RBSE Solutions Class 12 Chemistry Chapter 3 Electrochemistry

Rbse Class 12 Chemistry Chapter 3 Multiple Choice Questions

Question 1. Which of the following is not a conductor?
(a) Cu(metal)
(b) NaCl(aq)
(c) NaCl(molten)
(d) NaCl(solid)
Answer: (d) NaCl(solid)
In simple words: Solid sodium chloride (salt) does not conduct electricity because its ions are fixed in a crystal lattice and cannot move freely. It becomes a conductor when dissolved in water or melted, as the ions can then move.

🎯 Exam Tip: Remember that for ionic compounds, conductivity relies on the mobility of ions. In solid form, ions are held in place, but in molten or aqueous states, they are free to move.

Question 2. If conductance and conductivity of a cell are equivalent then cell constant will be :
(a) 1
(b) 0
Answer: (a) 1
In simple words: The cell constant is a value that relates conductivity to conductance. If these two values are the same, it means the cell constant must be 1. It helps in measuring the conductivity of a solution accurately.

🎯 Exam Tip: Recall the formula: Conductivity = Conductance × Cell Constant. If Conductivity = Conductance, then the Cell Constant must be unity (1).

Question 4. The unit of conductivity (specific conductance) is:
(a) \( \text{ohm}^{-1} \)
(b) \( \text{ohm}^{-1} \text{ cm}^{-1} \)
(c) \( \text{ohm}^{-1} \text{ cm}^2 \text{ equiv}^{-1} \)
(d) \( \text{ohm}^{-1} \text{ cm}^2 \)
Answer: (b) \( \text{ohm}^{-1} \text{ cm}^{-1} \)
In simple words: Conductivity measures how well a material conducts electricity. Its unit is typically ohm inverse centimeter inverse, which is often written as Siemen per centimeter. This shows how efficiently a substance allows current to flow through a specific volume.

🎯 Exam Tip: Distinguish between conductance (unit: ohm⁻¹ or Siemens) and conductivity (unit: ohm⁻¹ cm⁻¹ or S cm⁻¹). Conductivity is an intensive property that describes the material itself, while conductance depends on the sample's dimensions.

Question 5. If redox reaction takes place in cell then electromotive force (e.m.f.) of cell will be:
(a) positive
(b) negative
(c) zero
(d) one
Answer: (a) positive
In simple words: For a reaction to happen by itself in an electrochemical cell, the electromotive force (EMF) must be positive. A positive EMF means the cell can produce electrical energy spontaneously.

🎯 Exam Tip: A spontaneous redox reaction in a galvanic cell always has a positive cell potential (E°cell > 0) and a negative Gibbs free energy change (ΔG° < 0).

Question 6. Which statement will be correct for the cell made up of zinc and copper on the basis of electro chemical series?
(a) Zinc will act as cathode and copper as anode.
(b) Zinc will act as anode and copper as cathode.
(c) The flow of electrons takes place from copper to zinc.
(d) Copper electrode dissolves and zinc is deposited at zinc electrode.
Answer: (b) Zinc will act as anode and copper as cathode.
In simple words: In a zinc-copper cell, zinc is more reactive, so it loses electrons and becomes the anode. Copper is less reactive, so it gains electrons and acts as the cathode. Electrons always flow from the anode to the cathode.

🎯 Exam Tip: Remember the mnemonic "An Ox Red Cat" - Anode is for Oxidation, Cathode is for Reduction. In a galvanic cell, the more reactive metal typically undergoes oxidation at the anode.

Question 7. How many coulomb charge will required for oxidation of one mole \( \text{H}_2\text{O} \) into \( \text{O}_2 \)
(a) \( 1.93 \text{ x } 10^5 \text{ C} \)
(b) \( 9.65 \text{ x } 10^4 \text{ C} \)
(c) \( 6.023 \text{ x } 10^{23} \text{ C} \)
(d) \( 4.825 \text{ x } 10^4 \text{ C} \)
Answer: (b) \( 9.65 \text{ x } 10^4 \text{ C} \)
In simple words: This amount of charge is equal to one Faraday. One Faraday represents the charge of one mole of electrons, which is \( 96500 \) Coulombs. It is used in electrochemistry to relate charge to the amount of substance reacting.

🎯 Exam Tip: Recall Faraday's constant: \( 1 \text{ F} = 96485 \text{ C mol}^{-1} \). This constant is crucial for calculations involving electrolysis and electrochemical reactions.

Question 9. Which of the following mixture represents rusting of iron?
(a) \( \text{FeO} \) and \( \text{Fe(OH)}_3 \)
(b) \( \text{FeO} \) and \( \text{Fe(OH)}_2 \)
(c) \( \text{Fe}_2\text{O}_3 \) and \( \text{Fe(OH)}_3 \)
(d) \( \text{Fe}_3\text{O}_4 \) and \( \text{Fe(OH)}_2 \)
Answer: (c) \( \text{Fe}_2\text{O}_3 \) and \( \text{Fe(OH)}_3 \)
In simple words: Rust is mainly made up of hydrated iron(III) oxide. This chemical compound is formed when iron reacts with oxygen and water, causing the metal to break down over time.

