frac{T_2 - T_1}{T_1 T_2} \right) \]
Substitute the values:
\[ \log(2.727) = \frac{E_a}{19.147} \left( \frac{308 - 298}{298 \times 308} \right) \]
\[ 0.4357 = \frac{E_a}{19.147} \left( \frac{10}{91784} \right) \]
\[ E_a = 0.4357 \times 19.147 \times 9178.4 \approx 76569 \text{ J/mol} \approx 76.57 \text{ kJ/mol} \]
The activation energy is \( 76.57 \text{ kJ/mol} \).
In simple words: Since the reaction rates at both temperatures lead to a rate ratio of 2.727, using the Arrhenius equation gives an activation energy of 76.57 kJ/mol.

Exam Tip: Be sure to write the Arrhenius equation fully before solving, and state the final answer in kilojoules per mole.

Question 35. 2HI(g) → H2(g) +I2(g) Ea at 581K is 209.5KJ/mole. Determine the fraction of molecules having energy equal to or greater than Ea.
Answer:
The fraction of molecules (\( f \)) having energy equal to or greater than \( E_a \) is given by the Boltzmann factor:
\[ f = e^{-E_a/RT} \implies \log_{10} f = -\frac{E_a}{2.303 R T} \]
Given:
* \( E_a = 209.5 \text{ kJ/mol} = 209500 \text{ J/mol} \)
* \( T = 581 \text{ K} \)
* \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \)
Substitute the values:
\[ \log_{10} f = -\frac{209500}{2.303 \times 8.314 \times 581} = -\frac{209500}{11124.6} \approx -18.831 \]
\[ f = 10^{-18.831} \approx 1.48 \times 10^{-19} \]
The fraction of molecules is \( 1.48 \times 10^{-19} \).
In simple words: Only a tiny fraction of molecules (roughly \( 1.48 \times 10^{-19} \)) have enough kinetic energy to overcome this high activation energy barrier at 581 K.

Exam Tip: Remember to use activation energy in Joules (\( \text{J/mol} \)) rather than kilojoules in the exponent calculation.

Question 36. Ea of a reaction is 75 KJ/mole in the absence of a catalyst and 50KJ/mole in the presence of a Catalyst at 300K. Determine the extent to which the rate of reaction is increased.
Answer:
Let \( k_u \) be the rate constant of the uncatalyzed reaction and \( k_c \) be the rate constant of the catalyzed reaction:
\[ k_u = A e^{-E_{a,u}/RT}, \quad k_c = A e^{-E_{a,c}/RT} \]
Taking their ratio:
\[ \frac{k_c}{k_u} = e^{(E_{a,u} - E_{a,c})/RT} \implies \log_{10} \frac{k_c}{k_u} = \frac{E_{a,u} - E_{a,c}}{2.303 R T} \]
Substitute the given values:
* \( E_{a,u} - E_{a,c} = 75000 - 50000 = 25000 \text{ J/mol} \)
* \( T = 300 \text{ K} \)
* \( R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1} \)
\[ \log_{10} \frac{k_c}{k_u} = \frac{25000}{2.303 \times 8.314 \times 300} = \frac{25000}{5744.1} \approx 4.352 \]
\[ \frac{k_c}{k_u} = 10^{4.352} \approx 22490 \]
The rate of reaction is increased by approximately 22,490 times (or \( 2.25 \times 10^4 \) times).
In simple words: Lowering the activation energy by 25 kJ/mol using a catalyst allows the reaction rate to increase by more than 22,000 times at room temperature.

Exam Tip: Be sure to subtract the two activation energy values and use the difference to solve for the rate ratio.

Question 37. The rate constant for the first order decomposition of H2O2 is given as logK = 14.34 – 1.25x 104 K/T. Calculate Ea for this reaction. At what temperature will its half life be 256 minutes?
Answer:
Step 1: Calculate the activation energy (\( E_a \)):
Comparing the given equation \( \log K = 14.34 - \frac{1.25 \times 10^4}{T} \) with the log form of the Arrhenius equation:
\[ \log K = \log A - \frac{E_a}{2.303 R T} \]
By matching the slope terms:
\[ \frac{E_a}{2.303 R} = 1.25 \times 10^4 \]
\[ E_a = 1.25 \times 10^4 \times 2.303 \times 8.314 \approx 239339 \text{ J/mol} \approx 239.3 \text{ kJ/mol} \]

