Question

At 298 K the standard free energy of formation of H${}_2$O$\left(l\right)$ is -237.20 kJ/mole while that of its ionisation into H${}^{+}$ ion and hydroxyl ions is 80 kJ/mole, then the emf of the following cell at 298 K will be: (take F = 96500 C)

$H_2(g,1bar)|H^{+}\left(1M\right)||O{H}^{-}\left(1M\right)|O_2(g,1bar)$

• A. $0.40V$
• B. $0.81V$
• C. $1.23V$
• D. $-0.40V$

Answer 1

The correct option is A $0.40V$

Cell reaction:
Cathode: $H_{2}O\left(l\right)\frac{1}{2}{O}_2\left(g\right)+2e^{-}\to 2O{H}^{-}(aq.)$
Anode: $H_2\left(g\right)\to 2H^{+}(aq.)+2e^{-}$
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$H_{2}O\left(l\right)+\frac{1}{2}{O}_2\left(g\right)+H_2\left(g\right)\to 2H^{+}(aq.)+2O{H}^{-}(aq.)$
Also we have
$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)$ $\Delta G_t^0=-237.2kJ/mole$........................1
$H_2\left(l\right)\to H^{+}(aq.)+O{H}^{-}(aq.)$ $\Delta G^0=80kJ/mole$........................................2
Hence for cell reaction
$\Delta G^0=-77.20kJ/mole$ ($2\times 2+1$)
So, $E^0=-\frac{\Delta G^0}{nF}=\frac{77200}{2\times 96500}=0.40V$