🎯 Exam Tip: Rust is chemically known as hydrated ferric oxide, \( \text{Fe}_2\text{O}_3 \cdot \text{nH}_2\text{O} \). It is important to know the composition of common chemical phenomena like rusting.

Question 10. When lead storage cell discharge, then
(a) \( \text{SO}_2 \) evolves
(b) \( \text{Pb SO}_4 \) destroys
(c) \( \text{Pb} \) forms
(d) \( \text{H}_2\text{SO}_4 \) destroys
Answer: (d) \( \text{H}_2\text{SO}_4 \) destroys
In simple words: When a lead storage battery is used, the sulfuric acid \( \text{H}_2\text{SO}_4 \) is consumed in the chemical reactions. This causes its concentration to decrease, which is why the density of the electrolyte drops during discharge.

🎯 Exam Tip: During discharge of a lead-acid battery, sulfuric acid is consumed, leading to a decrease in its concentration and density. The reverse happens during charging.

Rbse Class 12 Chemistry Chapter 3 Very Short Answer Type Questions

Question 1. Can you store copper sulphate solution in a zinc pot?
Answer: No, copper sulphate solution cannot be stored in a zinc pot. This is because zinc is more reactive than copper. When copper sulphate solution is placed in a zinc pot, zinc will displace copper from the solution according to the reaction:
\( \text{Zn(s)} + \text{CuSO}_4\text{(aq)} \rightarrow \text{ZnSO}_4\text{(aq)} + \text{Cu(s)} \)
This reaction occurs because zinc has a lower reduction potential (or higher oxidation potential) than copper. For this cell, the cell potential is positive:
\( \text{E}^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \)
\( \text{E}^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V} \)
\( \text{E}^\circ_{\text{Cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}} \)
Here, copper ions are reduced at the cathode, and zinc is oxidized at the anode.
\( \text{E}^\circ_{\text{Cell}} = (0.34 \text{ V}) - (-0.76 \text{ V}) = 0.34 \text{ V} + 0.76 \text{ V} = 1.10 \text{ V} \)
Since the standard EMF of the cell is positive, the reaction will spontaneously take place, and the zinc pot will react with the copper sulphate solution. This shows how knowing the electrochemical series helps predict chemical reactions.
In simple words: No, you cannot store copper sulphate in a zinc pot. Zinc is more reactive than copper, so it will react with the copper sulphate solution, displacing copper and damaging the pot.

🎯 Exam Tip: To determine if a solution can be stored in a particular metal container, compare the reactivity of the container metal with the metal ions in the solution. If the container metal is more reactive, it will displace the ions, and storage is not possible.

Question 3. Why does the conductivity of solution decreases with dilution?
Answer: The conductivity of a solution depends on the number of ions present in a unit volume. When a solution is diluted, more solvent is added, but the total number of ions remains the same. As a result, the number of ions per unit volume decreases. Even though the mobility of individual ions might increase slightly due to less inter-ionic attraction, the overall reduction in ion concentration per unit volume leads to a decrease in the solution's conductivity. This is a key concept in understanding electrolyte behavior.
In simple words: Conductivity drops when you add more water to a solution because there are fewer ions in each bit of liquid. Even if ions move a little faster, having fewer of them in the same space makes the solution conduct less electricity.

🎯 Exam Tip: While molar conductivity increases with dilution (due to increased ion mobility and dissociation), specific conductivity decreases because the concentration of ions per unit volume decreases significantly.

Question 4. Suggest a list of metals that are extracted electrolytically.
Answer: Metals that are highly reactive and have very negative standard electrode potentials are typically extracted electrolytically. These metals are difficult to reduce by chemical reducing agents. Common examples include sodium, calcium, magnesium, and aluminium. These processes often involve molten salts to achieve the high temperatures needed.
In simple words: Metals like sodium, calcium, magnesium, and aluminium are taken out from their compounds using electricity. This method is needed because these metals are very reactive and hard to separate using normal chemical reactions.

🎯 Exam Tip: Electrolytic extraction is primarily used for highly electropositive metals (Group 1, Group 2, and Aluminium) which cannot be reduced by carbon or other common reducing agents.

Question 5. Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer: Besides hydrogen, two other materials that can serve as fuels in fuel cells are methane and methanol. These substances undergo oxidation within the fuel cell to produce electricity, much like hydrogen. Fuel cells are an important technology for clean energy generation. The efficiency of these cells is high because they convert chemical energy directly to electrical energy.
In simple words: Methane and methanol can be used in fuel cells, just like hydrogen. They react inside the cell to make electricity.

🎯 Exam Tip: Fuel cells are highly efficient devices that convert chemical energy directly into electrical energy, making them promising for future energy needs. Remember that a variety of carbon-containing fuels can be reformed to produce hydrogen or used directly in specific fuel cell designs.

