At $298 \mathrm{~K}$ the standard free energy of formation of $\mathrm{H}_{2} \mathrm{O}$ (l) is $-237.20…

At $298 \mathrm{~K}$ the standard free energy of formation of $\mathrm{H}_{2} \mathrm{O}$ (l) is $-237.20 \mathrm{~kJ} / \mathrm{mol}$ while that of its
ionisation into $\mathrm{H}^{+}$ ion and hydroxyl ions is $80 \mathrm{~kJ} /$ mol, then the emf of the following cell at $298 \mathrm{~K}$
will be
[Take Faraday constant $\mathrm{F}=96500 \mathrm{C}$ ] $\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{bar})\left|\mathrm{H}^{+}(1 \mathrm{M})ight|\left|\mathrm{OH}^{-}(1 \mathrm{M})ight| \mathrm{O}_{2}(\mathrm{~g}, 1 \mathrm{bar})$
  1. $0.40 \mathrm{~V}$
  2. $0.81 \mathrm{~V}$
  3. $1.23 \mathrm{~V}$
  4. $-0.40 \mathrm{~V}$

Solution

Cell reaction cathode : $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}(\mathrm{aq})$ anode: $\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-}$ $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})$ Also we have $\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ $\Delta \mathrm{G}_{\mathrm{f}}^{\circ}=-237.2 \mathrm{~kJ} / \mathrm{mole}$ $\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})$ $\Delta \mathrm{G}^{\circ}=80 \mathrm{~kJ} / \mathrm{mol}$ Hence for cell reaction $\Delta \mathrm{G}^{\circ}=-237.2+(2 \times 80)=-77.20 \mathrm{~kJ} / \mathrm{mol}$ $\therefore \quad \mathrm{E}^{\circ}=-\frac{\Delta \mathrm{G}^{\circ}}{\mathrm{nF}}=\frac{77200}{2 \times 96500}=0.40 \mathrm{~V}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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