Q12 / 180 · KCET 2014 · Chemistry

For hydrogen - oxygen fuel cell at one atm and \( 298 \mathrm{~K} \)
\( \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_{2} \mathrm{O}(l) ; \Delta \mathrm{G}^{\circ}=-240 \mathrm{~kJ} \)
\( \mathrm{E}^{\circ} \) for the cell is approximately,
(Given \( \mathrm{F}=96,500 \mathrm{C} \) )

  • Correct answer
    A
    \( 1.24 \mathrm{~V} \)
  • B
    \( 1.26 \mathrm{~V} \)
  • C
    \( 2.48 \mathrm{~V} \)
  • D
    \( 2.5 \mathrm{~V} \)

Solution

At Anode : $2 \mathrm{H}_{2}(g)+4 \mathrm{OH}^{-}(a q) \rightarrow 4 \mathrm{H}_{2} \mathrm{O}(l)+4 e^{-}$ At Cathode : $\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)+4 e^{-} \rightarrow 4 \mathrm{OH}^{-}(a q)$ $\quad$ Overall : $2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(1) ; \Delta G^{\circ}$ $\quad-240 \times 2=-480 \mathrm{~kJ}=-480000 \mathrm{~J}$ Now, $\Delta G^{\circ}=-n F E_{\text {cell }}{ }^{\circ}$ $\Rightarrow-480 \mathrm{~kJ}=-4 \times 96500 E_{\text {cell }}$ $E_{\text {cell }}^{\circ}=\frac{-480000}{-4 \times 96500}=1.24 \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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