Question 6. Arrange the following metals in the order in which they displace each other from the solution of their salts. Ag, Al, Cu, Fe, Mg and Zn
Answer: The ability of a metal to displace another from its salt solution depends on its reactivity, which is related to its position in the electrochemical series. A more reactive metal (one with a lower standard reduction potential) can displace a less reactive metal. Arranging these metals from most reactive to least reactive (which is the order of increasing reducing power, or ability to displace others) gives:
\( \text{Mg} > \text{Al} > \text{Zn} > \text{Fe} > \text{Cu} > \text{Ag} \)
This means Magnesium can displace Aluminum, Aluminum can displace Zinc, and so on. Understanding this order helps predict displacement reactions.
In simple words: The order of metals from most able to displace others to least able is: Magnesium, Aluminum, Zinc, Iron, Copper, Silver. More reactive metals can push out less reactive ones from their compounds.

🎯 Exam Tip: Remember that metals higher in the activity series (or with more negative standard reduction potentials) are stronger reducing agents and can displace metals lower in the series from their salt solutions.

Rbse Class 12 Chemistry Chapter 3 Short Answer Type Questions

Question 1. How would you determine the standard electrode potential of \( \text{Mg}^{2+}/\text{Mg} \)?
Answer: To determine the standard electrode potential of \( \text{Mg}^{2+}/\text{Mg} \), we would set up an electrochemical cell by connecting a magnesium electrode to a Standard Hydrogen Electrode (SHE). The SHE acts as a reference electrode with a standard potential of 0.00 V. The cell can be represented as:
\( \text{Mg(s)} / \text{Mg}^{2+}\text{(1 M)} \parallel \text{H}^{+}(\text{1 M}) / \text{H}_2\text{(1 atm), Pt(s)} \)
In this cell, magnesium acts as the anode (where oxidation occurs) because it is more reactive than hydrogen. The SHE acts as the cathode (where reduction occurs). A voltmeter is used to measure the cell potential.
The reactions are:
Anode: \( \text{Mg(s)} \rightarrow \text{Mg}^{2+}\text{(aq)} + 2\text{e}^{-} \)
Cathode: \( 2\text{H}^{+}(\text{aq}) + 2\text{e}^{-} \rightarrow \text{H}_2\text{(g)} \)
The standard cell potential \( \text{E}^\circ_{\text{cell}} \) is given by:
\( \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}} \)
\( \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{H}^{+}/\frac{1}{2}\text{H}_2} - \text{E}^\circ_{\text{Mg}^{2+}/\text{Mg}} \)
Since \( \text{E}^\circ_{\text{H}^{+}/\frac{1}{2}\text{H}_2} = 0.00 \text{ V} \), then:
\( \text{E}^\circ_{\text{cell}} = 0.00 \text{ V} - \text{E}^\circ_{\text{Mg}^{2+}/\text{Mg}} \)
\( \text{E}^\circ_{\text{Mg}^{2+}/\text{Mg}} = -\text{E}^\circ_{\text{cell}} \)
The measured \( \text{E}^\circ_{\text{cell}} \) would be positive, and its negative value would give the standard electrode potential for magnesium. This method is standard for measuring unknown electrode potentials.
In simple words: To find the standard potential of magnesium, you connect a magnesium electrode to a special hydrogen electrode. You measure the voltage between them. Because the hydrogen electrode's voltage is zero, the measured voltage tells you the magnesium electrode's potential.

🎯 Exam Tip: Always remember that the Standard Hydrogen Electrode (SHE) serves as the reference point (0.00 V) for measuring other electrode potentials. The unknown electrode is always coupled with SHE to form a galvanic cell.

Question 2. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Answer: The potential of a hydrogen electrode in a solution with a specific pH can be calculated using the Nernst equation. Given: \( \text{pH} = 10 \)
We know that \( \text{pH} = -\log[\text{H}^{+}] \)
So, \( [\text{H}^{+}] = 10^{-\text{pH}} \)
\( [\text{H}^{+}] = 10^{-10} \text{ mol L}^{-1} \)
For a hydrogen electrode, the half-reaction is:
\( \text{H}^{+}(\text{aq}) + \text{e}^{-} \rightarrow \frac{1}{2}\text{H}_2\text{(g)} \)
The Nernst equation for this electrode potential is:
\( \text{E}_{\text{H}^{+}/1/2\text{H}_2} = \text{E}^\circ_{\text{H}^{+}/1/2\text{H}_2} + \frac{0.059}{\text{n}} \log \frac{[\text{H}^{+}]}{(\text{P}_{\text{H}_2})^{1/2}} \)
For the standard hydrogen electrode, \( \text{E}^\circ_{\text{H}^{+}/1/2\text{H}_2} = 0.00 \text{ V} \) and we assume \( \text{P}_{\text{H}_2} = 1 \text{ atm} \). The number of electrons involved, \( \text{n} = 1 \).
So, the equation simplifies to:
\( \text{E}_{\text{H}^{+}/1/2\text{H}_2} = 0.00 + \frac{0.059}{1} \log [\text{H}^{+}] \)
\( \text{E}_{\text{H}^{+}/1/2\text{H}_2} = 0.059 \log (10^{-10}) \)
\( \text{E}_{\text{H}^{+}/1/2\text{H}_2} = 0.059 \times (-10) \)
\( \text{E}_{\text{H}^{+}/1/2\text{H}_2} = -0.59 \text{ V} \)
Thus, the potential of the hydrogen electrode in this solution is \( -0.59 \text{ V} \). This calculation is vital for understanding how pH affects electrode potentials.
In simple words: First, find the hydrogen ion concentration from the pH value. Then, use the Nernst equation for a hydrogen electrode. Since standard hydrogen potential is zero, you just multiply 0.059 by the logarithm of the hydrogen ion concentration to get the electrode potential.

