Class 12 Chemistry Chapter 2 - Electrochemistry PYQs with Solutions

Class 12 Chemistry Previous Year Questions

Chapter 2: Electrochemistry (Fully Solved Board PYQs)
1. Electrochemical Cells and Electrode Potential
Question 1 (Assertion-Reason) 2024
Assertion (A): For a Daniell cell, $\text{Zn}|\text{Zn}^{2+}(1\text{M})||\text{Cu}^{2+}(1\text{M})|\text{Cu}$ with $E_{\text{cell}}^{\circ} = 1.1\text{ V}$, if the external opposing potential is more than $1.1\text{ V}$, the electrons flow from $\text{Cu}$ to $\text{Zn}$.
Reason (R): Under this condition, the cell acts like an electrolytic cell.
  • (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
  • (b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
  • (c) (A) is true but (R) is false.
  • (d) (A) is false but (R) is true.
View Solution

Correct Option: (a)

Explanation: In a Daniell cell, the standard cell potential is $1.1\text{ V}$. If an external opposing potential $E_{\text{ext}}$ greater than $1.1\text{ V}$ is applied, the chemical reaction is forced in the reverse direction. Consequently, electrons flow in the opposite direction (from copper to zinc, i.e., from the positive cathode to the negative anode), and the cell begins to act as an electrolytic cell (converting external electrical energy into chemical energy).

Question 2 (Very Short Answer) 2020
What will happen to the cell potential and the overall function of a Galvanic cell if the salt bridge is suddenly removed? Justify.
View Solution

Immediate Effect: The cell potential drops immediately to **zero** and current stops flowing.

Scientific Reason: The salt bridge serves two vital functions in an electrochemical cell:

  • It completes the electrical circuit by connecting the two half-cell electrolytes internally.
  • It maintains electrical neutrality in both half-cell solutions by allowing the migration of ions.

If the salt bridge is removed, the circuit is broken, charge immediately accumulates in both compartments (preventing further electrode reactions), and electricity cannot flow.

Question 3 (Numerical) 2020
Calculate the e.m.f. of the following cell at $298\text{ K}$:
$\text{Zn}(s) | \text{Zn}^{2+}(0.1\text{ M}) || \text{Ag}^{+}(0.01\text{ M}) | \text{Ag}(s)$
Given: $E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$, $E^{\circ}_{\text{Ag}^{+}/\text{Ag}} = +0.80\text{ V}$, and $\log(10) = 1$.
View Solution

Step 1: Write the half-cell reactions and determine the number of transferred electrons ($n$).

  • At Anode (Oxidation): $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^{-}$
  • At Cathode (Reduction): $2\text{Ag}^{+}(aq) + 2e^{-} \rightarrow 2\text{Ag}(s)$
  • Overall Reaction: $\text{Zn}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Zn}^{2+}(aq) + 2\text{Ag}(s)$

Thus, $n = 2$.

Step 2: Calculate the standard cell potential ($E^{\circ}_{\text{cell}}$).

$$ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = E^{\circ}_{\text{Ag}^{+}/\text{Ag}} - E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} $$ $$ E^{\circ}_{\text{cell}} = 0.80\text{ V} - (-0.76\text{ V}) = 1.56\text{ V} $$

Step 3: Set up the Nernst equation at $298\text{ K}$.

$$ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Zn}^{2+}]}{[\text{Ag}^{+}]^2} $$

Step 4: Substitute the concentration values and solve.

$$ E_{\text{cell}} = 1.56 - \frac{0.0591}{2} \log \frac{0.1}{(0.01)^2} $$ $$ E_{\text{cell}} = 1.56 - 0.02955 \log \frac{10^{-1}}{(10^{-2})^2} $$ $$ E_{\text{cell}} = 1.56 - 0.02955 \log \frac{10^{-1}}{10^{-4}} $$ $$ E_{\text{cell}} = 1.56 - 0.02955 \log(10^3) $$ $$ E_{\text{cell}} = 1.56 - 0.02955 \times 3 = 1.56 - 0.08865 = 1.47135\text{ V} \approx 1.47\text{ V} $$

Answer: The e.m.f. of the cell is $1.47\text{ V}$.

