Maharashtra Board Class 12 Chemistry Chapter 5 Electrochemistry Solutions

Electrochemistry Class 12 Exercise Question Answers Solutions Maharashtra Board

Class 12 Chemistry Chapter 5 Exercise Solutions Maharashtra Board

Chemistry Class 12 Chapter 5 Exercise Solutions

1. Choose the Most Correct Option.

Question i.Two solutions have the ratio of their concentrations 0.4 and ratio of their conductivities 0.216. The ratio of their molar conductivities will be
(a) 0.54
(b) 11.574
(c) 0.0864
(d) 1.852
Answer: (a) 0.54
In simple words: Molar conductivity is calculated by dividing conductivity by concentration. For two solutions, the ratio of molar conductivities is the ratio of conductivities divided by the ratio of concentrations.

🎯 Exam Tip: Remember the formula for molar conductivity and how to use ratios for comparison in such problems.

Question ii.On diluting the solution of an electrolyte,
(a) both \( \Lambda \) and \( \kappa \) increase
(b) both \( \Lambda \) and \( \kappa \) decrease
(c) \( \Lambda \) increases and \( \kappa \) decreases
(d) \( \Lambda \) decreases and \( \kappa \) increases
Answer: (c) \( \Lambda \) increases and \( \kappa \) decreases
In simple words: Dilution increases molar conductivity (\( \Lambda \)) because total ions move more freely, but decreases conductivity (\( \kappa \)) as the number of ions per unit volume reduces.

🎯 Exam Tip: Distinguish carefully between molar conductivity and specific conductivity (or simply conductivity) and their respective behaviors upon dilution.

Question iii.1 S m² mol-¹ is equal to
(a) 10-4 S m² mol-1
(b) 104 Ω-¹ cm² mol-1
(c) 10-2 S cm² mol-1
(d) 102 Ω-1 cm² mol-1
Answer: (b) 104 Ω-¹ cm² mol-1
In simple words: This question involves unit conversion. Since 1 S = 1 Ω-¹ and 1 m² = 104 cm², then 1 S m² mol-¹ converts to 104 Ω-¹ cm² mol-¹.

🎯 Exam Tip: Be proficient in unit conversions, especially between SI and CGS units for conductivity and molar conductivity.

Question iv.The standard potential of the cell in which the following reaction occurs H2+ (g, 1 atm) + Cu²+ (1 M) → 2H (1 M) + Cu(s), (E⁰Cu = 0.34 V) is
(a) - 0.34 V
(b) 0.34 V
(c) 0.17 V
(d) -0.17 V
Answer: (b) 0.34 V
In simple words: The standard cell potential for this reaction, involving hydrogen and copper, is determined by the standard reduction potential of copper, as hydrogen's standard potential is 0V.

🎯 Exam Tip: Remember that the standard electrode potential for the standard hydrogen electrode (SHE) is assigned as 0V, simplifying calculations for cells involving it.

Question v.For the cell, Pb(s)|Pb2+ (1 M)|| Ag+ (1 M)|Ag(s), if concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will
(a) increase by 10 V
(b) increase by 0.0296 V
(c) decrease by 10 V
(d) decrease by 0.0296 V
Answer: (d) decrease by 0.0296 V
In simple words: Increasing the concentration of the anode ion (Pb²+) shifts the equilibrium, based on the Nernst equation, reducing the cell's potential by 0.0296 V for a tenfold increase.

🎯 Exam Tip: The Nernst equation is crucial for understanding how concentration changes affect cell potential. Pay attention to which compartment's concentration is changing and its role in the reaction quotient.

Question vi.Consider the half reactions with standard potentials
(i) Ag\(^{ + }_{(aq)}\) + e- \( \longrightarrow \) Ag\(_{(s)}\) Eº = 0.8 V
(ii) I2\(_{(s)}\) + 2e- \( \longrightarrow \) 2I\(^{-}_{(aq)}\) Eº = 0.53 V
(iii) Pb\(^{2+}_{(aq)}\) + 2e- \( \longrightarrow \) Pb\(_{(s)}\) Eº = -0.13 V
(iv) Fe\(^{2+}_{(aq)}\) + 2e- \( \longrightarrow \) Fe\(_{(s)}\) Eº = -0.44 V
The strongest oxidising and reducing agents respectively are
(a) Ag and Fe\(^{2+}\)
(b) Ag\(^{+}\) and Fe
(c) Pb\(^{2+}\) and I\(^{-}\)
(d) I2 and Fe\(^{2+}\)
Answer: (b) Ag\(^{+}\) and Fe
In simple words: The strongest oxidizing agent has the highest positive standard reduction potential (Ag\(^{+}\)), and the strongest reducing agent has the lowest (most negative) standard reduction potential (Fe).

🎯 Exam Tip: A higher (more positive) standard reduction potential indicates a stronger oxidizing agent, while a lower (more negative) potential indicates a stronger reducing agent.

Question vii.For the reaction Ni(s) + Cu²+ (1 M) → Ni2+ (1 M) + Cu(s), E⁰cell = 0.57 V. Hence \( \Delta \)Gº of the reaction is
(a) 110 kJ
(b) -110 kJ
(c) 55 kJ
(d) -55 kJ
Answer: (b) -110 kJ
In simple words: The standard Gibbs free energy change (\( \Delta \)Gº) can be calculated from the standard cell potential (Eºcell) using the formula \( \Delta \)Gº = -nFEºcell, where n is the number of electrons transferred and F is Faraday's constant.

🎯 Exam Tip: Remember that for a spontaneous reaction, \( \Delta \)Gº is negative. Ensure correct values for n (number of electrons) and F (Faraday constant = 96500 C/mol) are used in the calculation.

Question viii.Which of the following is not correct ?
(a) Gibbs energy is an extensive property
(b) Electrode potential or cell potential is an intensive property.
(c) Electrical work = -\( \Delta \)G
(d) If half reaction is multiplied by a numerical factor, the corresponding Eº value is also multiplied by the same factor.
Answer: (d) If half reaction is multiplied by a numerical factor, the corresponding Eº value is also multiplied by the same factor.
In simple words: Standard electrode potential (Eº) is an intensive property, meaning it does not depend on the amount of substance, so multiplying a half-reaction by a factor does not change its Eº value.

🎯 Exam Tip: Differentiate between extensive properties (dependent on amount, like Gibbs energy) and intensive properties (independent of amount, like electrode potential).

Question ix.The oxidation reaction that takes place in lead storage battery during discharge is
(a) Pb\(_{(aq)}\) + SO\(_{4(aq)}\)\(^{2-}\) \( \longrightarrow \) PbSO\(_{4(s)}\)
(b) PbSO\(_{4(s)}\) + 2H\(_{2}\)O\(_{(l)}\) \( \longrightarrow \) PbO\(_{2(s)}\) + 4H\(^{+}_{(aq)}\) + SO\(_{4(aq)}\)\(^{2-}\) + 2e-
(c) Pb\(_{(s)}\) + SO\(_{4(aq)}\)\(^{2-}\) \( \longrightarrow \) PbSO\(_{4(s)}\) + 2e-
(d) PbSO\(_{4(s)}\) + 2e- \( \longrightarrow \) Pb\(_{(s)}\) + SO\(_{4(aq)}\)\(^{2-}\)
Answer: (c) Pb\(_{(s)}\) + SO\(_{4(aq)}\)\(^{2-}\) \( \longrightarrow \) PbSO\(_{4(s)}\) + 2e\(^{-}\)
In simple words: During discharge in a lead storage battery, lead metal (Pb) is oxidized to lead sulfate (PbSO4) at the anode, releasing two electrons.

🎯 Exam Tip: Memorize the anode and cathode reactions for common batteries like the lead storage battery, especially during both discharge and recharge cycles.

Question x.Which of the following expressions represent molar conductivity of Al2(SO4)3 ?
(a) 3\( \lambda ^{0}_{Al^{3+}}\) + 2\( \lambda ^{0}_{SO_{4}^{2-}}\)
(b) 2\( \lambda ^{0}_{Al^{3+}}\) + 3\( \lambda ^{0}_{SO_{4}^{2-}}\)
(c) \( \frac{1}{3} \lambda ^{0}_{Al^{3+}}\) + \( \frac{1}{2} \lambda ^{0}_{SO_{4}^{2-}}\)
(d) \( \lambda ^{0}_{Al^{3+}}\) + \( \lambda ^{0}_{SO_{4}^{2-}}\)
Answer: (b) 2\( \lambda ^{0}_{Al^{3+}}\) + 3\( \lambda ^{0}_{SO_{4}^{2-}}\)
In simple words: Molar conductivity at infinite dilution for Al2(SO4)3 is the sum of the molar ionic conductivities of its constituent ions, multiplied by their respective stoichiometric coefficients (2 for Al\(^{3+}\) and 3 for SO\(_{4}^{2-}\)).

🎯 Exam Tip: Kohlrausch's law states that at infinite dilution, each ion contributes independently to the total molar conductivity. Ensure to correctly apply the stoichiometric coefficients of the ions.

2. Answer the Following in One Or Two Sentences.

Question i.What is a cell constant ?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.
(B) .\. Cell constant = b = \( \frac{l}{a} \) cm\(^{-1}\).
In SI units it is expressed as m-1.
In simple words: A cell constant is a specific value for a conductivity cell, defined by the ratio of the distance between its electrodes to their cross-sectional area. It's used to relate measured resistance to conductivity.

🎯 Exam Tip: Understand the physical meaning of the cell constant (geometry of the cell) and its units (cm⁻¹ or m⁻¹).

Question ii.Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity.
Answer:If k is conductivity and \( \Lambda \)m is molar conductivity then, \( \Lambda \)m = \( \frac{\kappa \times 1000}{C} \)
Unit of molar conductivity is, Ω\(^{-1}\) cm² mol\(^{-1}\) or S cm² mol\(^{-1}\).
In simple words: Molar conductivity is calculated by dividing the specific conductivity of a solution by its molar concentration, often multiplied by 1000 for unit consistency, and its units reflect conductivity per mole.

🎯 Exam Tip: Know the formula connecting molar conductivity, conductivity, and concentration, along with the standard units for each term.