🎯 Exam Tip: Remember the Nernst equation formula and how to apply it, especially for hydrogen electrodes. Pay close attention to the value of 'n' (number of electrons transferred) and the terms inside the logarithm for the reaction quotient.

Question 4. The cell in which the following reaction occurs: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^{-}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2\text{(s)} \) has \( \text{E}^\circ_{\text{cell}} = 0.236 \text{ V} \) at \( 298 \text{ K} \). Calculate the standard Gibb's energy and the equilibrium constant for the cell reaction.
Answer: For the given reaction, we need to calculate the standard Gibbs energy \( (\Delta \text{G}^\circ) \) and the equilibrium constant \( (\text{K}_{\text{c}}) \).
The cell reaction is: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^{-}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2\text{(s)} \)
From the reaction, we can identify the half-reactions:
Reduction: \( 2\text{Fe}^{3+}(\text{aq}) + 2\text{e}^{-} \rightarrow 2\text{Fe}^{2+}(\text{aq}) \)
Oxidation: \( 2\text{I}^{-}(\text{aq}) \rightarrow \text{I}_2\text{(s)} + 2\text{e}^{-} \)
The number of electrons transferred \( (\text{n}) \) in the balanced reaction is 2.
Given: \( \text{E}^\circ_{\text{cell}} = 0.236 \text{ V} \), \( \text{T} = 298 \text{ K} \), \( \text{F} = 96500 \text{ C mol}^{-1} \).
First, calculate the standard Gibbs energy \( (\Delta \text{G}^\circ) \):
\( \Delta \text{G}^\circ = -\text{nF}\text{E}^\circ_{\text{cell}} \)
\( \Delta \text{G}^\circ = -(2 \text{ mol}) \times (96500 \text{ C mol}^{-1}) \times (0.236 \text{ V}) \)
\( \Delta \text{G}^\circ = -45548 \text{ J} \)
\( \Delta \text{G}^\circ = -45.55 \text{ kJ mol}^{-1} \)
Next, calculate the equilibrium constant \( (\text{K}_{\text{c}}) \). The relationship between \( \Delta \text{G}^\circ \) and \( \text{K}_{\text{c}} \) is:
\( \Delta \text{G}^\circ = -2.303 \text{ RT} \log \text{K}_{\text{c}} \)
So, \( \log \text{K}_{\text{c}} = \frac{-\Delta \text{G}^\circ}{2.303 \text{ RT}} \)
Given \( \text{R} = 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \).
\( \log \text{K}_{\text{c}} = \frac{-(-45548 \text{ J mol}^{-1})}{2.303 \times 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \times 298 \text{ K}} \)
\( \log \text{K}_{\text{c}} = \frac{45548}{5705.84} \)
\( \log \text{K}_{\text{c}} = 7.983 \)
\( \text{K}_{\text{c}} = \text{Antilog}(7.983) \)
\( \text{K}_{\text{c}} = 9.616 \times 10^7 \)
The positive standard cell potential means the reaction is spontaneous, which is reflected in the large positive equilibrium constant. This indicates that products are greatly favored at equilibrium.
In simple words: First, figure out how many electrons are moving in the reaction. Then, use that number with the given cell voltage and Faraday's constant to find the Gibbs energy change. After that, use the Gibbs energy to calculate the equilibrium constant, which tells you how much product forms.

🎯 Exam Tip: Always identify the number of electrons (n) correctly from the balanced redox reaction. Remember the formulas \( \Delta \text{G}^\circ = -\text{nF}\text{E}^\circ_{\text{cell}} \) and \( \Delta \text{G}^\circ = -2.303 \text{ RT} \log \text{K}_{\text{c}} \) and use consistent units for all values.

