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The value of $E_{H_2O|H_2(1\;bar)|Pt}$ at 298K would be

$\begin{array}{1 1}-0.207V\\-0.414V\\0.207V\\0.414V\end{array} $

1 Answer

Answer : $-0.414V$
$E=-\large\frac{RT}{F}$$\ln \large\frac{(p_{H_2}/bar)^{1/2}}{[H^+]/mol\;dm^{-3}}$
$\Rightarrow -(0.059 V)\log \large\frac{1}{(10^{-7})}$$=-7\times 0.059V=-0.413V$
answered Jul 21, 2014 by sreemathi.v
 

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