Question

At what $pH,$ the oxidation potential of Hydrogen gas electrode is $0.27V$ (Given: Pressure of $H_2(g)=0.1bar,\frac{2.203RT}{F}=0.06V$ )

Answer 1

Answer: 5

$\frac{1}{2}{H}_2(g)\to H_{(aq)}^{+}+e^{-}$
$E_{H_2/H^{+}}=E_{H_2/H+}^{\circ }-\frac{0.06}{1}log\frac{\left[H^{+}\right]}{{\left(P_{H_2}\right)}^{1/2}}$
$0.27=0-\frac{0.06}{1}log\frac{\left[H^{+}\right]}{[0.1{]}^{1/2}}$
$\frac{9}{2}=-log\left[H^{+}(aq)\right]-\frac{1}{2}log\frac{1}{0.1},pH=5$