Question 5. Suggest a way to determine the \( \Lambda^\circ_m(\text{HCI}) - \Lambda^\circ_m(\text{NaCl}) \)
Answer: To determine the difference between the limiting molar conductivity of \( \text{HCl} \) and \( \text{NaCl} \), you can use Kohlrausch's Law of Independent Migration of Ions. This law states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the limiting molar conductivities of its individual ions.
We can use the following approach:
\( \Lambda^\circ_m(\text{HCl}) = \lambda^\circ_{\text{H}^{+}} + \lambda^\circ_{\text{Cl}^{-}} \)
\( \Lambda^\circ_m(\text{NaCl}) = \lambda^\circ_{\text{Na}^{+}} + \lambda^\circ_{\text{Cl}^{-}} \)
So, if we subtract the two:
\( \Lambda^\circ_m(\text{HCl}) - \Lambda^\circ_m(\text{NaCl}) = (\lambda^\circ_{\text{H}^{+}} + \lambda^\circ_{\text{Cl}^{-}}) - (\lambda^\circ_{\text{Na}^{+}} + \lambda^\circ_{\text{Cl}^{-}}) \)
\( \Lambda^\circ_m(\text{HCl}) - \Lambda^\circ_m(\text{NaCl}) = \lambda^\circ_{\text{H}^{+}} - \lambda^\circ_{\text{Na}^{+}} \)
This relationship means you need to measure the limiting molar conductivities of \( \text{HCl} \) and \( \text{NaCl} \) separately, which can be done by extrapolating conductivity measurements of their dilute solutions to infinite dilution. It highlights how ionic contributions combine in solution.
In simple words: To find this difference, you can use Kohlrausch's Law. This law says you can find the total conductivity by adding up the conductivities of the individual ions. By subtracting the limiting molar conductivity of NaCl from that of HCl, you effectively find the difference between the limiting molar conductivities of H+ and Na+ ions.

🎯 Exam Tip: Kohlrausch's Law is crucial for calculating limiting molar conductivities, especially for weak electrolytes, by using data from strong electrolytes. It states that each ion contributes independently to the total molar conductivity at infinite dilution.

Question 7. Consider the reaction \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \). What is the quantity of electricity in coulombs needed to reduce 1 mol of \( \text{Cr}_2\text{O}_7^{2-} \)?
Answer: To find the quantity of electricity required, we look at the balanced chemical equation for the reduction of dichromate ion.
The reaction is: \( \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^{+} + 6\text{e}^{-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} \)
From the equation, it is clear that 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \) requires 6 moles of electrons \( (\text{6e}^{-}) \) for its complete reduction. Each mole of electrons carries a charge equivalent to one Faraday (F), which is approximately \( 96500 \) Coulombs.
So, the total quantity of electricity needed \( (\text{Q}) \) is:
\( \text{Q} = \text{n} \times \text{F} \)
Where \( \text{n} = 6 \) (moles of electrons) and \( \text{F} = 96500 \text{ C mol}^{-1} \).
\( \text{Q} = 6 \times 96500 \text{ C} \)
\( \text{Q} = 579000 \text{ C} \)
Therefore, \( 579000 \) Coulombs of electricity are needed to reduce 1 mole of \( \text{Cr}_2\text{O}_7^{2-} \). This shows the direct link between stoichiometry and electrical charge in redox reactions.
In simple words: The reaction shows that you need 6 electrons to reduce one dichromate ion. Since one mole of electrons carries about \( 96500 \) Coulombs of charge, you multiply \( 6 \) by \( 96500 \) to get the total charge needed, which is \( 579000 \) Coulombs.

🎯 Exam Tip: For any redox reaction, identify the number of moles of electrons transferred (n) per mole of reactant. Then, multiply 'n' by Faraday's constant (96500 C/mol) to get the total charge in Coulombs.

Question 8. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer: Recharging a lead storage battery involves reversing the chemical reactions that occur during discharge. This process requires an external electric current to be passed through the battery in the opposite direction. Essentially, electrical energy is converted back into chemical energy and stored.
During recharging, lead sulfate \( (\text{PbSO}_4) \), which formed on both electrodes during discharge, is converted back into lead \( (\text{Pb}) \) at the anode and lead dioxide \( (\text{PbO}_2) \) at the cathode. Simultaneously, sulfuric acid \( (\text{H}_2\text{SO}_4) \) is regenerated, leading to an increase in the density of the electrolyte.
The half-reactions during charging are:
At Anode (where oxidation occurs): \( \text{PbSO}_4\text{(s)} + 2\text{e}^{-} \rightarrow \text{Pb(s)} + \text{SO}_4^{2-}\text{(aq)} \)
At Cathode (where reduction occurs): \( \text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow \text{PbO}_2\text{(s)} + \text{SO}_4^{2-}\text{(aq)} + 4\text{H}^{+}(\text{aq}) + 2\text{e}^{-} \)
The overall reaction during recharging is:
\( 2\text{PbSO}_4\text{(s)} + 2\text{H}_2\text{O(l)} \rightarrow \text{Pb(s)} + \text{PbO}_2\text{(s)} + 2\text{H}_2\text{SO}_4\text{(aq)} \)
This process restores the active materials on the electrodes and the concentration of the sulfuric acid electrolyte, preparing the battery for another discharge cycle. Understanding this cycle is key to battery maintenance.
In simple words: To recharge a lead battery, you push electricity through it backwards. This changes the lead sulfate back into pure lead at one side and lead dioxide at the other. Also, the sulfuric acid, which got used up, is reformed, making the battery ready to work again.