Question 4 (Numerical) 2020
Write electrode reactions and calculate the $E_{\text{cell}}$ of the following cell at $298\text{ K}$:
$\text{Ni}(s) | \text{Ni}^{2+}(0.001\text{ M}) || \text{Ag}^{+}(0.1\text{ M}) | \text{Ag}(s)$
Given: $E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = -0.25\text{ V}$, $E^{\circ}_{\text{Ag}^{+}/\text{Ag}} = +0.80\text{ V}$.
View Solution

Step 1: Write half-cell and net reactions.

  • Anode (Oxidation): $\text{Ni}(s) \rightarrow \text{Ni}^{2+}(aq) + 2e^{-}$
  • Cathode (Reduction): $2\text{Ag}^{+}(aq) + 2e^{-} \rightarrow 2\text{Ag}(s)$
  • Overall cell reaction: $\text{Ni}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$

Number of electrons transferred ($n$) = $2$.

Step 2: Compute standard cell potential ($E^{\circ}_{\text{cell}}$).

$$ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{Ag}^{+}/\text{Ag}} - E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = 0.80\text{ V} - (-0.25\text{ V}) = 1.05\text{ V} $$

Step 3: Apply the Nernst equation.

$$ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Ni}^{2+}]}{[\text{Ag}^{+}]^2} $$ $$ E_{\text{cell}} = 1.05 - \frac{0.0591}{2} \log \frac{0.001}{(0.1)^2} $$ $$ E_{\text{cell}} = 1.05 - 0.02955 \log \frac{10^{-3}}{10^{-2}} $$ $$ E_{\text{cell}} = 1.05 - 0.02955 \log(10^{-1}) $$ $$ E_{\text{cell}} = 1.05 - 0.02955 \times (-1) = 1.05 + 0.02955 = 1.07955\text{ V} \approx 1.08\text{ V} $$

Note on potential ambiguity in past papers: If standard potentials are taken as $E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = +0.25\text{ V}$ (as printed in some board exam copies), standard cell potential behaves as $E^{\circ}_{\text{cell}} = 0.55\text{ V}$, which changes the final output to $E_{\text{cell}} = 0.5795\text{ V}$.

Answer: The e.m.f. of the cell is $1.08\text{ V}$ (or $0.58\text{ V}$ depending on the electrode potential sign specified in your question sheet).

Question 5 (Numerical) 2020
Represent the cell in which the following reaction takes place:
$2\text{Al}(s) + 3\text{Cd}^{2+}(0.1\text{ M}) \rightarrow 3\text{Cd}(s) + 2\text{Al}^{3+}(0.01\text{ M})$
If the value of $E^{\circ}_{\text{cell}}$ is $1.26\text{ V}$, calculate the value of $E_{\text{cell}}$.
View Solution

Step 1: Represent the cell layout.

Anode is Aluminum (Oxidation) and Cathode is Cadmium (Reduction):

$$ \text{Al}(s) | \text{Al}^{3+}(0.01\text{ M}) || \text{Cd}^{2+}(0.1\text{ M}) | \text{Cd}(s) $$

Step 2: Find the electron change ($n$).

Since $\text{Al} \rightarrow \text{Al}^{3+} + 3e^{-}$ and $\text{Cd}^{2+} + 2e^{-} \rightarrow \text{Cd}$, the lowest common multiple is $6$. Hence, $n = 6$.