Question iii.Write the electrode reactions during electrolysis of molten KCl.
Answer:KCl \(_{(molten)}\) \( \longrightarrow \) K\(^{+}_{(l)}\) + Cl\(^{-}_{(l)}\)
Reaction at cathode :
2K\(^{+}_{(l)}\) + 2e- \( \longrightarrow \) 2K\(_{(s)}\) Reduction
Reaction at anode :
2Cl\(^{-}_{(l)}\) \( \longrightarrow \) 2Cl\(_{(g)}\) + 2e-
2Cl\(_{(g)}\) \( \longrightarrow \) Cl\(_{2(g)}\)
Overall reaction : 2K\(^{+}_{(l)}\) + 2Cl\(^{-}_{(l)}\) \( \longrightarrow \) 2K\(_{(s)}\) + Cl\(_{2(g)}\)
In simple words: During the electrolysis of molten KCl, potassium ions gain electrons at the cathode to form potassium metal, and chloride ions lose electrons at the anode to form chlorine gas.

🎯 Exam Tip: For molten salts, only the ions present in the melt participate in the electrode reactions. Reduction occurs at the cathode, and oxidation at the anode.

Question iv.Write any two functions of salt bridge.
Answer:The functions of a salt bridge are :
• It maintains the electrical contact between the two electrode solutions of the half cells.
• It prevents the mixing of electrode solutions.
• It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
• It eliminates the liquid junction potential.
In simple words: A salt bridge connects two half-cells, maintaining electrical neutrality by allowing ion flow and preventing solution mixing, which is essential for continuous current.

🎯 Exam Tip: Focus on the primary roles: completing the circuit and maintaining charge balance by preventing accumulation of charges in half-cells.

Question v.What is standard cell potential for the reaction 3Ni(s) + 2Al3+ (1M) → 3NI²+ (1M) + 2Al(s) if E⁰_Ni = - 0.25 V and E⁰_Al = -1.66V?
Answer:Solution : Given : EºNi\(^{2+}\)/Ni = -0.25 V EºAl\(^{3+}\)/Al = - 1.66 V; Eºcell = ? Since Ni is oxidised and Al\(^{3+}\) is reduced, Eºcell = EºAl\(^{3+}\)/Al - EºNi\(^{2+}\)/Ni = - 1.66 - (-0.25) = - 1.41 V Ans. Eºcell = -1.41 V [Note: Since Eºcell is negative, the given reaction is not possible but reverse reaction is possible.]
In simple words: The standard cell potential is calculated by subtracting the standard potential of the anode (oxidation) from the standard potential of the cathode (reduction). A negative value means the reaction is not spontaneous as written.

🎯 Exam Tip: When calculating Eºcell, always remember Eºcell = Eºcathode - Eºanode. A positive Eºcell indicates a spontaneous reaction.

Question vi.Write Nernst equation. What part of it represents the correction factor for nonstandard state conditions ?
Answer:(1) Nernst equation for cell potential is, \[ E_{cell} = E^{0}_{cell} - \frac{2.303 RT}{nF} \log_{10} \frac{[Products]}{[Reactants]} \] (2) The part of equation namely, \[ \frac{2.303 RT}{nF} \log_{10} \frac{[Products]}{[Reactants]} \] represents the correction factor for nonstandard state conditions.
In simple words: The Nernst equation shows how cell potential changes from its standard value under non-standard conditions, with the logarithmic term representing the correction factor based on reactant and product concentrations.

🎯 Exam Tip: Know the Nernst equation and be able to identify the terms for standard cell potential and the reaction quotient. Understand how this equation allows for calculations at non-standard conditions.

Question vii.Under what conditions the cell potential is called standard cell potential ?
Answer:In the standard cell, the active masses of the substances taking part in the electrochemical reaction have unit value, i.e., 1 M solution or ions and 1 atm gas.
In simple words: Standard cell potential is measured when all solutions are at 1 M concentration and all gases are at 1 atm pressure, typically at 25°C.

🎯 Exam Tip: Clearly define standard conditions: 1 M concentration for solutions, 1 atm pressure for gases, and usually a specified temperature (often 298 K or 25°C).

Question viii.Formulate a cell from the following electrode reactions : Au\(^{3+}_{(aq)}\) + 3e- \( \longrightarrow \) Au\(_{(s)}\) Mg\(_{(s)}\) \( \longrightarrow \) Mg\(^{2+}_{(aq)}\) + 2e-
Answer:An electrochemical cell from above electrode reactions is, Au\(^{3+}_{(aq)}\) + 3e- \( \longrightarrow \) Au\(_{(s)}\) (Cathode reduction reaction) Mg\(_{(s)}\) \( \longrightarrow \) Mg\(^{2+}_{(aq)}\) + 2e- (Anode oxidation reaction) Mg\(_{(s)}\) | Mg\(^{2+}_{(aq)}\) 1M || Au\(^{3+}_{(aq)}\) 1 M | Au
In simple words: To formulate the cell, magnesium (oxidation) is placed on the anode side, and gold (reduction) on the cathode side, with a double vertical line representing the salt bridge.

🎯 Exam Tip: When formulating a galvanic cell, remember the "Anode | Anode ion || Cathode ion | Cathode" notation. Oxidation occurs at the anode (left side), and reduction at the cathode (right side).

Question ix.How many electrons would have a total charge of 1 coulomb ?
Answer:Given: 1 Faraday = charge on 1 mol of electrons = 6.022 × 10\(^{23}\) electrons and 1 Faraday = 96500 C Therefore: 96500 C = 6.022 × 10\(^{23}\) electrons So, 1 C = \( \frac{6.022 \times 10^{23}}{96500} \) = 6.24 × 10\(^{18}\) electrons Ans. Number of electrons = 6.24 × 10\(^{18}\)
In simple words: One coulomb of charge is equivalent to the charge carried by approximately 6.24 x 10\(^{18}\) electrons, calculated using Faraday's constant and Avogadro's number.

🎯 Exam Tip: Remember Faraday's constant (96500 C/mol) represents the charge of one mole of electrons, and Avogadro's number (6.022 x 10\(^{23}\) particles/mol) relates moles to individual particles.

Question x.What is the significance of the single vertical line and double vertical line in the formulation galvanic cell.
Answer:(i) Consider representation of Daniell cell, Zn\(_{(s)}\) | Zn\(^{2+}\) 1M || Cu\(^{2+}\) 1M | Cu\(_{(s)}\) Single vertical line represents separation of two phases, solid Zn\(_{(s)}\) and solution of ions. (ii) Double vertical lines represent a salt bridge.
In simple words: In galvanic cell notation, a single vertical line signifies a phase boundary between an electrode and its electrolyte, while a double vertical line indicates a salt bridge, connecting two half-cells and preventing their mixing.

🎯 Exam Tip: Cell notation concisely represents the components and arrangement of an electrochemical cell. Understanding the meaning of single and double vertical lines is fundamental to interpreting these notations.

3. Answer the Following in Brief

Question i.Explain the effect of dilution of solution on conductivity ?
Answer:
• The conductance of a solution is due to the presence of ions in the solution. More the ions, higher is the conductance of the solution.
• Conductivity or the specific conductance is the conductance of unit volume (1 cm³) of the electrolytic solution.
• The conductivity of the electrolytic solution always decreases with the decrease in the concentration of the electrolyte or the increase in dilution of the solution.
• On dilution, the concentration of the solution decreases, hence the number of (current carrying) ions per unit volume decreases. Therefore the conductivity of the solution decreases, with the decrease concentration or increase in dilution. (It is to be noted here that, molar conductivity increases with dilution.)
In simple words: Dilution reduces the number of ions per unit volume, which decreases the specific conductivity of the solution, even though the molar conductivity might increase due to increased ionic mobility.

🎯 Exam Tip: Clearly differentiate between specific conductivity (or simply conductivity) and molar conductivity. Their responses to dilution are opposite, and this distinction is a common point of confusion.

Question ii.What is a salt bridge ?
Answer:A salt bridge is a U-shaped glass tube containing a saturated solution of a strong electrolyte, like KCl, NH4NO3, Na2SO4 in a solidified agar-agar gel. A hot saturated solution of these electrolytes in 5% agar solution is filled in the U-shaped tube and allowed it to cool and solidify forming a gel.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सॉल्ट ब्रिज को दर्शाता है, जो एक U-आकार की ट्यूब है जिसमें संतृप्त KCl समाधान और अगर-अगर जेल भरा होता है। इसके दोनों सिरे ग्लास वूल प्लग से बंद होते हैं, जो आयनों को पारित होने देते हैं लेकिन विलयनों को मिलने से रोकते हैं। यह एक गैल्वेनिक सेल के दो आधे-सेलों को जोड़ता है। Fig. 5.9: Salt bridge It is used to connect two half cells or electrodes forming a galvanic or voltaic cell.
In simple words: A salt bridge is a U-tube containing an inert electrolyte gel that connects two half-cells of a galvanic cell, maintaining electrical contact and charge neutrality without mixing the solutions.

🎯 Exam Tip: Describe the composition (U-tube, inert electrolyte, agar-agar gel) and primary functions (electrical contact, charge neutrality, prevention of mixing) of a salt bridge.

Question iii.Write electrode reactions for the electrolysis of aqueous NaCl.
Answer:Reactions in electrolytic cell : (i) Reduction half reaction at cathode : There are Na\(^{+}\) and H\(^{+}\)ions but since H\(^{+}\) are more reducible than Na\(^{+}\), they undergo reduction liberating hydrogen and Na\(^{+}\) are left in the solution. 2H\(_{2}\)O\(_{(l)}\) + 2e\(^{-}\) \( \longrightarrow \) H\(_{2(g)}\) + 2OH\(^{-}_{(aq)}\) (reduction) E° = -0.83 V (ii) Oxidation half reaction at anode : At anode there are Cl\(^{-}\) and OH\(^{-}\). But Cl\(^{-}\) ions are preferably oxidised due to less decomposition potential. 2Cl\(^{-}_{(aq)}\) \( \longrightarrow \) 2Cl\(_{(g)}\) + 2e- 2Cl\(_{(g)}\) \( \longrightarrow \) Cl\(_{2(g)}\) 2Cl\(^{-}_{(g)}\) \( \longrightarrow \) Cl\(_{2(g)}\) + 2e\(^{-}\) (overall reaction) E° = - 1.36 V Net cell reaction : Since two electrons are gained at cathode and two electrons are released at anode for each redox step, the electrical neutrality is maintained. Hence we can write, 2H\(_{2}\)O\(_{(l)}\) + 2e\(^{-}\) \( \longrightarrow \) H\(_{2(g)}\) + 2OH\(^{-}_{(aq)}\) (cathodic reduction) 2Cl\(^{-}_{(aq)}\) \( \longrightarrow \) Cl\(_{2(g)}\) + 2e\(^{-}\) (anodic oxidation) 2H\(_{2}\)O\(_{(l)}\) + 2Cl\(^{-}_{(aq)}\) \( \longrightarrow \) H\(_{2(g)}\) + 2OH\(^{-}_{(aq)}\) + Cl\(_{2(g)}\) (Overall cell reaction) Since Na\(^{+}\) and OH\(^{-}\) are left in the solution, they form NaOH\(_{(aq)}\).
In simple words: In aqueous NaCl electrolysis, water is reduced to hydrogen gas and hydroxide at the cathode, and chloride ions are oxidized to chlorine gas at the anode, leaving sodium hydroxide in solution.

