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  • For hydrogen -oxygen fuel cell with reaction: <img alt="\begin{array}{l}{2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{\: H}_{2} \mathrm{O}(\mathrm{l})

For hydrogen -oxygen fuel cell with reaction:

\begin{array}{l}{2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{\: H}_{2} \mathrm{O}(\mathrm{l})} \\\\ {\Delta \mathrm{G}_{\mathrm{f}}^{\mathrm{-}}\left(\mathrm{H}_{2} \mathrm{O}\right)=-237.2 \mathrm{\: kJ} \mathrm{\: mol}^{-1}}\end{array}

The EMF (in V) of fuel cell is?

Option: 1

1.23


Option: 2

-2.46


Option: 3

2.46


Option: 4

-1.23


Answers (1)

best_answer

Fuel cell reaction is:

\begin{array}{l}{2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})} \\\\ {\text {At anode: }2 \mathrm{H}_{2}(\mathrm{g})+4 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow 4 \mathrm{H}_{2} \mathrm{O}+4 e^{-}}\: \: (Half\: oxidation) \\\\ {\text {At cathode: }} \\ {\mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})+4 e^{-} \longrightarrow 4 \mathrm{OH}^{-} \text { (aq) }} {\text { \: \: \: \: \: \: \: (Half reduction) }}\end{array}

\text { Cell reaction: } 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})
\begin{array}{l}{\therefore\mathrm n_{\text {cell }}=4} \\\\ {\Delta_{r} G^{-}=2\: \Delta_{f} \: G_{\left(\mathrm{H}_{2\: } \mathrm{O}\right)}^{-}-2 \: \Delta_{\mathrm{f}} \: G_{\left(\mathrm{H}_{2}\right)}^{-}-2 \: \Delta_{\mathrm{f}} \: G_{\left(\mathrm{O}_{2}\right)}^{\mathrm{-}}}\end{array}
\begin{aligned} &=2(-237.2) \mathrm\: {kJ\: } \mathrm{mol}^{-1}-2 \times 0-0 \\ &=2(-237.2) \mathrm{\: kJ\: } \mathrm{mol}^{-1} \\ \Delta_{r} G^{-} &=-n F E_{\text {cell }}^{-} \\ 2(-237.2) \mathrm{\: kJ\: } \mathrm{mol}^{-1} &=-4 \times 96500 \mathrm{C} \times E_{\text {cell }}^{\Theta} \\ \therefore E_{\text {cell }}^{0}=& 1.23 \mathrm{\: V} \end{aligned}

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