Step 3: Write the Nernst equation.

$$ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log \frac{[\text{Al}^{3+}]^2}{[\text{Cd}^{2+}]^3} $$

Step 4: Substitute values and compute.

$$ E_{\text{cell}} = 1.26 - \frac{0.0591}{6} \log \frac{(0.01)^2}{(0.1)^3} $$ $$ E_{\text{cell}} = 1.26 - 0.00985 \log \frac{10^{-4}}{10^{-3}} $$ $$ E_{\text{cell}} = 1.26 - 0.00985 \log(10^{-1}) $$ $$ E_{\text{cell}} = 1.26 - 0.00985 \times (-1) = 1.26 + 0.00985 = 1.26985\text{ V} \approx 1.27\text{ V} $$

Answer: $E_{\text{cell}} = \mathbf{1.27\text{ V}}$.

Question 6 (Numerical) 2020
Calculate $\Delta G^{\circ}$ for the reaction:
$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$
Given: $E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}$ and $E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$, $1\text{ F} = 96500\text{ C mol}^{-1}$.
View Solution

Step 1: Find the number of electrons transferred ($n$) and standard potential ($E^{\circ}_{\text{cell}}$).

The reaction involves a 2-electron transfer, so $n = 2$.

$$ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.34\text{ V} - (-0.76\text{ V}) = 1.10\text{ V} $$

Step 2: Compute standard Gibbs energy ($\Delta G^{\circ}$).

$$ \Delta G^{\circ} = -n F E^{\circ}_{\text{cell}} $$ $$ \Delta G^{\circ} = -2 \times 96500\text{ C mol}^{-1} \times 1.10\text{ V} $$ $$ \Delta G^{\circ} = -212300\text{ J mol}^{-1} = -212.3\text{ kJ mol}^{-1} $$

Answer: The standard Gibbs energy change ($\Delta G^{\circ}$) is $-212.3\text{ kJ mol}^{-1}$.

Question 7 (Numerical) 2023, 2020
Calculate the maximum work and $\log K_c$ for the given reaction at $298\text{ K}$:
$\text{Ni}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Ni}^{2+}(aq) + 2\text{Ag}(s)$
Given: $E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = -0.25\text{ V}$, $E^{\circ}_{\text{Ag}^{+}/\text{Ag}} = +0.80\text{ V}$, $1\text{ F} = 96500\text{ C mol}^{-1}$.
View Solution

Step 1: Calculate the standard cell potential ($E^{\circ}_{\text{cell}}$) with $n = 2$.

$$ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{Ag}^{+}/\text{Ag}} - E^{\circ}_{\text{Ni}^{2+}/\text{Ni}} = 0.80\text{ V} - (-0.25\text{ V}) = 1.05\text{ V} $$

Step 2: Calculate the maximum work ($W_{\text{max}}$).

The maximum work obtainable equals the negative change of standard Gibbs energy ($W_{\text{max}} = -\Delta G^{\circ}$):

$$ W_{\text{max}} = n F E^{\circ}_{\text{cell}} = 2 \times 96500 \times 1.05 = 202650\text{ J mol}^{-1} = 202.65\text{ kJ mol}^{-1} $$

Step 3: Compute the equilibrium constant ($\log K_c$).

$$ E^{\circ}_{\text{cell}} = \frac{0.0591}{n} \log K_c \implies \log K_c = \frac{E^{\circ}_{\text{cell}} \times n}{0.0591} $$ $$ \log K_c = \frac{1.05 \times 2}{0.0591} \approx 35.53 $$

Answer: Maximum work is $202.65\text{ kJ}$ and $\log K_c \approx \mathbf{35.53}$.

Question 8 (Numerical) 2022
What is the pH of $\text{HCl}$ solution when the hydrogen gas electrode shows a potential of $-0.59\text{ V}$ at standard temperature and pressure ($P_{\text{H}_2} = 1\text{ bar}$)?
View Solution

Step 1: Write the reduction reaction for the hydrogen electrode.

$$ \text{H}^{+}(aq) + e^{-} \rightarrow \frac{1}{2}\text{H}_2(g) $$

Here, $n = 1$, and $E^{\circ}_{\text{H}^{+}/\text{H}_2} = 0.00\text{ V}$.

