(a) Solution

(i) Why is the Equilibrium Constant ($K_c$) related to $E^\circ_{\text{cell}}$ and not to $E_{\text{cell}}$? [1 Mark]

• Step 1 (Governing Principle): At equilibrium, the cell reaction stops, meaning the potential difference between the electrodes becomes zero ($E_{\text{cell}} = 0$).
• Step 2 (Mathematical Relation): The relationship between Gibbs energy and cell potential is given by $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ and $\Delta G^\circ = -RT \ln K_c$.
• Step 3 (Conclusion): Since $E_{\text{cell}}$ is a variable that depends on the concentration of reactants and products, it becomes zero at equilibrium. $E^\circ_{\text{cell}}$ is a constant characteristic of the cell at a standard state, which uniquely defines the equilibrium position ($K_c$).

(ii) Which metal will liberate hydrogen gas from dil. $\text{H}_2\text{SO}_4$? [1 Mark]

• Step 1 (Governing Principle): A metal can displace hydrogen from an acid if its standard reduction potential ($E^\circ_{\text{red}}$) is less than that of the hydrogen electrode ($E^\circ_{\text{H}^+/ \text{H}_2} = 0.00\text{ V}$).
• Step 2 (Data Analysis): Metal A has $E^\circ = -0.24\text{ V}$ and Metal B has $E^\circ = +0.80\text{ V}$.
• Step 3 (Conclusion): Since $-0.24\text{ V} < 0.00\text{ V}$, Metal A will liberate hydrogen gas.

(iii) Cell reaction in lead storage battery during charging: [1 Mark]

• Step 1 (Reaction Equation): During charging, the discharge reaction is reversed by applying an external potential.
• Step 2 (Final Equation): $$\text{PbSO}_4(s) + 2\text{H}_2\text{O}(l) \xrightarrow{\text{charging}} \text{Pb}(s) + \text{PbO}_2(s) + 2\text{H}_2\text{SO}_4(aq)$$

(b) OR Solution

(i) Type of battery: [1 Mark]

• The Mercury cell is a primary cell (non-rechargeable).

(ii) Advantage over dry cell: [1 Mark]

• The cell potential of a Mercury cell remains constant throughout its life because the overall reaction does not involve any ion whose concentration changes in the solution.

(iii) Overall reaction in Mercury cell: [1 Mark]

• Step 1 (Reaction Equation): The reaction involves the reduction of mercuric oxide by zinc.
• Step 2 (Final Equation): $$\text{Zn}(Hg) + \text{HgO}(s) \longrightarrow \text{ZnO}(s) + \text{Hg}(l)$$