Pairs of reactants (R) and product (P) are given below. The pair which can be used in a fuel cell is :

1. $R\mathit{=}{H}_{\mathit{2}\mathit{\left(}\mathit{g}\mathit{\right)}}\mathit{,}{O}_{\mathit{2}\mathit{\left(}\mathit{g}\mathit{\right)}}\mathit{;}P\mathit{=}{H}_{\mathit{2}}{O}_{\mathit{2}\mathit{\left(}\mathit{g}\mathit{\right)}}$

2. $\mathrm{R}={\mathrm{H}}_{2\left(\mathrm{g}\right)},{\mathrm{O}}_{2\left(\mathrm{g}\right)};\mathrm{P}={\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{\ell }\right)}$

3. $\mathrm{R}={\mathrm{H}}_{2\left(\mathrm{g}\right)},{\mathrm{O}}_{2\left(\mathrm{g}\right)},{\mathrm{Cl}}_{2\left(\mathrm{g}\right)};\mathrm{P}={\mathrm{HClO}}_{4\left(\mathrm{aq}\right)}$

4. $\mathrm{R}={\mathrm{H}}_{2\left(\mathrm{g}\right)},{\mathrm{N}}_{2\left(\mathrm{g}\right)};\mathrm{P}={\mathrm{NH}}_{3\left(\mathrm{aq}\right)}$

The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution  is represented by : 

1. Na⁺(aq) + e^- ⟶ Na(s) ; \(E_{cell}^{o} \ = \ -2.71 \ V \)

2. 2H₂O(l) ⟶ O₂(g) + 4H⁺(aq) + 4e^- ; \(E_{cell}^{o} \) = 1.23 V

3. H⁺(aq) + e^- ⟶ \(\frac{1}{2}\)H₂(g) ; \(E_{cell}^{o} \) = 0.00 V

4. Cl^-(aq) ⟶ \(\frac{1}{2}\)Cl₂(g) + e- ; \(E_{cell}^{o}\) $= 1.\mathrm{36 }\mathrm{V}$

On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:

1. Oxygen gas

2. ${\mathrm{H}}_2\mathrm{S}$ gas

3. ${\mathrm{SO}}_2$ gas

4. Hydrogen gas