Step 2: Apply the Nernst equation.

$$ E_{\text{H}^{+}/\text{H}_2} = E^{\circ}_{\text{H}^{+}/\text{H}_2} - \frac{0.0591}{n} \log \frac{(P_{\text{H}_2})^{1/2}}{[\text{H}^{+}]} $$

Since $P_{\text{H}_2} = 1\text{ bar}$ and $E_{\text{H}^{+}/\text{H}_2} = -0.59\text{ V}$:

$$ -0.59 = 0 - 0.0591 \log \frac{1}{[\text{H}^{+}]} $$ $$ -0.59 = -0.0591 \left(-\log[\text{H}^{+}]\right) $$

By definition, $\text{pH} = -\log[\text{H}^{+}]$, so:

$$ -0.59 = -0.0591 \times \text{pH} $$ $$ \text{pH} = \frac{-0.59}{-0.0591} \approx 10 $$

Answer: The pH of the hydrochloric acid solution is $10$.

2. Conductivity of Electrolytic Solutions
Question 9 (MCQ) 2024
Dilution affects both conductivity ($\kappa$) as well as molar conductivity ($\Lambda_m$). The effect of dilution on both is as follows:
  • (a) both increase with dilution.
  • (b) both decrease with dilution.
  • (c) conductivity increases whereas molar conductivity decreases on dilution.
  • (d) conductivity decreases whereas molar conductivity increases on dilution.
View Solution

Correct Option: (d)

Explanation:

  • Conductivity ($\kappa$) decreases with dilution: Conductivity is defined as the conductance of ions present in a unit volume of solution. Upon dilution, the total volume increases, which causes the number of current-carrying ions per unit volume to decrease.
  • Molar Conductivity ($\Lambda_m$) increases with dilution: $\Lambda_m$ is defined as the conducting power of all ions produced by dissolving 1 mole of electrolyte. As dilution increases, the degree of dissociation ($\alpha$) increases (for weak electrolytes) and interionic attraction forces decrease (for strong electrolytes), causing ionic mobility to rise.
Question 10 (MCQ) 2020
Which of the following represents the limiting molar conductivity ($\Lambda^{\circ}_m$) of acetic acid ($\text{CH}_3\text{COOH}$) if the limiting molar conductivity of $\text{CH}_3\text{COONa}$ is $91\text{ S cm}^2\text{ mol}^{-1}$?
The limiting ionic conductivities of individual ions are given as: $\lambda^{\circ}(\text{H}^{+}) = 349.6\text{ S cm}^2\text{ mol}^{-1}$ and $\lambda^{\circ}(\text{Na}^{+}) = 50.1\text{ S cm}^2\text{ mol}^{-1}$.
  • (a) $350\text{ S cm}^2\text{ mol}^{-1}$
  • (b) $375.3\text{ S cm}^2\text{ mol}^{-1}$
  • (c) $390.5\text{ S cm}^2\text{ mol}^{-1}$
  • (d) $340.4\text{ S cm}^2\text{ mol}^{-1}$
View Solution

Correct Option: (c)

Derivation via Kohlrausch's Law:

$$ \Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = \lambda^{\circ}(\text{CH}_3\text{COO}^{-}) + \lambda^{\circ}(\text{H}^{+}) $$

We can obtain this from the available values as follows:

$$ \Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = \Lambda^{\circ}_m(\text{CH}_3\text{COONa}) - \lambda^{\circ}(\text{Na}^{+}) + \lambda^{\circ}(\text{H}^{+}) $$ $$ \Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = 91\text{ S cm}^2\text{ mol}^{-1} - 50.1\text{ S cm}^2\text{ mol}^{-1} + 349.6\text{ S cm}^2\text{ mol}^{-1} $$ $$ \Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = 40.9\text{ S cm}^2\text{ mol}^{-1} + 349.6\text{ S cm}^2\text{ mol}^{-1} = 390.5\text{ S cm}^2\text{ mol}^{-1} $$
Question 11 (MCQ) 2024
The limiting molar ionic conductivities of $\text{Ca}^{2+}$ and $\text{Cl}^{-}$ are $119.0$ and $76.3\text{ S cm}^2\text{ mol}^{-1}$ respectively. The value of limiting molar conductivity ($\Lambda^{\circ}_m$) of $\text{CaCl}_2$ will be:
  • (a) $195.3\text{ S cm}^2\text{ mol}^{-1}$
  • (b) $271.6\text{ S cm}^2\text{ mol}^{-1}$
  • (c) $314.3\text{ S cm}^2\text{ mol}^{-1}$
  • (d) $43.3\text{ S cm}^2\text{ mol}^{-1}$
View Solution