🎯 Exam Tip: For aqueous solutions, compare the reduction potentials of cations/water and oxidation potentials of anions/water to determine which species react at the electrodes.

Question iv.How many moles of electrons are passed when 0.8 ampere current is passed for 1 hour through molten CaCl2 ?
Answer:Given: I = 0.8 A; t = 1 × 60 × 60 = 3600 s Number of moles of electrons = ? Q = I x t = 0.8 x 3600 = 2880 C 1 Faraday = 1 mol electrons 1 Faraday = 96500 C Therefore: 96500 C = 1 mol electrons
In simple words: To find moles of electrons, first calculate the total charge passed (Q=It), then divide this charge by Faraday's constant (96500 C/mol).

🎯 Exam Tip: Remember the relationship Q = It (charge = current × time) and the definition of Faraday's constant to convert charge into moles of electrons.

4. Answer The Following

Question i. What is Kohrausch law of independent migration of ions? How is it useful in obtaining molar conductivity at zero concentration of a weak electrolyte ? Explain with an example.
Answer:
(A) Statement of Kohlrausch's law : This states that at infinite dilution of the solution, each ion of an electrolyte migrates independently of its co-ions and contributes independently to the total molar conductivity of the electrolyte, irrespective of the nature of other ions present in the solution.
(B) Explanation : Both the ions, cation and anion of the electrolyte make a definite contribution to the molar conductivity of the electrolyte at infinite dilution or zero concentration (\( \Lambda_0 \)).
If \( \lambda_+^0 \) and \( \lambda_-^0 \) are the molar ionic conductivities of cation and anion respectively at infinite dilution, then
\( \Lambda_0 = \lambda_+^0 + \lambda_-^0 \).
This is known as Kohlrausch's law of independent migration of ions.
For an electrolyte, BxAy giving x number of cations and y number of anions,
\( \Lambda_0 = x\lambda_+^0 + y\lambda_-^0 \).
(C) Applications of Kohlrausch's law :
(1) With this law, the molar conductivity of a strong electrolyte at zero concentration can be determined. For example,
\( \Lambda_0(KCl) = \lambda_{K^+}^0 + \lambda_{Cl^-}^0 \)
(2) \( \Lambda \) values of weak electrolyte with those of strong electrolytes can be obtained. For example,
\( \Lambda_0(CH_3COOH) = \Lambda_0(HCl) + \Lambda_0(CH_3COONa) - \Lambda_0(NaCl) \)
\( = (\lambda_{H^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{CH_3COO^-}^0 + \lambda_{Na^+}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) \)
Molar conductivity of a weak electrolyte at infinite dilution or zero concentration cannot be measured experimentally.
Consider the molar conductivity (\( \Lambda_0 \)) of a weak acid, CH3COOH at zero concentration. By Kohlrausch's law,
\( \Lambda_0CH_3COOH = \lambda_{CH_3COO^-}^0 + \lambda_{H^+}^0 \)
where \( \lambda_{CH_3COO^-}^0 \) and \( \lambda_{H^+}^0 \) are the molar ionic conductivities of \( CH_3COO^- \) and \( H^+ \) ions respectively.
If \( \Lambda_0CH_3COONa \), \( \Lambda_0HCl \) and \( \Lambda_0NaCl \) are molar conductivities of \( CH_3COONa \), HCl and NaCl respectively at zero concentration, then by Kohlrausch's law,
\( \Lambda_0CH_3COONa = \lambda_{CH_3COO^-}^0 + \lambda_{Na^+}^0 \)
\( \Lambda_0HCl = \lambda_{H^+}^0 + \lambda_{Cl^-}^0 \)
\( \Lambda_0NaCl = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0 \)
Now,
\( \Lambda_0CH_3COONa + \Lambda_0HCl - \Lambda_0NaCl = [\lambda_{CH_3COO^-}^0 + \lambda_{Na^+}^0] + [\lambda_{H^+}^0 + \lambda_{Cl^-}^0] - [\lambda_{Na^+}^0 + \lambda_{Cl^-}^0] \)
\( = \lambda_{CH_3COO^-}^0 + \lambda_{H^+}^0 \)
\( = \Lambda_0CH_3COOH \)
Hence, from \( \Lambda_0 \) values of strong electrolytes, \( \Lambda_0 \) of a weak electrolyte \( CH_3COOH \), at infinite dilution can be calculated.
In simple words: Kohlrausch's law states that at infinite dilution, each ion contributes independently to the total molar conductivity of an electrolyte. This law is crucial for calculating the molar conductivity of weak electrolytes at zero concentration, which cannot be measured directly.

🎯 Exam Tip: Understanding Kohlrausch's law and its applications, especially for weak electrolytes, is a frequently tested concept. Pay attention to how the conductivities of strong electrolytes are used to derive that of a weak one.

Question ii. Explain electrolysis of molten NaCl.
Answer:
(1) Construction of an electrolytic cell : It consists of a vessel containing molten (fused) NaCl. Two graphite (carbon) inert electrodes are dipped in it, and connected to an external source of direct electric current (battery). The electrode connected to a negative terminal of the battery is a cathode and that connected to a positive terminal is an anode.
(2) Working of the cell :
(A) In the external circuit, the electrons flow through the wires from anode to cathode of the cell.
(B) The fused NaCl dissociates to form cations (Na⁺) and anions (Cl⁻).
NaCl(fused)
\( \implies \) Na⁺(l) + Cl⁻(l)
Na⁺ migrate towards cathode and Cl⁻ migrate towards anode.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक इलेक्ट्रोलाइटिक सेल को दर्शाता है जिसका उपयोग पिघले हुए सोडियम क्लोराइड के इलेक्ट्रोलिसिस के लिए किया जाता है। इसमें एक डी.सी. सप्लाई से जुड़े दो ग्रेफाइट इलेक्ट्रोड होते हैं, एक एनोड (नकारात्मक) और एक कैथोड (सकारात्मक)। सेल के अंदर पिघला हुआ NaCl होता है, जहाँ Na+ आयन कैथोड की ओर और Cl- आयन एनोड की ओर बढ़ते हैं, जिससे Na धातु और Cl2 गैस बनती है।
(C) Reactions in electrolytic cell :
(i) Reduction half reaction at cathode : The Na⁺ ions get reduced by accepting electrons from a cathode supplied by a battery and form metallic sodium.
Na⁺ + e-
\( \implies \) Na(s) (reduction)
(ii) Oxidation half reaction at anode : The Cl⁻ ions get oxidised by giving up electrons to the anode forming neutral Cl atoms in the primary process, and these Cl atoms combine forming Cl2 gas in the secondary process.
2Cl⁻
\( \implies \) Cl(g) + Cl(g) + 2e-
(Primary oxidation half reaction)
2Cl(g)
\( \implies \) Cl2(g)
(Secondary process)
2Cl⁻
\( \implies \) Cl2(g) + 2e-
(Overall oxidation at anode)
The released electrons in the anodic oxidation half reaction return to battery through the metallic wires.
Net cell reaction : In order to maintain the electrical neutrality, the number of electrons gained at cathode must be equal to the number of electrons released at anode. Hence the reduction half reaction is multiplied by 2 and both reactions, oxidation half reaction and reduction half reaction are added to obtain a net cell reaction.
2Na⁺(l) + 2e-
\( \implies \) 2Na(s) (Reduction half reaction)
2Cl⁻(l)
\( \implies \) Cl2(g) + 2e- (Oxidation half reaction)
2Na⁺(l) + 2Cl⁻(l)
\( \implies \) 2Na(s) + Cl2(g) (Overall reaction)
Results of electrolysis :
(i) A molten silvery white Na is formed at cathode which floats on the surface of molten NaCl.
(ii) A pale green Cl2 gas is liberated at anode.
In simple words: Electrolysis of molten NaCl involves passing electricity through it, causing Na+ ions to gain electrons at the cathode and form sodium metal, while Cl- ions lose electrons at the anode to form chlorine gas. This process converts electrical energy into chemical energy, leading to the decomposition of NaCl.

🎯 Exam Tip: For electrolysis questions, clearly identify the anode and cathode reactions, including oxidation and reduction processes, and the overall cell reaction. State the products formed at each electrode.

Question iii. What current strength in amperes will be required to produce 2.4g of Cu from CuSO4 solution in 1 hour ? Molar mass of Cu = 63.5 g mol⁻¹.
Answer:
Given : WCu = 2.4 g; t = 1 hr = 1 \( \times \) 60 \( \times \) 60 s = 3600 s
MCu = 63.5 g mol⁻¹; I = ?
The reaction for copper deposition is: \( Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \)
Mole ratio of Cu to electrons = \(\frac{1 \, mole \, of \, Cu}{2 \, moles \, of \, electrons}\)
We know that, \( W = \frac{I \times t}{96500} \times \) mole ratio \( \times \) molar mass of Cu
Therefore, \( I = \frac{W_{Cu} \times 96500}{t \times \text{mole ratio} \times \text{molar mass of Cu}} \)
\( I = \frac{2.4 \times 96500}{1 \times 60 \times 60 \times \frac{1}{2} \times 63.5} \)
\( I = \frac{231600}{1 \times 3600 \times 0.5 \times 63.5} \)
\( I = \frac{231600}{114300} \)
\( I = 2.026 \, A \)
Ans. Current strength = I = 2.026 A
In simple words: This problem involves using Faraday's laws of electrolysis to calculate the current needed to produce a specific amount of copper. The key is to relate the mass of copper to the moles of electrons, charge, and current over time.

🎯 Exam Tip: Problems involving Faraday's laws of electrolysis require careful application of formulas relating charge, current, time, moles of electrons, and molar mass. Ensure units are consistent (e.g., time in seconds).

