$H _{2} \rightarrow 2 H ^{+}+2 e ^{-}$
At $1\,atm$ pressure, $pH$ is equal to $10$ .
Therefore,
$\left[ H ^{+}\right]=10^{-10}$
The standard reduction potential of $S.H.E$ is equal to zero.
Apply Nernst equation,
$E_{ H _{2} / H ^{+}}=0-\frac{0.059}{2} \log \frac{\left(10^{-10}\right)^{2}}{1}$
$=0.0295 \times 20$
$=0.59 \,V$
$Zn\,(s)\,\, + \,\,C{u^{2 + }}\,(aq)\, \rightleftharpoons \,Z{n^{2 + \,}}\,(aq)\, + Cu\,(s)$
$(R = 8 \,JK^{-1}\,mol^{-1},\, F = 96000\,C\,mol^{-1})$
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