Question

Calculate reduction potential of hydrogen electrode at $298$K which is prepared with the help of aqueous solution of acetic acid with $0.1$M conc at $1$ atm pressure $K_a=$ $1.8\times {10}^{-8}$.

Answer 1

$C{H}_{3}COOH\rightleftharpoons C{H}_{3}CO{O}^{-}+H^{+}$

at $t=0$ $C$ - -

after $Eq{b}^m$ $C-C\alpha$ $C\alpha$ $C\alpha$

$K_a=\frac{\left[C\alpha \right]\left[\alpha \right]}{[C-C\alpha ]}=\frac{C{\alpha }^2}{[1-\alpha ]}$

as $C{H}_{3}COOH$ is a weak acid, thus $\alpha <<<1$

$1-\alpha \simeq 1$

$K_a=C{\alpha }^2$

$\Rightarrow \alpha =\sqrt{\frac{K_a}{C}}$

$\Rightarrow \left[H^{+}\right]=C\alpha =C\times \sqrt{\frac{K_a}{C}}=\sqrt{K_a\times C}$

$\Rightarrow \left[H^{+}\right]=\sqrt{1.8\times {10}^{-8}\times 0.1M}$

$\Rightarrow \left[H^{+}\right]=4.24\times {10}^{-5}M$

For hydrogen electrode, the cell reaction is

$H^{+}+e^{-}⟶1/2H_2$

$E_{cell}=E_{cell}^o-\frac{0.0591}{1}\log \frac{1}{\left[H^{+}\right]}$

as $E_{cell}^o$ for $H^{+}+C^{-}⟶1/2H_2$ is zero.

$\Rightarrow E_{cell}=-0.0591\log \frac{1}{\left[H^{+}\right]}$

$\Rightarrow E_{cell}=-0.0591\log \frac{1}{4.24\times {10}^{-5}}$

$\Rightarrow E_{cell}=-0.258V$

This is the reduction potential of the Hydrogen electrode.