Question iv. Equilibrium constant of the reaction, 2Cu⁺(aq) \( \rightarrow \) Cu²⁺(aq) + Cu(s) is 1.2 \( \times \) 10⁶. What is the standard potential of the cell in which the reaction takes place ?
Answer:
For the cell reaction, n = 1
Given: K = 1.2 \( \times \) 10⁶; \( E^0_{cell} \) = ?
We know the relationship between standard cell potential and equilibrium constant:
\( E^0_{cell} = \frac{0.0592}{n} \log_{10} K \)
\( E^0_{cell} = \frac{0.0592}{1} \log_{10} (1.2 \times 10^6) \)
\( E^0_{cell} = 0.0592 \times (6 + \log_{10} 1.2) \)
\( E^0_{cell} = 0.0592 \times (6 + 0.07918) \)
\( E^0_{cell} = 0.0592 \times 6.07918 \)
\( E^0_{cell} = 0.35985 \)
\( E^0_{cell} \approx 0.36 \, V \)
Ans. \( E^0_{cell} \) = 0.36 V
In simple words: The standard cell potential is directly related to the equilibrium constant of a reaction through the Nernst equation. A larger equilibrium constant indicates a more spontaneous reaction and a higher positive standard cell potential.

🎯 Exam Tip: Remember the Nernst equation for calculating standard cell potential from the equilibrium constant. Be precise with the number of electrons (n) transferred and the logarithmic calculation.

Question v. Calculate emf of the cell Zn(s)|Zn²⁺ (0.2M)||H⁺(1.6M)|H2(g, 1.8 atm)|Pt at 25°C.
Answer:
Given : Cell notation Zn(s)|Zn²⁺(0.2M)||H⁺(1.6M)|H2(g, 1.8 atm)|Pt
Standard reduction potentials (from data tables, often implicit for common electrodes):
\( E^0_{Zn^{2+}/Zn} = -0.763 \, V \)
\( E^0_{H^+/H_2} = 0.00 \, V \)
[Zn²⁺] = 0.2 M; [H⁺] = 1.6 M; \( P_{H_2} = 1.8 \, atm \)
\( E_{cell} \) = ?
Cell reactions :
(LHE) Oxidation half reaction at anode: \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)
(RHE) Reduction half reaction at cathode: \( 2H^+(aq) + 2e^- \rightarrow H_2(g) \)
Overall reaction: \( Zn(s) + 2H^+(aq) \rightarrow Zn^{2+}(aq) + H_2(g) \)
Concentrations: [Zn²⁺] = 0.2 M, [H⁺] = 1.6 M, \( P_{H_2} = 1.8 \, atm \)
\( \therefore \, n = 2 \) (Number of electrons transferred)
Standard cell potential \( E^0_{cell} \):
\( E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{H^+/H_2} - E^0_{Zn^{2+}/Zn} \)
\( E^0_{cell} = 0.00 \, V - (-0.763 \, V) = +0.763 \, V \)
Nernst equation for cell potential at non-standard conditions (at 25°C):
\( E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log_{10} \frac{[Zn^{2+}] \times P_{H_2}}{[H^+]^2} \)
\( E_{cell} = 0.763 - \frac{0.0592}{2} \log_{10} \frac{0.2 \times 1.8}{(1.6)^2} \)
\( E_{cell} = 0.763 - 0.0296 \log_{10} \frac{0.36}{2.56} \)
\( E_{cell} = 0.763 - 0.0296 \log_{10} (0.140625) \)
\( E_{cell} = 0.763 - 0.0296 \times (-0.8521) \)
\( E_{cell} = 0.763 + 0.02522 \)
\( E_{cell} = 0.78822 \, V \)
Ans. \( E_{cell} \) = 0.7882 V
In simple words: The Nernst equation allows us to calculate the cell potential under non-standard conditions, taking into account the concentrations of reactants and products, and the partial pressure of gases. For a Zn-H cell, zinc is oxidized at the anode and hydrogen ions are reduced at the cathode.

🎯 Exam Tip: For Nernst equation calculations, correctly identify the anode and cathode, write the half-reactions, determine 'n' (number of electrons), and set up the reaction quotient 'Q' accurately. Standard reduction potentials are usually provided.

Question vi. Calculate emf of the following cell at 25°C. Zn(s)| Zn²⁺(0.08M)||Cr³⁺(0.1M)|Cr E°Zn = - 0.76 V, E°Cr = - 0.74 V
Answer:
Given : Cell notation Zn(s)|Zn²⁺(0.08M)||Cr³⁺(0.1M)|Cr
[Zn²⁺] = 0.08 M; [Cr³⁺] = 0.1 M
Standard potentials:
\( E^0_{Zn^{2+}/Zn} = -0.76 \, V \)
\( E^0_{Cr^{3+}/Cr} = -0.74 \, V \)
\( E_{cell} \) = ?
Cell reactions:
(LHE) Oxidation half reaction at anode: \( 3Zn(s) \rightarrow 3Zn^{2+}(aq) + 6e^- \)
(RHE) Reduction half reaction at cathode: \( 2Cr^{3+}(aq) + 6e^- \rightarrow 2Cr(s) \)
Net cell reaction: \( 3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s) \)
Concentrations: [Zn²⁺] = 0.08 M, [Cr³⁺] = 0.1 M
\( n = 6 \) (Number of electrons transferred)
Standard cell potential \( E^0_{cell} \):
\( E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Cr^{3+}/Cr} - E^0_{Zn^{2+}/Zn} \)
\( E^0_{cell} = -0.74 \, V - (-0.76 \, V) = 0.02 \, V \)
Nernst equation for cell potential at non-standard conditions (at 25°C):
\( E_{cell} = E^0_{cell} - \frac{0.0592}{n} \log_{10} \frac{[Zn^{2+}]^3}{[Cr^{3+}]^2} \)
\( E_{cell} = 0.02 - \frac{0.0592}{6} \log_{10} \frac{(0.08)^3}{(0.1)^2} \)
\( E_{cell} = 0.02 - 0.009867 \log_{10} \frac{0.000512}{0.01} \)
\( E_{cell} = 0.02 - 0.009867 \log_{10} (0.0512) \)
\( E_{cell} = 0.02 - 0.009867 \times (-1.2907) \)
\( E_{cell} = 0.02 + 0.01273 \)
\( E_{cell} = 0.03273 \, V \)
Ans. \( E_{cell} \) = 0.03273 V
In simple words: Similar to the previous question, this involves calculating the cell potential for a galvanic cell under non-standard conditions using the Nernst equation, given the standard electrode potentials and ion concentrations.

🎯 Exam Tip: When dealing with different number of electrons in half-reactions, ensure you balance the overall reaction to find the correct 'n' for the Nernst equation. Pay attention to the stoichiometry when setting up the reaction quotient.

Question vii. What is a cell constant ? What are its units? How is it determined experimentally?
Answer:
(A) Cell constant of a conductivity cell is defined as the ratio of the distance between the electrodes divided by the area of cross section of the electrodes.
(B) Formula for cell constant:
\( \therefore \) Cell constant = b = \(\frac{l}{a}\)
Units:
In C.G.S. units: \(\frac{cm}{cm^2}\)
\( \implies \) \( cm^{-1} \)
In SI units it is expressed as \( m^{-1} \).
Determination of cell constant experimentally:
The resistance of an electrolytic solution is measured by using a conductivity cell and Wheatstone bridge.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख व्हीटस्टोन ब्रिज विधि का उपयोग करके एक चालकता सेल के सेल स्थिरांक को निर्धारित करने की प्रक्रिया को दर्शाता है। इसमें एक परिवर्तनीय प्रतिरोध बॉक्स, एक चालकता सेल, एक जॉकी के साथ एक तार AB, एक हेडफ़ोन (करंट डिटेक्टर), और एक प्रत्यावर्ती धारा स्रोत शामिल है। यह सेटअप विभिन्न प्रतिरोधों को संतुलित करके इलेक्ट्रोलाइटिक समाधान की चालकता को मापने की अनुमति देता है।
The measurement of molar conductivity of a solution involves two steps as follows :
Step I : Determination of cell constant of the conductivity cell :
KCl solution (0.01 M) whose conductivity is accurately known (\( \kappa = 0.00141 \, \Omega^{-1} \, cm^{-1} \)) is taken in a beaker and the conductivity cell is dipped. The two electrodes of the cell are connected to one arm while the variable known resistance (R) is placed in another arm of Wheatstone bridge.
A current detector D' which is a head phone or a magic eye is used. J is the sliding jockey (contact) that slides on the arm AB which is a wire of uniform cross section. A source of A.C. power (alternating power) is used to avoid electrolysis of the solution.
By sliding the jockey on wire AB, a balance point (null point) is obtained at C. Let AC and BC be the lengths of wire.
If \( R_{solution} \) is the resistance of KCl solution and \( R_x \) is the known resistance then by Wheatstone's bridge principle,
\( \frac{R_{solution}}{BC} = \frac{R_x}{AC} \)
\( \therefore R_{solution} = BC \times \frac{R_x}{AC} \)
Then the cell constant 'b' of the conductivity cell is obtained by, \( b = \kappa_{KCl} \times R_{solution} \).
Step II : Determination of conductivity of the given solution :
KCl solution is replaced by the given electrolytic solution and its resistance (\( R_s \)) is measured by Wheatstone bridge method by similar manner by obtaining a null point at D.
The conductivity (\( \kappa \)) of the given solution is,
\( \kappa = \frac{\text{cell constant}}{R_s} = \frac{b}{R_s} \)
Step III: Calculation of molar conductivity :
The molar conductivity (\( \Lambda_m \)) is given by,
\( \Lambda_m = \frac{\kappa}{C} \) (or \( \Lambda_m = \frac{1000 \times \kappa}{C} \))
Since the concentration of the solution is known, \( \Lambda_m \) can be calculated.
In simple words: A cell constant is a unique geometric property of a conductivity cell that relates its resistance to the solution's conductivity. It's determined by measuring the resistance of a standard solution with known conductivity using a Wheatstone bridge.

🎯 Exam Tip: Clearly define the cell constant, state its units, and outline the experimental procedure for its determination using a standard solution and a Wheatstone bridge. Mention the purpose of using AC current.