Correct Option: (b)

Calculation:

According to Kohlrausch's Law, limiting molar conductivity is the sum of limiting ionic conductivities multiplied by the number of ions in the salt:

$$ \Lambda^{\circ}_m(\text{CaCl}_2) = \lambda^{\circ}(\text{Ca}^{2+}) + 2\lambda^{\circ}(\text{Cl}^{-}) $$ $$ \Lambda^{\circ}_m(\text{CaCl}_2) = 119.0\text{ S cm}^2\text{ mol}^{-1} + 2(76.3\text{ S cm}^2\text{ mol}^{-1}) $$ $$ \Lambda^{\circ}_m(\text{CaCl}_2) = 119.0 + 152.6 = 271.6\text{ S cm}^2\text{ mol}^{-1} $$
Question 12 (Very Short Answer) 2019, 2014
State Kohlrausch's law of independent migration of ions. Explain why limiting molar conductivity ($\Lambda^{\circ}_m$) of weak electrolytes cannot be calculated by direct extrapolation.
View Solution

Kohlrausch's Law: This law states that at infinite dilution (when concentration approaches zero), the limiting molar conductivity of an electrolyte can be expressed as the sum of the individual ionic contributions of its anions and cations.

$$ \Lambda^{\circ}_m = \nu_{+} \lambda^{\circ}_{+} + \nu_{-} \lambda^{\circ}_{-} $$

Where $\nu_{+}$ and $\nu_{-}$ represent the stoichiometric number of cations and anions produced per formula unit.

Why Extrapolation Fails: Strong electrolytes are completely dissociated, and their conductivity increases linearly with the square root of concentration ($\sqrt{C}$), allowing direct linear extrapolation to $C \rightarrow 0$. However, weak electrolytes have low dissociation at higher concentrations. As concentration decreases towards zero, their degree of dissociation ($\alpha$) increases steeply according to Ostwald's dilution law. This causes $\Lambda_m$ to shoot up non-linearly, making the curve nearly parallel to the vertical axis, preventing any direct extrapolation.

Question 13 (Numerical) 2024
The resistance of a conductivity cell filled with $0.2\text{ mol L}^{-1}$ $\text{KCl}$ solution is $200\ \Omega$. If the resistance of the same cell when filled with $0.05\text{ mol L}^{-1}$ $\text{KCl}$ solution is $620\ \Omega$, calculate the conductivity ($\kappa$) and molar conductivity ($\Lambda_m$) of the $0.05\text{ mol L}^{-1}$ $\text{KCl}$ solution. (The conductivity of $0.2\text{ mol L}^{-1}$ $\text{KCl}$ solution is $0.0248\text{ S cm}^{-1}$).
View Solution

Step 1: Calculate the cell constant ($G^{*}$) of the conductivity cell using the reference solution ($0.2\text{ M}$ KCl).

$$ G^{*} = \kappa \times R $$ $$ G^{*} = 0.0248\text{ S cm}^{-1} \times 200\ \Omega = 4.96\text{ cm}^{-1} $$

Step 2: Compute the conductivity ($\kappa$) of the $0.05\text{ M}$ solution.

$$ \kappa = \frac{G^{*}}{R} = \frac{4.96\text{ cm}^{-1}}{620\ \Omega} = 0.008\text{ S cm}^{-1} $$

Step 3: Calculate the molar conductivity ($\Lambda_m$) of the $0.05\text{ M}$ solution.