Step 2: Calculate the temperature for \( t_{1/2} = 256 \text{ minutes} \):
Convert the half-life to seconds:
\[ t_{1/2} = 256 \times 60 = 15360 \text{ s} \]
For a first-order decay:
\[ K = \frac{0.693}{t_{1/2}} = \frac{0.693}{15360 \text{ s}} \approx 4.51 \times 10^{-5} \text{ s}^{-1} \]
Taking the common logarithm:
\[ \log K = \log(4.51 \times 10^{-5}) \approx -4.346 \]
Substitute this back into the original equation:
\[ -4.346 = 14.34 - \frac{1.25 \times 10^4}{T} \]
\[ \frac{1.25 \times 10^4}{T} = 14.34 + 4.346 = 18.686 \]
\[ T = \frac{12500}{18.686} \approx 669 \text{ K} \]
The activation energy is \( 239.3 \text{ kJ/mol} \) and the required temperature is \( 669 \text{ K} \) (or \( 396\ ^\circ\text{C} \)).
In simple words: By matching slopes with the Arrhenius equation, we find the activation energy is 239.3 kJ/mol. Using the desired half-life to find the rate constant, we calculate the required temperature is 669 K.

Exam Tip: Be sure to convert minutes to seconds when calculating the rate constant \( K \) because the pre-exponential factor is given in per second.

Question 38. The decomposition of hydrocarbon follows the equation K = (4.5x 1011 sec-1) e-28000K/T. Calculate Ea.
Answer:
Comparing the given equation \( K = (4.5 \times 10^{11} \text{ s}^{-1}) e^{-28000\text{ K}/T} \) with the Arrhenius equation:
\[ K = A e^{-E_a/RT} \]
By matching the exponents:
\[ \frac{E_a}{R} = 28000 \text{ K} \]
\[ E_a = 28000 \text{ K} \times 8.314 \text{ J K}^{-1} \text{ mol}^{-1} = 232792 \text{ J/mol} \approx 232.8 \text{ kJ/mol} \]
The activation energy \( E_a \) is \( 232.8 \text{ kJ/mol} \).
In simple words: Exponent matching shows that \( E_a/R \) equals 28,000, which yields a calculated activation energy of 232.8 kJ/mol.

Exam Tip: The units of the exponent term must cancel out completely, so \( E_a/R \) has the unit of Kelvin.

Question 39. 2NO(g) + O2(g) → 2NO2(g) occurs in one step. What will happen to the rate when the Volume of the reaction vessel is reduced to 1/3 of the original volume?
Answer:
Since the reaction occurs in a single step, it is an elementary reaction, and its rate law matches its stoichiometry:
\[ \text{Rate}_1 = K [\text{NO}]^2 [\text{O}_2] \]
When the volume of the reaction vessel is reduced to \( 1/3 \), the molar concentration of each reactant gas becomes \( 3 \) times its original value because concentration is inversely proportional to volume (\( C = \frac{n}{V} \)).
The new rate is:
\[ \text{Rate}_2 = K (3[\text{NO}])^2 (3[\text{O}_2]) = 27 \cdot K [\text{NO}]^2 [\text{O}_2] \]
\[ \text{Rate}_2 = 27 \times \text{Rate}_1 \]
Therefore, the rate of the reaction increases by 27 times.
In simple words: Squeezing the container to 1/3 volume triples the concentration of all reactants. Because the reaction is second-order in NO and first-order in O2, the rate increases by 3 squared times 3, which is 27 times.

Exam Tip: Be sure to write out the inverse proportionality between volume and concentration to justify the three-fold increase in concentrations.

Question 40. Rate of a reaction becomes 1.414 times when concentration of the reactant is doubled. Determine the order of the reaction.
Answer:
Let the rate law be: \( \text{Rate} = k [A]^n \).
When the concentration is doubled:
\[ \text{Rate}' = k (2[A])^n = 2^n \cdot k [A]^n = 2^n \cdot \text{Rate} \]
We are given that \( \text{Rate}' = 1.414 \cdot \text{Rate} \):
\[ 2^n = 1.414 \]
Since \( 1.414 \approx \sqrt{2} = 2^{0.5} \):
\[ 2^n = 2^{0.5} \implies n = 0.5 \]
The reaction is of the 0.5 order (or half-order).
In simple words: Because doubling the reactant concentration increases the rate by a factor of the square root of 2 (1.414), the order of this reaction is exactly 0.5.

Exam Tip: Show that \( 1.414 \) can be written as \( 2^{1/2} \) to make the exponent comparison clear and easy to follow.