Question viii. How will you calculate the moles of electrons passed and mass of the substance produced during electrolysis of a salt solution using reaction stoichiometry.
Answer:
Calculation of moles of electrons passed : The charge carried by one mole of electrons is referred to as one faraday (F). If total charge passed is Q C, then moles of electrons passed = \(\frac{Q(C)}{F(C/mole^-)}\)
Calculation of mass of product : Mass, W of product formed is given by,
W = moles of product \( \times \) molar mass of product (M)
\( W = \frac{Q}{96500} \times \) mole ratio \( \times \) M
\( W = \frac{I \times t}{96500} \times \) mole ratio \( \times \) M
When two electrolytic cells containing different electrolytes are connected in series so that same quantity of electricity is passed through them, then the masses \( W_1 \) and \( W_2 \) of products produced are given by,
\( W_1 = \frac{Q}{96500} \times \) (mole ratio)\(_1 \times M_1\)
\( W_2 = \frac{Q}{96500} \times \) (mole ratio)\(_2 \times M_2\)
\( \therefore \frac{W_1}{W_2} = \frac{(\text{mole ratio})_1 \times M_1}{(\text{mole ratio})_2 \times M_2} \)
This shows that the ratio of mass to (mole ratio x molar mass) is constant for products in series electrolytic cells.
In simple words: This explains how to calculate the amount of substance produced during electrolysis using Faraday's laws. The amount of electricity passed (charge) directly determines the moles of electrons, which in turn dictates the mass of the product based on its molar mass and reaction stoichiometry.

🎯 Exam Tip: Master Faraday's laws of electrolysis. Understand the relationship between current, time, charge, moles of electrons, and the stoichiometric ratio of product to electrons. This is crucial for quantitative electrolysis problems.

Question ix. Write the electrode reactions when lead storage cell generates electricity. What are the anode and cathode and the electrode reactions during its recharging?
Answer:
Working of a lead accumulator :
(1) Discharging (when it generates electricity) :
When the electric current is withdrawn from lead accumulator, the following reactions take place :
Anode (negative electrode) reaction (Oxidation):
\( Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^- \)
Cathode (positive electrode) reaction (Reduction):
\( PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l) \)
Net cell reaction during discharging:
\( Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l) \)
Explanation for cell potential decrease: The cell potential or emf of the cell depends upon the concentration of sulphuric acid. During discharge, sulphuric acid is consumed, and its concentration decreases. This leads to a decrease in the specific gravity of the electrolyte (from 1.28 to 1.17) and thus, the emf of the cell decreases. The emf of a fully charged cell is about 2.0 V.
(2) Recharging (how cell potential can be increased):
When the discharged battery is connected to an external electric source and a higher external potential is applied, the cell reaction gets reversed, regenerating H2SO4.
Anode (positive electrode during recharging) reaction (Oxidation):
\( PbSO_4(s) + 2H_2O(l) \rightarrow PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \)
Cathode (negative electrode during recharging) reaction (Reduction):
\( PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}(aq) \)
Net cell reaction during charging:
\( 2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \)
The emf of the accumulator depends only on the concentration of H2SO4; by regenerating H2SO4, the cell potential is increased.
In simple words: In a lead storage cell, discharging involves lead and lead dioxide reacting with sulfuric acid to form lead sulfate, producing electricity and consuming acid. Recharging reverses this process, converting lead sulfate back to lead and lead dioxide, regenerating sulfuric acid, and storing energy.

🎯 Exam Tip: For lead storage cell questions, clearly differentiate between discharge and recharge reactions at both anode and cathode. Explain how sulfuric acid concentration changes and its impact on cell potential.

Question x. What are anode and cathode of H2-O2 fuel cell ? Name the electrolyte used in it. Write electrode reactions and net cell reaction taking place in the fuel cell.
Answer:
Construction :
(i) In fuel cell the anode and cathode are porous electrodes with suitable catalyst like finely divided platinum.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक हाइड्रोजन-ऑक्सीजन (H₂-O₂) ईंधन सेल की संरचना को दर्शाता है। इसमें दो छिद्रपूर्ण कार्बन इलेक्ट्रोड होते हैं- एक एनोड (-) और एक कैथोड (+), जो प्लेटिनम उत्प्रेरक के साथ लेपित होते हैं। इलेक्ट्रोलाइट के रूप में गर्म जलीय NaOH या KOH का उपयोग किया जाता है। हाइड्रोजन गैस एनोड में और ऑक्सीजन गैस कैथोड में बुदबुदाई जाती है, जिससे जल बनता है और विद्युत ऊर्जा उत्पन्न होती है।
(ii) H2 is continuously bubbled through anode while O2 gas is bubbled through cathode.
Working (cell reactions) :
(i) Oxidation at anode : At anode, hydrogen gas is oxidised to H2O.
\( 2H_2(g) + 4OH^-(aq) \rightarrow 4H_2O(l) + 4e^- \) (oxidation half reaction)
(ii) Reduction at cathode : The electrons released at anode travel to cathode through external circuit and reduce oxygen gas to OH⁻.
\( O_2(g) + 2H_2O(l) + 4e^- \rightarrow 4OH^-(aq) \) (reduction half reaction)
(iii) Net cell reaction: Addition of both the above reactions at anode and cathode gives a net cell reaction.
\( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \) (overall cell reaction)
In simple words: A hydrogen-oxygen fuel cell uses hydrogen and oxygen gases at porous electrodes with an electrolyte to directly convert chemical energy into electrical energy, producing water as a byproduct.

🎯 Exam Tip: Describe the anode, cathode, and electrolyte for fuel cells. Write the half-reactions for oxidation and reduction, and the overall cell reaction. Highlight the advantages of fuel cells.

Question xi. What are anode and cathode for Leclanche' dry cell ? Write electrode reactions and overall cell reaction when it generates electricity.
Answer:
A dry cell has zinc vessel as anode and graphite rod as cathode and moist paste of ZnCl2, MnO2, NH4Cl as electrolytes.
At anode :
\( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \) (Oxidation half reaction)
At graphite (c) cathode :
\( 2NH_4^+(aq) + 2e^- \rightarrow 2NH_3(aq) + H_2(g) \) (Reduction half reaction)
\( 2MnO_2(s) + H_2(g) \rightarrow Mn_2O_3(s) + H_2O(l) \)
There is a side reaction inside the cell, between Zn²⁺ ions and aqueous NH3.
\( Zn^{2+}(aq) + 4NH_3(aq) \rightarrow [Zn(NH_3)_4]^{2+}(aq) \)
The overall cell reaction is a complex combination of these processes, but primarily involves the oxidation of Zinc and reduction of Manganese dioxide in the presence of ammonium chloride.
In simple words: A Leclanche' dry cell uses a zinc container as the anode and a graphite rod as the cathode, with a moist paste of electrolytes. Zinc is oxidized, and manganese dioxide is reduced, generating electricity for common devices.

🎯 Exam Tip: Identify the components (anode, cathode, electrolyte) of a dry cell and list the key electrode reactions. Mention any significant side reactions that occur.

Question xii. Identify oxidising agents and arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis. Al(- 1.66 V), Cl2 (1.36 V), Cd2+ (-0.4 V), Fe (-0.44 V), I2 (0.54 V), Br- (1.09 V).
Answer:
The oxidising agents are I2, Br⁻ and Cl2. The increasing strength is
\( I_2 \)
\( \implies \) \( Br^- \)
\( \implies \) \( Cl_2 \)
\( +0.54 \, V \)
\( \implies \) \( +1.09 \, V \)
\( \implies \) \( +1.36 \, V \)
(Note: Actually Br2 acts as an oxidising agent but not Br⁻.)
In simple words: The strength of an oxidising agent is determined by its standard reduction potential; a higher (more positive) reduction potential means a stronger oxidising agent, as it has a greater tendency to accept electrons and get reduced.

🎯 Exam Tip: To compare oxidising agent strengths, always refer to their standard reduction potentials (E°). Stronger oxidising agents have higher E° values. Ensure you correctly identify which species are oxidising agents (undergo reduction).

Question xiii. Which of the following species are reducing agents? Arrange them in order of increasing strength under standard state conditions. The standard potentials are given in parenthesis. K (-2.93V), Br2(1.09V), Mg(-2.36V), Co³+(1.61V), Ti²+(-0.37V), Ag+(0.8V), Ni (-0.23V).
Answer:
Lower the standard reduction potential, higher is reducing power.
The reducing agents are Ni, Mg, K and Ti²⁺.
Their increasing strength is,
Ni (-0.23 V)
\( \implies \) Ti²⁺ (-0.37 V)
\( \implies \) Mg (-2.36 V)
\( \implies \) K (-2.93 V)
(Note: Cations don't act as reducing agent since they are already in oxidised state.)
In simple words: The strength of a reducing agent is inversely related to its standard reduction potential; a lower (more negative) reduction potential indicates a stronger reducing agent, as it has a greater tendency to lose electrons and get oxidized.

🎯 Exam Tip: To compare reducing agent strengths, always refer to their standard reduction potentials (E°). Stronger reducing agents have lower (more negative) E° values. Ensure you correctly identify which species are reducing agents (undergo oxidation).