Concentration $C = 0.05\text{ mol L}^{-1}$.

$$ \Lambda_m = \frac{\kappa \times 1000}{C} $$ $$ \Lambda_m = \frac{0.008\text{ S cm}^{-1} \times 1000}{0.05\text{ mol L}^{-1}} = 160\text{ S cm}^2\text{ mol}^{-1} $$

Answer: The conductivity of the $0.05\text{ M}$ KCl solution is $0.008\text{ S cm}^{-1}$ and its molar conductivity is $160\text{ S cm}^2\text{ mol}^{-1}$.

Question 14 (Numerical) 2022
The molar conductivity of an acetic acid solution is $39.0\text{ S cm}^2\text{ mol}^{-1}$. If limiting molar conductivities ($\Lambda^{\circ}_m$) of $\text{NaCl}$, $\text{HCl}$ and $\text{CH}_3\text{COONa}$ are $126.4$, $425.9$ and $91.0\text{ S cm}^2\text{ mol}^{-1}$ respectively, calculate:
(A) The limiting molar conductivity of acetic acid.
(B) The degree of dissociation ($\alpha$) and the percentage of acetic acid present in unionized form.
View Solution

Part (A): Limiting molar conductivity calculation.

$$ \Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = \Lambda^{\circ}_m(\text{CH}_3\text{COONa}) + \Lambda^{\circ}_m(\text{HCl}) - \Lambda^{\circ}_m(\text{NaCl}) $$ $$ \Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = 91.0 + 425.9 - 126.4 = 390.5\text{ S cm}^2\text{ mol}^{-1} $$

Part (B): Degree of dissociation ($\alpha$) and unionized form.

Degree of dissociation is the ratio of molar conductivity at concentration $C$ to that at infinite dilution:

$$ \alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{39.0\text{ S cm}^2\text{ mol}^{-1}}{390.5\text{ S cm}^2\text{ mol}^{-1}} \approx 0.10 \text{ (or } 10\% \text{)} $$

Since $10\%$ of the acetic acid is ionized in the solution, the portion remaining in the unionized state is:

$$ \text{Percentage unionized} = 100\% - 10\% = 90\% $$

Answer: (A) $\Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = \mathbf{390.5\text{ S cm}^2\text{ mol}^{-1}}$, (B) degree of dissociation is $0.10$ ($10\%$) and the amount in unionized form is $90\%$.

3. Electrolytic Cell and Electrolysis
Question 15 (Very Short Answer) 2024, 2020
Identify the type of electrochemical cell used in the Apollo space program. State its electrode reactions and list two distinct advantages of this cell over conventional batteries.
View Solution

Cell Type: The **Hydrogen-Oxygen ($\text{H}_2\text{-O}_2$) Fuel Cell**.

Electrode Reactions: Uses concentrated hot aqueous $\text{KOH}$ as electrolyte:

  • At Anode (Oxidation): $2\text{H}_2(g) + 4\text{OH}^{-}(aq) \rightarrow 4\text{H}_2\text{O}(l) + 4e^{-}$
  • At Cathode (Reduction): $\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^{-} \rightarrow 4\text{OH}^{-}(aq)$
  • Overall Reaction: $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(l)$

Advantages:

  1. Continuous Supply & High Efficiency: Conventional batteries run down once electrodes are consumed. Fuel cells operate continuously as long as reactants are supplied, achieving an efficiency of nearly $70\%$ (compared to $\approx 40\%$ for thermal power plants).
  2. Eco-Friendly & Dual Purpose: It is zero-emission. In the Apollo spacecraft, the product water vapor was condensed and safely used as drinking water by the astronauts.
Question 16 (Very Short Answer) 2017, 2015
The following reduction reactions occur at the cathode during the electrolysis of an aqueous copper (II) chloride ($\text{CuCl}_2$) solution:
(1) $\text{Cu}^{2+}(aq) + 2e^{-} \rightarrow \text{Cu}(s)$, $E^{\circ} = +0.34\text{ V}$
(2) $\text{H}^{+}(aq) + e^{-} \rightarrow \frac{1}{2}\text{H}_2(g)$, $E^{\circ} = 0.00\text{ V}$
Based on standard reduction potentials, which reaction occurs preferentially at the cathode and why?
View Solution

