JEE Advanced 2023 Paper 2 Chemistry: Paragraph 1

The cell reaction involves $n=4$ electrons for $2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O(l)}$. Pure liquid water has unit activity, so $Q=\dfrac{1}{p_{\mathrm{H_2}}^2\,p_{\mathrm{O_2}}}$ (pressures in bar). Nernst equation: $E=E^\circ-\dfrac{RT}{nF}\ln Q=E^\circ+\dfrac{RT}{4F}\ln\left(p_{\mathrm{H_2}}^2p_{\mathrm{O_2}}\right)$. Here $p_{\mathrm{H_2}}^2p_{\mathrm{O_2}}=25\times0.20=5.0$, so $\ln5.0=1.609$. Also $\dfrac{RT}{4F}=\dfrac{8.314\times298}{4\times96485}=6.420\times10^{-3}\ \mathrm{V}$. Therefore $E=1.23+6.420\times10^{-3}\times1.609=1.23+0.0103=1.2403\ \mathrm{V}\approx1.24\ \mathrm{V}$. Note that the pressure of $\mathrm{H_2}$ enters squared because the balanced equation uses $2\ \mathrm{H_2}$ per $\mathrm{O_2}$; using the one-mole equation would need $n=2$ and gives the same result.

Charge passed: $Q=It=0.50\ \mathrm{A}\times(2.0\times3600\ \mathrm{s})=3600\ \mathrm{C}$. Moles of electrons: $n_e=\dfrac{3600}{96485}=3.731\times10^{-2}\ \mathrm{mol}$. In the cell reaction $2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}$, four electrons are transferred for every two moles of water formed, that is, each mole of $\mathrm{H_2O}$ needs two moles of electrons (anode: $\mathrm{H_2\rightarrow2H^++2e^-}$). Hence $n(\mathrm{H_2O})=\dfrac{3.731\times10^{-2}}{2}=1.866\times10^{-2}\ \mathrm{mol}$ and the mass is $1.866\times10^{-2}\times18=0.336\ \mathrm{g}\approx0.34\ \mathrm{g}$. The result does not depend on the cell voltage or the gas pressures, which affect only the energy delivered.