Question xiv. Predict whether the following reactions would occur spontaneously under standard state conditions. (use information of Table 5.1)
a. Ca(s) + Cd²⁺(aq) \( \rightarrow \) Ca²⁺(aq) + Cd(s)
b. 2 Br⁻(s) + Sn²⁺(aq) \( \rightarrow \) Br2(l) + Sn(s)
c. 2Ag(s) + Ni²⁺(aq) \( \rightarrow \) 2 Ag⁺(aq) + Ni(s)
Answer:
A reaction is spontaneous if the standard cell potential (E°cell) is positive.
a. Ca(s) + Cd²⁺(aq) \( \rightarrow \) Ca²⁺(aq) + Cd(s)
Standard reduction potentials (from Table 5.1):
\( E^0_{Ca^{2+}/Ca} = -2.866 \, V \)
\( E^0_{Cd^{2+}/Cd} = -0.403 \, V \)
Anode (Oxidation): Ca(s) \( \rightarrow \) Ca²⁺(aq) + 2e⁻
Cathode (Reduction): Cd²⁺(aq) + 2e⁻ \( \rightarrow \) Cd(s)
\( E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Cd^{2+}/Cd} - E^0_{Ca^{2+}/Ca} \)
\( E^0_{cell} = -0.403 \, V - (-2.866 \, V) = +2.463 \, V \)
Since \( E^0_{cell} \) is positive, the reaction is spontaneous.
b. 2 Br⁻(s) + Sn²⁺(aq) \( \rightarrow \) Br2(l) + Sn(s)
Standard reduction potentials (from Table 5.1):
\( E^0_{Br_2/Br^-} = +1.09 \, V \)
\( E^0_{Sn^{2+}/Sn} = -0.136 \, V \)
Anode (Oxidation): 2Br⁻(s) \( \rightarrow \) Br2(l) + 2e⁻
Cathode (Reduction): Sn²⁺(aq) + 2e⁻ \( \rightarrow \) Sn(s)
\( E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Sn^{2+}/Sn} - E^0_{Br_2/Br^-} \)
\( E^0_{cell} = -0.136 \, V - (+1.09 \, V) = -1.226 \, V \)
Since \( E^0_{cell} \) is negative, the reaction is non-spontaneous. The reverse reaction is spontaneous.
c. 2Ag(s) + Ni²⁺(aq) \( \rightarrow \) 2 Ag⁺(aq) + Ni(s)
Standard reduction potentials (from Table 5.1):
\( E^0_{Ag^+/Ag} = +0.799 \, V \)
\( E^0_{Ni^{2+}/Ni} = -0.257 \, V \)
Anode (Oxidation): 2Ag(s) \( \rightarrow \) 2Ag⁺(aq) + 2e⁻
Cathode (Reduction): Ni²⁺(aq) + 2e⁻ \( \rightarrow \) Ni(s)
\( E^0_{cell} = E^0_{cathode} - E^0_{anode} = E^0_{Ni^{2+}/Ni} - E^0_{Ag^+/Ag} \)
\( E^0_{cell} = -0.257 \, V - (+0.799 \, V) = -1.056 \, V \)
Since \( E^0_{cell} \) is negative, the reaction is non-spontaneous. The reverse reaction is spontaneous.
In simple words: The spontaneity of a redox reaction under standard conditions can be predicted by calculating the standard cell potential (E°cell). If E°cell is positive, the reaction is spontaneous; if negative, it is non-spontaneous, and the reverse reaction is spontaneous.

🎯 Exam Tip: For predicting spontaneity, calculate \( E^0_{cell} = E^0_{cathode} - E^0_{anode} \). A positive \( E^0_{cell} \) indicates spontaneity. Carefully assign anode (oxidation) and cathode (reduction) based on the given reaction and standard potentials.

12th Chemistry Digest Chapter 5 Electrochemistry Intext Questions And Answers

Question 1. How does electrical resistance depend on the dimensions of an electronic (metallic) conductor?
Answer:
The electrical resistance of an electronic conductor is linearly proportional to its length (l) and inversely proportional to its cross section area a.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक इलेक्ट्रॉनिक (धात्विक) कंडक्टर को दर्शाता है, जिसमें उसकी लंबाई (l) और अनुप्रस्थ काट का क्षेत्रफल (a) स्पष्ट रूप से चिन्हित है। यह दिखाता है कि कंडक्टर का प्रतिरोध उसकी लंबाई के सीधे आनुपातिक होता है और उसके अनुप्रस्थ काट के क्षेत्रफल के व्युत्क्रमानुपातिक होता है।
Thus, \( R \propto l \); \( R \propto \frac{1}{a} \)
\( \therefore R \propto \frac{l}{a} \) or \( R = \rho \times \frac{l}{a} \)
where the proportionality constant \( \rho \) is called specific resistance. IUPAC recommends the term resistivity for specific resistance.
In simple words: The electrical resistance of a metallic conductor depends on its physical dimensions: it increases with length and decreases with cross-sectional area. This relationship introduces resistivity, an intrinsic property of the material.

🎯 Exam Tip: Remember the formula \( R = \rho \frac{l}{a} \). Understand the proportionality of resistance to length and inverse proportionality to area. Define resistivity and its significance.

Question 2. What are the units of resistivity ?
Answer:
For an electronic conductor of length l, and cross section area a, the resistance R is represented as
\( R = \rho \times \frac{l}{a} \)
where \( \rho \) is the resistivity of the conductor.
\( \therefore \rho = R \times \frac{a}{l} \)
If \( l = 1 \, m \), \( a = 1 \, m^2 \), then \( \rho = R \)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक इकाई आयतन (SI में 1 m³ और C.G.S. में 1 cm³) के धात्विक कंडक्टर को दर्शाता है। यह कंडक्टर की लंबाई (l), अनुप्रस्थ काट का क्षेत्रफल (a), और इकाई आयतन को स्पष्ट करता है, जो प्रतिरोधकता (ρ) की इकाई (ओम-मीटर या ओम-सेमी) को समझने में मदद करता है, जो इकाई आयतन के कंडक्टर का प्रतिरोध है।
Hence, resistivity is the resistance of a conductor of volume of 1 m³.
In simple words: Resistivity, which is an intrinsic property of a material, has units derived from the resistance, length, and area relationship. In SI units, it is ohm-meter (\( \Omega \, m \)), and in C.G.S. units, it is ohm-centimeter (\( \Omega \, cm \)).

🎯 Exam Tip: Be able to derive the units of resistivity from its defining formula (\( \rho = R \frac{a}{l} \)) in both SI and C.G.S. systems.

Question 3. Define resistivity. What are its units ?
Answer:
Resistivity (or specific resistance) : It is the resistance of a conductor that is 1 m in length and 1 m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) Hence, the resistivity is the resistance of a conductor of unit volume. (In case of electrolytic solution, \( \rho \) is the resistivity i.e., resistance of a solution of unit volume.)
It has SI units, ohm m and C.G.S. units, ohm cm.
In simple words: Resistivity is the fundamental electrical resistance offered by a material of unit length and unit cross-sectional area. It quantifies how strongly a material opposes the flow of electric current.

🎯 Exam Tip: Provide a clear definition of resistivity and specify its standard units in both SI and C.G.S. systems.

Question 4. Why is alternating current used in the measurement of conductivity of the solution ?
Answer:
If direct current (D.C.) by battery is used, there will be electrolysis and the concentration of the solution is changed. Hence alternating current (A.C.) with high frequency is used.
In simple words: Alternating current (AC) is used to measure solution conductivity to prevent electrolysis, which would alter the concentration of the electrolyte and give inaccurate conductivity readings.

🎯 Exam Tip: Explain why DC current is unsuitable (electrolysis, concentration change) and why AC current is preferred (prevents electrolysis, maintains concentration) for conductivity measurements.

Try this... (Textbook page No. 93)

Question 1. What must be the concentration of a solution of silver nitrate to have the molar conductivity of 121.4 Ω⁻¹ cm² mol⁻¹ and the conductivity of 2.428 \( \times \) 10⁻³ Ω⁻¹ cm⁻¹ at 25 °C ?
Answer:
Given : Molar conductivity \( \Lambda_m = 121.4 \, \Omega^{-1} \, cm^2 \, mol^{-1} \)
Conductivity \( \kappa = 2.428 \times 10^{-3} \, \Omega^{-1} \, cm^{-1} \)
Temperature = 25 °C
Concentration C = ?
We know the relationship:
\( \Lambda_m = \frac{\kappa \times 1000}{C} \) (where C is in mol dm⁻³)
Rearranging for C:
\( C = \frac{\kappa \times 1000}{\Lambda_m} \)
\( C = \frac{2.428 \times 10^{-3} \, \Omega^{-1} \, cm^{-1} \times 1000 \, cm^3 \, dm^{-3}}{121.4 \, \Omega^{-1} \, cm^2 \, mol^{-1}} \)
\( C = \frac{2.428}{121.4} \, mol \, dm^{-3} \)
\( C = 0.02 \, mol \, dm^{-3} \)
Ans. Concentration of a Solution = 0.02 M
In simple words: This problem uses the relationship between molar conductivity, specific conductivity, and concentration. By knowing the molar conductivity and specific conductivity, one can calculate the concentration of the solution.

🎯 Exam Tip: Know the formula linking molar conductivity (\( \Lambda_m \)), specific conductivity (\( \kappa \)), and concentration (C): \( \Lambda_m = \frac{\kappa \times 1000}{C} \). Ensure correct unit conversion, especially for volume.

Try this... (Textbook page No. 96)

Question 1. Obtain the expression for dissociation constant in terms of \( \Lambda_c \) and \( \Lambda_0 \) using Ostwald's dilution law.
Answer:
Consider a solution of a weak electrolyte, BA having concentration C mol dm⁻³. If \( \alpha \) is the degree of dissociation, then by Ostwald's theory of weak electrolytes,
Equilibrium:
\( BA(aq) \rightleftharpoons B^+(aq) + A^-(aq) \)
Initial concentration: C 0 0
At equilibrium: \( C(1-\alpha) \) \( C\alpha \) \( C\alpha \)
If K is the dissociation constant of the weak electrolyte, then by Ostwald's dilution law,
\( K = \frac{[B^+][A^-]}{[BA]} = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha} \)
If \( \Lambda_m \) is the molar conductivity of the electrolyte BA at the concentration C and \( \Lambda_0 \) is the molar conductivity at zero concentration or infinite dilution, then
\( \alpha = \frac{\Lambda_m}{\Lambda_0} \)
Substitute \( \alpha \) into the expression for K:
\( K = \frac{C \left(\frac{\Lambda_m}{\Lambda_0}\right)^2}{1 - \frac{\Lambda_m}{\Lambda_0}} \)
\( K = \frac{C \frac{\Lambda_m^2}{\Lambda_0^2}}{\frac{\Lambda_0 - \Lambda_m}{\Lambda_0}} \)
\( K = \frac{C \Lambda_m^2}{\Lambda_0^2} \times \frac{\Lambda_0}{\Lambda_0 - \Lambda_m} \)
\( K = \frac{C \Lambda_m^2}{\Lambda_0 (\Lambda_0 - \Lambda_m)} \)
Hence by measuring \( \Lambda_m \) at the concentration C and knowing \( \Lambda_0 \), the dissociation constant can be calculated.
If \( \lambda_+^0 \) and \( \lambda_-^0 \) are the ionic conductivities, then by Kohlrausch's law, \( \Lambda_0 = \lambda_+^0 + \lambda_-^0 \).
In simple words: Ostwald's dilution law relates the dissociation constant of a weak electrolyte to its degree of dissociation and concentration. The degree of dissociation can also be expressed using molar conductivity at a given concentration and infinite dilution.

🎯 Exam Tip: Be able to derive the expression for the dissociation constant (K) in terms of \( \Lambda_m \), \( \Lambda_0 \), and concentration (C) by combining Ostwald's dilution law and the relationship for the degree of dissociation.