🎯 Exam Tip: For lead storage batteries, remember that charging reverses the discharge reactions. Lead sulfate on both electrodes is converted back to lead and lead dioxide, and sulfuric acid is regenerated.

Question 9. Given the standard electrode potentials \( \text{K}^{+}/\text{K} = -2.93 \text{ V} \), \( \text{Ag}^{+}/\text{Ag} = 0.80 \text{ V} \), \( \text{Hg}_2^{2+}/\text{Hg} = 0.79 \text{ V} \), \( \text{Mg}^{2+}/\text{Mg} = 237\text{ V} \), \( \text{Cr}^{2+}/\text{Cr} = -0.74 \text{ V} \). Arrange these metals in their increasing order of reducing power.
Answer: The reducing power of a metal is inversely related to its standard reduction potential. A more negative (or less positive) standard reduction potential indicates a stronger reducing agent. We are given the following standard reduction potentials:
\( \text{E}^\circ_{\text{K}^{+}/\text{K}} = -2.93 \text{ V} \)
\( \text{E}^\circ_{\text{Ag}^{+}/\text{Ag}} = +0.80 \text{ V} \)
\( \text{E}^\circ_{\text{Hg}_2^{2+}/\text{Hg}} = +0.79 \text{ V} \)
\( \text{E}^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 \text{ V} \) (Corrected from \( 237 \text{ V} \) which is a likely OCR error, as \( \text{Mg} \) is a highly reactive metal with a very negative reduction potential.)
\( \text{E}^\circ_{\text{Cr}^{2+}/\text{Cr}} = -0.74 \text{ V} \)
Now, let's arrange these potentials from most positive (weakest reducing agent) to most negative (strongest reducing agent):
\( \text{Ag} (+0.80 \text{ V}) > \text{Hg} (+0.79 \text{ V}) > \text{Cr} (-0.74 \text{ V}) > \text{Mg} (-2.37 \text{ V}) > \text{K} (-2.93 \text{ V}) \)
Therefore, the increasing order of reducing power is:
\( \text{Ag} < \text{Hg} < \text{Cr} < \text{Mg} < \text{K} \)
This order demonstrates how the electrochemical series is used to compare the relative strengths of reducing agents.
In simple words: Metals with more negative standard electrode potentials are stronger reducing agents. Listing the given metals from the highest potential to the lowest potential gives their order of increasing reducing power, starting with Silver as the weakest and Potassium as the strongest.

🎯 Exam Tip: A metal's reducing power is strongest when its standard reduction potential is most negative. Always list elements in the order requested (e.g., increasing or decreasing) and double-check potential values, especially for common elements.

Rbse Class 12 Chemistry Chapter 3 Long Answer Type Questions

Question 1. Calculate the standard cell potentials of galvanic cells in which the following reactions take place :
(i) \( 2\text{Cr(s)} + 3\text{Cd}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{Cd(s)} \)
(ii) \( \text{Fe}^{2+}(\text{aq}) + \text{Ag}^{+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag(s)} \)
Given \( \text{E}^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V} \), \( \text{E}^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \text{ V} \), \( \text{E}^\circ_{\text{Ag}^{+}/\text{Ag}} = 0.80 \text{ V} \), \( \text{E}^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77 \text{ V} \). Also calculate \( \Delta \text{G}^\circ \) and equilibrium constants for the reactions.
Answer: We need to calculate the standard cell potential \( (\text{E}^\circ_{\text{cell}}) \), standard Gibbs energy \( (\Delta \text{G}^\circ) \), and the equilibrium constant \( (\text{K}_{\text{c}}) \) for both given reactions.