Preferential Reaction: The reduction of copper (Reaction 1) occurs at the cathode:

$$ \text{Cu}^{2+}(aq) + 2e^{-} \rightarrow \text{Cu}(s) $$

Reason: At the cathode of an electrolytic cell, multiple reduction reactions compete. The reaction with the higher standard reduction potential ($E^{\circ}$) is thermodynamically favored because it has a greater tendency to gain electrons. Since $E^{\circ}_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V}$ is more positive than $E^{\circ}_{\text{H}^{+}/\text{H}_2} = 0.00\text{ V}$, copper ions are reduced and deposited preferentially over hydrogen ions.

Question 17 (Numerical) 2022
Calculate the quantity of electricity in Coulombs required to produce $4.8\text{ g}$ of Magnesium ($\text{Mg}$) from molten $\text{MgCl}_2$. How much Calcium ($\text{Ca}$) will be produced if the same amount of electricity is passed through molten $\text{CaCl}_2$? (Atomic masses: $\text{Mg} = 24\text{ u}$, $\text{Ca} = 40\text{ u}$).
View Solution

Step 1: Set up the reduction equation for magnesium to determine charge.

$$ \text{Mg}^{2+} + 2e^{-} \rightarrow \text{Mg}(s) $$

This implies that $1\text{ mole}$ ($24\text{ g}$) of $\text{Mg}$ requires $2\text{ F}$ (Faradays) of charge.

Step 2: Calculate the charge in Coulombs required for $4.8\text{ g}$ of $\text{Mg}$.

$$ \text{Moles of Mg produced} = \frac{\text{Given mass}}{\text{Atomic mass}} = \frac{4.8\text{ g}}{24\text{ g mol}^{-1}} = 0.2\text{ mol} $$ $$ \text{Charge required (Q)} = 0.2\text{ mol} \times 2\text{ F mol}^{-1} = 0.4\text{ F} $$ $$ Q = 0.4 \times 96500\text{ C} = 38600\text{ C} $$

Step 3: Calculate calcium deposit using the same charge ($0.4\text{ F}$).

The reduction reaction of Calcium is:

$$ \text{Ca}^{2+} + 2e^{-} \rightarrow \text{Ca}(s) $$

Thus, $2\text{ F}$ is required to deposit $1\text{ mole}$ ($40\text{ g}$) of Calcium.

$$ \text{Mass of Ca deposited} = 0.4\text{ F} \times \frac{40\text{ g}}{2\text{ F}} = 0.2\text{ mol} \times 40\text{ g mol}^{-1} = 8.0\text{ g} $$

Answer: The charge required is $38600\text{ C}$, and the mass of Calcium deposited is $8.0\text{ g}$.

Question 18 (Numerical) 2019
A steady current of $2\text{ amperes}$ was passed through two electrolytic cells X and Y connected in series containing solutions of $\text{FeSO}_4$ and $\text{ZnSO}_4$ respectively, until $2.8\text{ g}$ of iron ($\text{Fe}$) was deposited at the cathode of cell X. How long did the current flow? Calculate the mass of zinc ($\text{Zn}$) deposited at the cathode of cell Y. (Molar masses: $\text{Fe} = 56\text{ g mol}^{-1}$, $\text{Zn} = 65.3\text{ g mol}^{-1}$)
View Solution

Step 1: Determine current flow duration ($t$) in cell X.

Reduction of iron is: $\text{Fe}^{2+} + 2e^{-} \rightarrow \text{Fe}(s)$ (transfer of $n = 2$ electrons).