Learn This As Well...

Question 1. How is the cell constant of a conductivity cell determined?
Answer:The cell constant of a given conductivity cell is obtained by measuring the resistance (R) (or the conductance) of a standard solution whose conductivity (\( \kappa \)) is accurately known by using Wheatstone's bridge (discussed in Q. 37). For this purpose, KCl solution of accurately known conductivity is used. \( \kappa_{KCl} = \frac{1}{R_{KCl}} \times \frac{l}{a} \) where \( \frac{l}{a} \) is a cell constant, represented by b.
\( \therefore \kappa_{KCl} = \frac{b}{R_{KCl}} \)
\( \text{or } b = \kappa_{KCl} \times R_{KCl} \) For example, the conductivity of 0.01 M KCl is \( 0.00141 \Omega^{-1} \text{ cm}^{-1} \text{ (S cm}^{-1}) \). Hence by measuring R KCl the cell constant b can be obtained.
In simple words: The cell constant, a measure of the cell's geometry, is determined by measuring the resistance of a standard solution (like KCl) with known conductivity using a Wheatstone's bridge. This allows calculating the cell constant from the known conductivity and measured resistance.

🎯 Exam Tip: Understanding the relationship between cell constant, conductivity, and resistance is crucial for numerical problems. Remember the formula \( \kappa = b/R \).

Try this... (Textbook page No. 95)

Question 1. Calculate \( \Lambda_0 (\text{CH}_2\text{ClCOOH}) \) if \( \Lambda \) values for HCl, KCl and \( \text{CH}_2\text{ClCOOK} \) are respectively, \( 4.261, 1.499 \) and \( 1.132 \Omega^{-1} \text{ m}^2 \text{ mol}^{-1} \).
Solution:Given : \( \Lambda_0(\text{HCl}) = 4.261 \Omega^{-1} \text{ m}^2 \text{ mol}^{-1} \) \( \Lambda_0(\text{KCl}) = 1.499 \Omega^{-1} \text{ m}^2 \text{ mol}^{-1} \) \( \Lambda_0(\text{CH}_2\text{ClCOOK}) = 1.132 \Omega^{-1} \text{ m}^2 \text{ mol}^{-1} \) \( \Lambda_0(\text{CH}_2\text{ClCOOH}) = ? \) \( \Lambda_0(\text{CH}_2\text{ClCOOH}) = \lambda^0_{\text{CH}_2\text{ClCOO}^-} + \lambda^0_{\text{H}^+} \) ... (I)
\( \therefore \Lambda_0(\text{CH}_2\text{ClCOOK}) = \lambda^0_{\text{CH}_2\text{ClCOO}^-} + \lambda^0_{\text{K}^+} \) ... (i)
\( \Lambda_0(\text{HCl}) = \lambda^0_{\text{H}^+} + \lambda^0_{\text{Cl}^-} \) ... (ii)
\( \Lambda_0(\text{KCl}) = \lambda^0_{\text{K}^+} + \lambda^0_{\text{Cl}^-} \) ... (iii) Adding equations (i) and (ii) and subtracting equation (iii) we get equation (I).
\( \therefore \Lambda_0(\text{CH}_2\text{ClCOOH}) = \Lambda_0(\text{CH}_2\text{ClCOOK}) + \Lambda_0(\text{HCl}) - \Lambda_0(\text{KCl}) \) \( = 1.132 + 4.261 - 1.499 \) \( = 3.894 \Omega^{-1} \text{ m}^2 \text{ mol}^{-1} \)
In simple words: Using Kohlrausch's law, the molar conductivity of a weak electrolyte like chloroacetic acid at infinite dilution can be calculated by combining the molar conductivities of strong electrolytes that share its component ions. By adding the conductivities of potassium chloroacetate and hydrochloric acid and subtracting that of potassium chloride, we derive the required value.

🎯 Exam Tip: Kohlrausch's law is fundamental for calculating molar conductivities of weak electrolytes. Pay close attention to how the ionic conductivities combine (addition and subtraction) to yield the desired result.

Can You Tell? (Textbook Page No. 103)

Question 1. You have learnt Daniel cell in XIth standard. Write notations for anode and cathode. Write the cell formula.
Answer:Daniell cell is represented as, \( \text{Zn(s)} | \text{Zn}^{2+}\text{(aq)} || \text{Cu}^{2+}\text{(aq)} | \text{Cu(s)} \)
(Anode) 1 M (Cathode)
In simple words: The Daniell cell notation shows the anode (zinc solid and its ions), followed by the salt bridge, and then the cathode (copper ions and copper solid), with concentrations usually indicated.

🎯 Exam Tip: Correctly writing cell notation is essential. Remember the order: Anode | Anode ion || Cathode ion | Cathode, and use single and double vertical lines correctly.

Try This... (Textbook Page No. 104)

Question 1. Write electrode reactions and overall cell reaction for Daniel cell you learnt in standard XI.
Answer:Reactions for Daniell cell: Oxidation at Zn anode : \( \text{Zn(s)} \longrightarrow \text{Zn}^{2+}\text{(aq)} + 2\text{e}^- \)
(Oxidation half reaction) Reduction at Cu cathode : \( \text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \longrightarrow \text{Cu(s)} \) (Reduction half reaction)
\( \text{Zn(s)} + \text{Cu}^{2+}\text{(aq)} \longrightarrow \text{Zn}^{2+}\text{(aq)} + \text{Cu(s)} \)
(1 M) (1 M)
In simple words: In a Daniell cell, zinc metal is oxidized at the anode to zinc ions, releasing electrons. Copper ions at the cathode gain these electrons and are reduced to copper metal, resulting in an overall spontaneous redox reaction.

🎯 Exam Tip: Distinguish between oxidation and reduction half-reactions and correctly sum them to get the net cell reaction, ensuring electron balance.

Question 1. Describe different types of reversible electrodes with examples. (1 mark for each type)
Answer:A reversible electrochemical cell or a galvanic cell consists of two reversible half cells or electrodes. There are four types of reversible electrodes according to their compositions. (1) Metal-metal ion electrode : This electrode is set up by dipping a metal in a solution containing its own ions, e.g. Zn rod dipped into \( \text{ZnSO}_4 \) solution containing \( \text{Zn}^{++} \) ions of concentration C. It is represented as, \( \text{Zn}^{2+}\text{(aq)} | \text{Zn(s)} \) The reduction reaction at the electrode is, \( \text{Zn}^{++}\text{(aq)} + 2\text{e}^- \longrightarrow \text{Zn(s)} \)
(2) Metal-sparingly soluble salt electrode : This electrode consists of a metal coated with one of its sparingly soluble salts and immersed in a solution containing an electrolyte having a common anion as that of the salt. For example, silver electrode coated with sparingly soluble AgCl dipped in KCl solution with common anion \( \text{Cl}^- \). This electrode is represented as, \( \text{Cl}^-\text{(aq)} | \text{AgCl(s)} | \text{Ag(s)} \) The reduction reaction is, \( \text{AgCl(s)} + \text{e} \longrightarrow \text{Ag(s)} + \text{Cl}^-\text{(aq)} \)
(3) Gas electrode : This is developed by bubbling pure and dry gas around a platinised platinum foil dipped in the solution containing ions (of the gas) reversible with respect to the gas bubbled. The gas is adsorbed on the surface of platinum foil and establishes an equilibrium with its ions in the solution. Pt electrode provides electrical contact and also acts as a catalyst. Some of the gas electrodes are represented as follows : (i) Hydrogen gas electrode : \( \text{H}^+\text{(aq)} | \text{H}_2\text{(g, P}_{\text{H}_2}) | \text{Pt} \) Reduction reaction: \( \text{H}^+\text{(aq)} + \text{e}^- \longrightarrow \frac{1}{2}\text{H}_2\text{(g)} \) (ii) Chlorine gas electrode : \( \text{Cl}^-\text{(aq)} | \text{Cl}_2\text{(g, P}_{\text{Cl}_2}) | \text{Pt} \) Reduction reaction : \( \frac{1}{2}\text{Cl}_2\text{(g)} + \text{e}^- \longrightarrow \text{Cl}^-\text{(aq)} \)
(4) Redox electrode (Oxidation reduction electrode) : This electrode consists of a platinum wire dipped in a solution containing the ions of the same metal (or a substance) in two different oxidation states, like \( \text{Fe}^{2+} - \text{Fe}^{3+}, \text{Sn}^{2+} - \text{Sn}^{4+}, \text{Mn}^{++} - \text{MnO}_4 \), etc. A platinum electrode which provides an electrical contact and acts as catalyst aquires an equilibrium between two ions in the solution, due to their tendency to undergo a change from one oxidation state to another. The electrodes are represented as, \( \text{Fe}^{2+}\text{(aq)}, \text{Fe}^{3+}\text{(aq)} | \text{Pt} \) Reduction reaction : \( \text{Fe}^{3+}\text{(aq)} + \text{e}^- \longrightarrow \text{Fe}^{2+}\text{(aq)} \)
\( \text{SnCl}_2\text{(aq)}, \text{SnCl}_4\text{(aq)} | \text{Pt} \) Reduction reaction : \( \text{Sn}^{4+}\text{(aq)} + 2\text{e}^- \longrightarrow \text{Sn}^{2+}\text{(aq)} \)
In simple words: Reversible electrodes facilitate redox reactions and can be categorized into metal-metal ion (like zinc in zinc sulfate), metal-sparingly soluble salt (like silver-silver chloride), gas electrodes (like hydrogen or chlorine gas with platinum), and redox electrodes (like iron(II)/iron(III) with platinum), each allowing for equilibrium between different oxidation states.

🎯 Exam Tip: For each type of reversible electrode, remember a specific example and its corresponding reduction half-reaction. This will help illustrate the concept clearly during exams.