(i) Reaction: \( 2\text{Cr(s)} + 3\text{Cd}^{2+}(\text{aq}) \rightarrow 2\text{Cr}^{3+}(\text{aq}) + 3\text{Cd(s)} \)
Given standard reduction potentials:
\( \text{E}^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V} \)
\( \text{E}^\circ_{\text{Cd}^{2+}/\text{Cd}} = -0.40 \text{ V} \)
In this reaction, Chromium \( (\text{Cr}) \) is oxidized to \( \text{Cr}^{3+} \), so it acts as the anode. Cadmium ions \( (\text{Cd}^{2+}) \) are reduced to \( \text{Cd} \), so they act as the cathode.
\( \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}} \)
\( \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{Cd}^{2+}/\text{Cd}} - \text{E}^\circ_{\text{Cr}^{3+}/\text{Cr}} \)
\( \text{E}^\circ_{\text{cell}} = (-0.40 \text{ V}) - (-0.74 \text{ V}) \)
\( \text{E}^\circ_{\text{cell}} = -0.40 \text{ V} + 0.74 \text{ V} \)
\( \text{E}^\circ_{\text{cell}} = +0.34 \text{ V} \)
To calculate \( \Delta \text{G}^\circ \), we need the number of electrons transferred \( (\text{n}) \). From the balanced reaction, \( 2 \text{Cr} \rightarrow 2\text{Cr}^{3+} + 6\text{e}^{-} \) and \( 3\text{Cd}^{2+} + 6\text{e}^{-} \rightarrow 3\text{Cd} \). So, \( \text{n} = 6 \).
\( \Delta \text{G}^\circ = -\text{nF}\text{E}^\circ_{\text{cell}} \)
\( \Delta \text{G}^\circ = -(6 \text{ mol}) \times (96500 \text{ C mol}^{-1}) \times (0.34 \text{ V}) \)
\( \Delta \text{G}^\circ = -196860 \text{ J} \)
\( \Delta \text{G}^\circ = -196.86 \text{ kJ mol}^{-1} \)
Now, calculate the equilibrium constant \( (\text{K}_{\text{c}}) \):
\( \Delta \text{G}^\circ = -2.303 \text{ RT} \log \text{K}_{\text{c}} \)
\( \log \text{K}_{\text{c}} = \frac{-\Delta \text{G}^\circ}{2.303 \text{ RT}} \)
\( \log \text{K}_{\text{c}} = \frac{-(-196860 \text{ J mol}^{-1})}{2.303 \times 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \times 298 \text{ K}} \)
\( \log \text{K}_{\text{c}} = \frac{196860}{5705.84} \)
\( \log \text{K}_{\text{c}} = 34.50 \)
\( \text{K}_{\text{c}} = \text{Antilog}(34.50) \)
\( \text{K}_{\text{c}} = 3.173 \times 10^{34} \)

(ii) Reaction: \( \text{Fe}^{2+}(\text{aq}) + \text{Ag}^{+}(\text{aq}) \rightarrow \text{Fe}^{3+}(\text{aq}) + \text{Ag(s)} \)
Given standard reduction potentials:
\( \text{E}^\circ_{\text{Ag}^{+}/\text{Ag}} = 0.80 \text{ V} \)
\( \text{E}^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.77 \text{ V} \)
In this reaction, \( \text{Fe}^{2+} \) is oxidized to \( \text{Fe}^{3+} \), so it acts as the anode. \( \text{Ag}^{+} \) is reduced to \( \text{Ag} \), so it acts as the cathode.
\( \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{cathode}} - \text{E}^\circ_{\text{anode}} \)
\( \text{E}^\circ_{\text{cell}} = \text{E}^\circ_{\text{Ag}^{+}/\text{Ag}} - \text{E}^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} \)
\( \text{E}^\circ_{\text{cell}} = 0.80 \text{ V} - 0.77 \text{ V} \)
\( \text{E}^\circ_{\text{cell}} = 0.03 \text{ V} \)
To calculate \( \Delta \text{G}^\circ \), we need the number of electrons transferred \( (\text{n}) \). From the balanced reaction, \( \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^{-} \) and \( \text{Ag}^{+} + \text{e}^{-} \rightarrow \text{Ag} \). So, \( \text{n} = 1 \).
\( \Delta \text{G}^\circ = -\text{nF}\text{E}^\circ_{\text{cell}} \)
\( \Delta \text{G}^\circ = -(1 \text{ mol}) \times (96500 \text{ C mol}^{-1}) \times (0.03 \text{ V}) \)
\( \Delta \text{G}^\circ = -2895 \text{ J} \)
\( \Delta \text{G}^\circ = -2.895 \text{ kJ mol}^{-1} \)
Now, calculate the equilibrium constant \( (\text{K}_{\text{c}}) \):
\( \log \text{K}_{\text{c}} = \frac{-\Delta \text{G}^\circ}{2.303 \text{ RT}} \)
\( \log \text{K}_{\text{c}} = \frac{-(-2895 \text{ J mol}^{-1})}{2.303 \times 8.314 \text{ J K}^{-1}\text{ mol}^{-1} \times 298 \text{ K}} \)
\( \log \text{K}_{\text{c}} = \frac{2895}{5705.84} \)
\( \log \text{K}_{\text{c}} = 0.5074 \)
\( \text{K}_{\text{c}} = \text{Antilog}(0.5074) \)
\( \text{K}_{\text{c}} = 3.22 \)
These calculations are essential for predicting reaction spontaneity and the extent to which products are formed at equilibrium.
In simple words: For each reaction, first, find the cell voltage by subtracting the anode potential from the cathode potential. Then, use this voltage with the number of electrons and Faraday's constant to get the Gibbs energy. Finally, use the Gibbs energy to calculate the equilibrium constant, which tells you how much product is formed at balance.

🎯 Exam Tip: Remember to correctly identify the anode (oxidation) and cathode (reduction) to calculate \( \text{E}^\circ_{\text{cell}} \). Pay attention to the stoichiometry to determine 'n' for \( \Delta \text{G}^\circ \) and ensure consistent units for R and T in the \( \text{K}_{\text{c}} \) calculation.

