Question

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of $10^{-6}\, M$ hydrogen ions. The emf of the cell is $0.118 \,V$ at $25^\circ C.$ Calculate the concentration of hydrogen ions at the positive electrode.

Answer 1

$\because$ Emf $ =0.118 V >0$, it is galvanic cell and anode is negative electrode :
At anode : $ H _{2}(g) \longrightarrow 2 H ^{+}\left(10^{-6} M \right)+2 e^{-}$
At cathode $: 2 H ^{+}(x)+2 e^{-} \longrightarrow H _{2}$
Cell reaction : $ H ^{+}(x) \longrightarrow H ^{+}\left(10^{-6} M \right)$
Emf $=0.118 V =0-0.0592 \log \frac{10^{-6}}{x}$
$\Rightarrow x=10^{-4} M$