ElectrochemistryMedium
Question
The electrode potential of hydrogen electrode in neutral solution and 298 K is
Options
A.−0.413 V
B.zero
C.−0.826 V
D.+0.413 V
Solution
$2H^{+}(aq) + 2e^{-} \rightleftharpoons H_{2}(g)$
$E = E^{o} - \frac{0.059}{2}.\log\frac{1}{\left( H^{+} \right)^{2}} = 0 - \frac{0.059}{2}.\log\frac{1}{\left( 10^{- 7} \right)^{2}} = - 0.0413$
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