Question

Answer the following :
(i) Why is the Equilibrium Constant (\(K_c\)) related to \(E^{\circ}_{cell}\) and not to \(E_{cell}\)?
(ii) Two metals ‘A’ and ‘B’ have standard electrode potential values of \(-0.24\) V and \(+0.80\) V respectively. Which of these will liberate hydrogen gas from dil. \(H_2SO_4\)?
(iii) Write the cell reaction which occurs in lead storage battery when it is in charging.

Hint:

Standard potential \(E^{\circ}\) is independent of concentration and only changes with temperature, making it a reliable predictor of the equilibrium constant.

Answer 1

Step 1: Understanding the Concept:
Electrochemistry relates chemical energy to electrical potential. Standard potential (\(E^{\circ}\)) defines the inherent drive of a reaction, while \(E_{cell}\) reflects current state. Equilibrium is reached when this drive (\(E_{cell}\)) becomes zero.
Step 2: Detailed Explanation:
(i) Equilibrium Relation: When a chemical system reaches equilibrium, the rates of forward and backward reactions are equal, and the net chemical change is zero.
Consequently, the potential of the cell (\(E_{cell}\)) drops to zero because the system can no longer perform electrical work.
Substituting these equilibrium conditions into the Nernst equation gives:
\[ 0 = E^{\circ}_{cell} - \frac{2.303 RT}{nF} \log K_c \]
\[ E^{\circ}_{cell} = \frac{2.303 RT}{nF} \log K_c \]
This shows that \(K_c\) is determined by the standard potential (\(E^{\circ}_{cell}\)), which is constant for a given reaction.
Since \(E_{cell}\) is always zero at equilibrium for any reaction, it cannot distinguish between different values of \(K_c\).
(ii) Hydrogen Liberation: For a metal to liberate \(H_2\) from an acid, it must reduce \(H^+\) ions to \(H_2\).
\[ 2H^+ + 2e^- \rightarrow H_2 \quad (E^{\circ} = 0.00 \text{ V}) \]
This requires the metal to have a lower (more negative) standard reduction potential than hydrogen.
Metal ‘A’ has \(E^{\circ} = -0.24\) V, which is less than \(0.00\) V.
Metal ‘B’ has \(E^{\circ} = +0.80\) V, which is greater than \(0.00\) V.
Therefore, only metal ‘A’ is a strong enough reducing agent to liberate hydrogen gas.
(iii) Lead Storage Battery (Charging): During charging, an external power source reverses the chemical changes that occurred during discharge.
The lead sulfate (\(PbSO_4\)) produced at both electrodes is converted back to lead (\(Pb\)) at the anode and lead dioxide (\(PbO_2\)) at the cathode.
The overall chemical equation is:
\[ 2PbSO_4(s) + 2H_2O(l) \xrightarrow{\text{Electrical energy}} Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \]
Step 3: Final Answer:
(i) \(E_{cell}\) is zero at equilibrium, leaving \(K_c\) related to the constant \(E^{\circ}_{cell}\).
(ii) Metal A (\(-0.24\) V) liberates \(H_2\).
(iii) The overall reaction for charging is provided in Step 2.