Question

The reduction potential of hydrogen electrode is $118mV$. The concentration of $H^{+}$ in the solution is

• A. $0.01$ M
• B. $2$ M
• C. $10^{-4}$ M
• D. $1$ M

Answer 1

Answer: (A)

$E_{cell}=\frac{0.0591}{n}log\left[H^{+}\right]$
$-0.118=\frac{0.0591}{2}log\left[H^{+}\right]$
$\frac{0.118}{\frac{0.0591}{2}}=-log\left[H^{+}\right]$
$\Rightarrow pH=0.01M$