A hydrogen gas electrode is made by dipping platinum wire in a solution of $\mathrm{HCl}$ of…
- $0.59 \mathrm{~V}$
- $0.118 \mathrm{~V}$
- $1.18 \mathrm{~V}$
- $0.059 \mathrm{~V}$
Solution
$1 \mathrm{~atm} \quad 10^{-10}$
$\mathrm{E}_{\mathrm{H}_{2} / \mathrm{H}^{+}}=0-\frac{0.059}{2} \log \frac{\left(10^{-10}ight)^{2}}{1}$
$\mathrm{E}_{\mathrm{H}_{2} / \mathrm{H}^{+}}=+0.59 \mathrm{~V}$ .
Asked in: JEE-TOPICTESTS-CHEMISTRY