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Given that $E_{H_2O|H_2(1\;bar)|Pt}=0$ at 298K.The pressure of $H_2$ gas would be

$\begin{array}{1 1}10^{-7}bar\\10^{-10}bar\\10^{-12}bar\\10^{-14}bar\end{array} $

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Answer : $10^{-14}bar$
We have
$E=-\large\frac{RT}{F}$$\ln \large\frac{(P_{H_2}^{1/2}/bar)}{([H^+]/mol\;dm^{-3})}$
Since $E=0$,we must have $(p_{H_2}/bar)^{1/2}/([H^+]/mol\;dm^{-3})=1$
Thus $\large\frac{p_{H_2}}{bar}=\frac{[H^+]^2}{(mol\;dm^{-3})^2}=$$(10^{-7})^2$.
Hence $p_{H_2}=10^{-14}bar$
answered Jul 21, 2014 by sreemathi.v

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