Solution for (a) Fuel Cell

Step 1 (Definition): [1 Mark]

A Fuel cell is a galvanic cell that converts the chemical energy of combustion of fuels (like $\text{H}_2$, $\text{CH}_4$, $\text{CH}_3\text{OH}$) directly into electrical energy.

Step 2 (Advantages): [1 Mark]

• High Efficiency: Fuel cells are much more efficient than conventional methods (like burning fuels to produce heat, then steam, then electricity) as they are not limited by Carnot cycle efficiency.
• Pollution-free: Since the byproduct is usually water ($\text{H}_2\text{O}$), they are eco-friendly and do not cause environmental pollution.

Solution for (b) Electrolysis Products

Step 1 (Governing Principles): [0.5 Mark]

The products of electrolysis depend on the standard electrode potentials of the ions present and the nature of the electrodes. For dilute $\text{H}_2\text{SO}_4$, water undergoes oxidation and reduction. For aqueous $\text{CuCl}_2$, $\text{Cu}^{2+}$ ions are preferentially discharged over $\text{H}^+$ ions at the cathode.

Step 2 (Step-by-Step Analysis): [1 Mark]

(i) Dilute $\text{H}_2\text{SO}_4$ with Pt electrodes:

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

(ii) Aqueous $\text{CuCl}_2$ with Pt electrodes:

• At Cathode: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$
• At Anode: $2\text{Cl}^-(aq) \rightarrow \text{Cl}_2(g) + 2e^-$

Step 3 (Final Concluded Answer): [0.5 Mark]

• (i) Products: $\text{H}_2$ gas at cathode and $\text{O}_2$ gas at anode.
• (ii) Products: $\text{Cu}$ metal at cathode and $\text{Cl}_2$ gas at anode.