Using Faraday's First Law:

$$ w = z \cdot I \cdot t = \frac{M}{n \cdot F} I \cdot t $$ $$ 2.8\text{ g} = \frac{56\text{ g mol}^{-1}}{2 \times 96500\text{ C mol}^{-1}} \times 2\text{ A} \times t $$ $$ 2.8 = \frac{56}{96500} \times t \implies t = \frac{2.8 \times 96500}{56} = 4825\text{ s} $$

Step 2: Compute mass of deposited zinc ($w_{\text{Zn}}$) using Faraday's Second Law.

Since the cells are connected in series, the same charge flows through both:

$$ \frac{w_{\text{Zn}}}{w_{\text{Fe}}} = \frac{\text{Equivalent mass of Zn}}{\text{Equivalent mass of Fe}} $$
  • Equivalent weight of $\text{Fe} = M/2 = 56/2 = 28$
  • Equivalent weight of $\text{Zn} = M/2 = 65.3/2 = 32.65$
$$ \frac{w_{\text{Zn}}}{2.8\text{ g}} = \frac{32.65}{28} $$ $$ w_{\text{Zn}} = \frac{32.65 \times 2.8}{28} = 3.265\text{ g} $$

Answer: The current passed for $4825\text{ seconds}$ (approx. 1 hour 20 minutes) and the mass of zinc deposited is $3.265\text{ g}$.

4. Commercial Cells, Batteries and Corrosion
Question 19 (Very Short Answer) 2017
Identify the type of cell generally used in automotive engines and household inverters. Write the balanced chemical reactions occurring at both the anode and cathode of this cell during discharge.
View Solution

Cell Type: The **Lead-Acid Storage Battery** (a secondary, rechargeable electrochemical cell).

Discharge Reactions: Uses spongy lead as anode, lead dioxide ($\text{PbO}_2$) as cathode, and $\approx 38\% \, w/w$ sulfuric acid ($\text{H}_2\text{SO}_4$) as electrolyte:

  • At Anode (Oxidation):
    $$ \text{Pb}(s) + \text{SO}_4^{2-}(aq) \rightarrow \text{PbSO}_4(s) + 2e^{-} $$
  • At Cathode (Reduction):
    $$ \text{PbO}_2(s) + \text{SO}_4^{2-}(aq) + 4\text{H}^{+}(aq) + 2e^{-} \rightarrow \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) $$
  • Net Overall Cell Reaction during discharge:
    $$ \text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq) \rightarrow 2\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) $$
Question 20 (Very Short Answer) 2017, 2015
Which primary cell is commonly utilized in hearing aids and electronic watches? Write its chemical equations and explain why its cell potential remains constant throughout its entire operational lifetime.
View Solution

Cell Type: The **Mercury Cell** (a small, button-sized primary cell).

Cell Reactions: Uses zinc-mercury amalgam as anode, mercuric oxide ($\text{HgO}$) mixed with carbon as cathode, and a paste of $\text{KOH}$ and $\text{ZnO}$ as electrolyte:

  • At Anode (Oxidation): $\text{Zn(Hg)} + 2\text{OH}^{-}(aq) \rightarrow \text{ZnO}(s) + \text{H}_2\text{O}(l) + 2e^{-}$
  • At Cathode (Reduction): $\text{HgO}(s) + \text{H}_2\text{O}(l) + 2e^{-} \rightarrow \text{Hg}(l) + 2\text{OH}^{-}(aq)$
  • Overall Reaction:
    $$ \text{Zn(Hg)} + \text{HgO}(s) \rightarrow \text{ZnO}(s) + \text{Hg}(l) $$

Reason for Constant Voltage: Unlike dry cells, the overall reaction of a mercury cell involves only solid substances ($\text{Zn(Hg)}$, $\text{HgO}$, $\text{ZnO}$) and liquid mercury. Since there are no ionic species in solution whose concentration can change during discharge, the chemical potential quotient remains unchanged. Thus, the mercury cell maintains a very steady output potential of $\approx 1.35\text{ V}$ throughout its life.