Use Your Brain Power! (Textbook Page No. 98)

Question 1. Distinguish between electrolytic and galvanic cells.
Answer:Electrolytic cell:
1. This device is used to bring about a non-spontaneous chemical reaction by passing an electric current.
2. It is used to bring about a chemical reaction generally for the dissociation (electrolysis) of compounds.
3. In this cell, electrical energy is converted into chemical energy.
4. In this cell, the cathode is negative and the anode is positive.
5. Electrolytic cells are irreversible.
6. Oxidation takes place at the positive electrode and reduction at the negative electrode.
7. The electrons are supplied by the external source and enter through cathode and come out through anode.
8. It is used for electroplating, electrorefining, etc. Electrochemical cell (Galvanic cell or Voltaic cell):
1. This device is used to produce electrical energy by a spontaneous chemical reaction.
2. It is used to generate electricity.
3. In this cell, chemical energy is converted into electrical energy.
4. In this cell, the cathode is positive and the anode is negative.
5. Electrochemical cells are reversible.
6. Oxidation takes place at the negative electrode and reduction at the positive electrode.
7. The electrons move from anode to cathode in the external circuit.
8. It is used as a source of electric current.
In simple words: Electrolytic cells use electricity to drive non-spontaneous chemical reactions (e.g., electroplating), converting electrical energy to chemical energy, with a negative cathode and positive anode. Galvanic cells, conversely, produce electricity from spontaneous chemical reactions (e.g., batteries), converting chemical energy to electrical energy, with a positive cathode and negative anode.

🎯 Exam Tip: Focus on the energy conversion (electrical to chemical vs. chemical to electrical) and the spontaneity of the reaction as key distinguishing factors. Also, remember the polarity of anode and cathode for each type.

Try This... (Textbook Page No. 107)

Question 1. Write expressions to calculate equilibrium constant from
i. Concentration data
ii. Thermochemical data
iii. Electrochemical data
Answer:(i) Consider following a reversible cell reaction. \( \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \) If [A], [B], [C] and [D] represent concentrations of reactants and products then the equilibrium constant K is, \( K = \frac{\text{[C]}^c \times \text{[D]}^d}{\text{[A]}^a \times \text{[B]}^b} \) (ii) If \( \Delta \text{G}^0 \) is the standard Gibbs free energy change at temperature T then, \( \Delta \text{G}^0 = -\text{RTlnK} = -2.303 \text{ RTlog}_{10}\text{K} \) (iii) From electrochemical data, if \( \text{E}^0_{\text{cell}} \) is the standard cell potential and K is the equilibrium constant for the cell reaction at a temperature T, then, \( \text{E}^0_{\text{cell}} = \frac{0.0592}{\text{n}} \text{log}_{10} \text{K} \)
In simple words: The equilibrium constant (K) can be calculated from reactant and product concentrations at equilibrium, from the standard Gibbs free energy change (\( \Delta \text{G}^0 \)) using thermodynamic relations, or from the standard cell potential (\( \text{E}^0_{\text{cell}} \)) using the Nernst equation at equilibrium.

🎯 Exam Tip: Be sure to distinguish between the three methods (concentration, thermochemical, and electrochemical) and remember the specific formula for each, especially the Nernst equation for electrochemical data.

Learn This As Well...

Question 1. The construction and working of the calomel electrode.
Answer:(1) Since standard hydrogen electrode (SHE) is not convenient for experimental use, a secondary reference electrode like calomel electrode is used. (2) Construction : It consists of a glass vessel with side arm B for dipping in a desired solution of another electrode like, \( \text{ZnSO}_4\text{(aq)} \) for an electric contact. The vessel is filled with mercury, a paste of Hg and \( \text{Hg}_2\text{Cl}_2 \) (calomel) and saturated KCl solution.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक कैलोमेल इलेक्ट्रोड के उपयोग से मानक इलेक्ट्रोड विभव के निर्धारण को दर्शाता है। इसमें एक पोटेंशियोमीटर, एक जिंक एनोड, 1M जिंक सल्फेट घोल, एक कैलोमेल इलेक्ट्रोड, और एक कैलोमेल इलेक्ट्रोड से जुड़ा एक जंक्शन होता है जो मरकरी, मरक्यूरस क्लोराइड पेस्ट और संतृप्त KCl घोल से बना होता है, जिससे संदर्भ इलेक्ट्रोड के रूप में कार्य करता है। (3) The potential developed depends upon the concentration of \( \text{Cl}^- \) or KCl solution. When saturated KCl solution is used, its reduction potential is 0.242 V. (4) Consider following cell : \( \text{Zn(s)} | \text{ZnSO}_4\text{(aq)} || \text{KCl(aq)} | \text{Hg}_2\text{Cl}_2\text{(s)} | \text{Hg} \) OR \( \text{Zn(s)} | \text{ZnSO}_4\text{(aq)} || \text{Calomel electrode} \) Reduction reaction for calomel electrode : \( \text{Hg}_2\text{Cl}_2\text{(s)} + 2\text{e}^- \longrightarrow 2\text{Hg(l)} + 2\text{Cl}^-\text{(aq)} \) Hence potential of calomel electrode depends on the concentration of \( \text{Cl}^- \) or KCl solution.
In simple words: The calomel electrode is a secondary reference electrode made of mercury, mercurous chloride paste, and saturated KCl solution, offering a stable and reproducible potential that depends on the chloride ion concentration, making it a convenient alternative to the standard hydrogen electrode for measuring cell potentials.

🎯 Exam Tip: Remember the components of a calomel electrode (Hg, \( \text{Hg}_2\text{Cl}_2 \), KCl solution) and its reduction half-reaction. Its function as a stable reference electrode is a key concept.

Can You Tell? (Textbook Page No. 114)

Question 1. In what ways are fuel cells and galvanic cells similar and in what ways are they different ?
Answer:Similarity between fuel cells and galvanic cells :
• In both the cells, there is oxidation at anode and reduction at cathode.
• The cell potential is developed due to net redox reactions.
• Both are galvanic cells. Difference in fuel cells and galvanic cells :
• Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with compact surface area.
• Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases.
• In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions.
• In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.
In simple words: Both fuel cells and galvanic cells are electrochemical cells that generate electricity via spontaneous redox reactions, involving oxidation at the anode and reduction at the cathode. However, fuel cells continuously consume external reactants (often gases) and produce energy from combustion-like reactions, making them more efficient for long-term power generation, while traditional galvanic cells use a finite amount of internal reactants.

🎯 Exam Tip: Key differences lie in reactant supply (continuous for fuel cells vs. finite for galvanic), the type of reactions (combustion-like vs. normal redox), and the electrode materials/surface area. Similarities include fundamental redox processes and being galvanic in nature.

Use Your Brain Power (Textbook Page No. 114)

Question 1. Indentify the strongest and the weakest oxidizing agents from the electrochemical series.
Answer:From the electrochemical series, (a) The strongest oxidising agent is fluorine since it has the highest standard reduction potential \( (E^0_{F_2/F^-} = + 2.87 \text{ V}) \). (b) The weakest oxidising agent (or the strongest reducing agent) is lithium since it has the lowest standard reduction potential, \( (E^0_{Li^+/Li} = -3.045 \text{ V}) \).
In simple words: The electrochemical series helps identify the strength of oxidizing and reducing agents. A species with a high positive standard reduction potential (like Fluorine) is a strong oxidizing agent, while one with a low negative standard reduction potential (like Lithium) is a weak oxidizing agent (and thus a strong reducing agent).

🎯 Exam Tip: Remember that a higher (more positive) standard reduction potential indicates a stronger oxidizing agent, and a lower (more negative) potential indicates a weaker oxidizing agent (or stronger reducing agent).

Use Your Brain Power (Textbook Page No. 115)

Question 1. Identify the strongest and the weakest reducing agents from the electrochemical series.
Answer:(a) From the electrochemical series, the strongest reducing agent is lithium since it has the lowest standard reduction potential \( (E^0_{Li^+/Li} = -3.045 \text{ V}) \). (b) The weakest reducing agent is fluorine since it has the highest standard reduction potential, \( (E^0_{F_2/F^-} = +2.87 \text{ V}) \).
In simple words: In the electrochemical series, a species with the most negative standard reduction potential (like Lithium) is the strongest reducing agent, meaning it is most easily oxidized. Conversely, a species with the most positive standard reduction potential (like Fluorine) is the weakest reducing agent, being very difficult to oxidize.

🎯 Exam Tip: A stronger reducing agent corresponds to a more negative standard reduction potential, and a weaker reducing agent corresponds to a more positive standard reduction potential. This is the inverse relationship to oxidizing agents.

Question 2. From \( \text{E}^0 \) values given in Table 5.1, predict whether Sn can reduce \( \text{I}_2 \) or \( \text{Ni}^{2+} \).
Answer:From electrochemical series, \( \text{E}^0_{\text{I}_2/\text{I}^-} = +0.535 \text{ V} \), \( \text{E}^0_{\text{Ni}^{2+}/\text{Ni}} = -0.257 \text{ V} \) and \( \text{E}^0_{\text{Sn}^{2+}/\text{Sn}} = -0.136 \text{ V} \).
\( \text{Sn(s)} + \text{I}_2\text{(aq)} \longrightarrow \text{Sn}^{2+}\text{(aq)} + 2\text{I}^-\text{(aq)} \) reduced oxidised
\( \text{E}^0_{\text{cell}} = \text{E}^0_{\text{I}_2/\text{I}^-} - \text{E}^0_{\text{Sn}^{2+}/\text{Sn}} = 0.535 - (-0.136) = +0.671 \text{ V} \) Since \( \text{E}^0_{\text{cell}} > 0 \), Sn can reduce \( \text{I}_2 \).
For \( \text{Sn(s)} + \text{Ni}^{2+}\text{(aq)} \longrightarrow \text{Sn}^{2+}\text{(aq)} + \text{Ni(s)} \) \( \text{E}^0_{\text{cell}} = \text{E}^0_{\text{Ni}^{2+}/\text{Ni}} - \text{E}^0_{\text{Sn}^{2+}/\text{Sn}} = -0.257 - (-0.136) = -0.121 \text{ V} \) Since \( \text{E}^0_{\text{cell}} < 0 \), Sn will not reduce \( \text{Ni}^{2+} \).
In simple words: Tin (Sn) can reduce iodine (\( \text{I}_2 \)) because the standard cell potential for the reaction is positive, indicating spontaneity. However, tin cannot reduce nickel ions (\( \text{Ni}^{2+} \)) because that reaction yields a negative standard cell potential, meaning it's non-spontaneous. This is because tin is a stronger reducing agent than \( \text{I}^- \) but weaker than Ni.

🎯 Exam Tip: To predict if a reaction is spontaneous, calculate \( \text{E}^0_{\text{cell}} = \text{E}^0_{\text{cathode}} - \text{E}^0_{\text{anode}} \). A positive \( \text{E}^0_{\text{cell}} \) indicates spontaneity, meaning the reducing agent can reduce the oxidizing agent. Always ensure the correct reduction potentials are used for the oxidizing and reducing species.