Question 1. Explain how rusting of iron is considered as setting up of an electrochemical cell.
Answer: Rusting of iron is an electrochemical process, similar to the functioning of a tiny galvanic cell. It happens in the presence of both oxygen and moisture.
Here's how it works:
When iron is exposed to rain or humid air, a layer of water forms on its surface. This water absorbs acidic gases from the air, such as \( \text{CO}_2 \) and \( \text{SO}_2 \), creating an acidic solution \( (\text{e.g., H}_2\text{CO}_3) \). This acid then dissociates to produce \( \text{H}^{+} \) ions.
\( \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{H}_2\text{CO}_3 \rightleftharpoons 2\text{H}^{+} + \text{CO}_3^{2-} \)
At certain spots on the iron surface, the iron acts as the anode, losing electrons and forming ferrous ions \( (\text{Fe}^{2+}) \). This is the oxidation half-reaction:
Anode: \( \text{Fe(s)} \rightarrow \text{Fe}^{2+}(\text{aq}) + 2\text{e}^{-} \)
The electrons released at the anodic spots travel through the iron metal to other spots. These spots act as the cathode, where dissolved oxygen and \( \text{H}^{+} \) ions from the acidic solution gain these electrons, undergoing reduction:
Cathode: \( \text{O}_2\text{(g)} + 4\text{H}^{+}(\text{aq}) + 4\text{e}^{-} \rightarrow 2\text{H}_2\text{O(l)} \)
The overall reaction for the formation of \( \text{Fe}^{2+} \) is:
\( 2\text{Fe(s)} + \text{O}_2\text{(g)} + 4\text{H}^{+}(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + 2\text{H}_2\text{O(l)} \)
The \( \text{Fe}^{2+} \) ions then diffuse away from the anodic spots and are further oxidized by atmospheric oxygen to ferric ions \( (\text{Fe}^{3+}) \). These ferric ions then combine with water molecules to form hydrated ferric oxide \( (\text{Fe}_2\text{O}_3 \cdot \text{xH}_2\text{O}) \), which is commonly known as rust. This entire process mirrors the electron flow and distinct anode/cathode regions of an electrochemical cell.
In simple words: Rusting happens like a tiny battery on the iron surface. Iron loses electrons at some points (anode), and oxygen in water gains these electrons at other points (cathode). This movement of electrons causes iron to change into rust.

🎯 Exam Tip: When explaining rusting as an electrochemical process, ensure you clearly identify the anodic (oxidation of iron) and cathodic (reduction of oxygen) regions, the electron flow, and the role of water and dissolved gases as the electrolyte.

Question 2. Depict the galvanic cell in which the reaction \( \text{Zn(s)} + 2\text{Ag}^{+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Ag(s)} \) takes place. Further, show,
(i) Which of the electrodes is negatively charged?
(ii) The carriers of the current in the cell.
Answer: A galvanic cell is a device that converts chemical energy into electrical energy through a spontaneous redox reaction. For the reaction \( \text{Zn(s)} + 2\text{Ag}^{+}(\text{aq}) \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{Ag(s)} \), the cell can be depicted as shown in the diagram below.

The cell notation for this galvanic cell is: \( \text{Zn(s)} / \text{Zn}^{2+}(\text{aq}) \parallel \text{Ag}^{+}(\text{aq}) / \text{Ag(s)} \)

Here's an explanation of the components and current carriers:
(i) **Which of the electrodes is negatively charged?**
In this galvanic cell, zinc \( (\text{Zn}) \) undergoes oxidation \( (\text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^{-}) \). Oxidation always occurs at the anode. By convention in galvanic cells, the anode is the negatively charged electrode because it is the source of electrons. Therefore, the **zinc electrode** is negatively charged. These electrons then flow through the external circuit to the other electrode.

(ii) **The carriers of the current in the cell.**
There are two types of current carriers in the cell:
* **In the external circuit:** The carriers of the current are **electrons**. They flow from the negatively charged zinc electrode (anode) to the positively charged silver electrode (cathode) through the external wire, often passing through a voltmeter or ammeter.
* **In the internal circuit (solution and salt bridge):** The carriers of the current are **ions**. Cations (positive ions) move towards the cathode, and anions (negative ions) move towards the anode. In the salt bridge, ions move to maintain electrical neutrality in the half-cells. This completes the circuit and allows continuous current flow.

(iii) **Reactions on each electrode:**
At Anode (Zinc electrode): \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2\text{e}^{-} \)
At Cathode (Silver electrode): \( \text{Ag}^{+}(\text{aq}) + \text{e}^{-} \rightarrow \text{Ag(s)} \)
The entire setup generates electricity through a controlled chemical reaction. The diagram provides a visual representation of how these components are arranged.
In simple words: This cell uses zinc and silver to make electricity. The zinc electrode is negative because it releases electrons. Electrons move through the wires outside the cell, while ions move through the liquid and salt bridge inside the cell to carry the current.

🎯 Exam Tip: For galvanic cell diagrams, clearly label the anode, cathode, electron flow (from anode to cathode in external circuit), ion flow (in solution and salt bridge), and the electrolyte solutions. Remember that the anode is negative and the cathode is positive in a galvanic cell.