Question

Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at ${25}^{\circ }C$:
$H_2\left(1atm\right)|0.4MHCOOH\parallel 1MC{H}_{3}COOH|\left(1atm\right)H_2$
$K_a$ for $HCOOH$ and $C{H}_{3}COOH$ are $1.77\times {10}^{-4}$ and $1.8\times {10}^{-5}$ respectively.

Answer 1

$\left[H^{+}\right]$ in $HCOOH=\sqrt{C\times K_a}=\sqrt{0.5\times 1.77\times {10}^{-4}}$
$=0.9407\times {10}^{-2}M$
$\left[H^{+}\right]$ in $C{H}_{3}COOH=\sqrt{C\times K_a}=\sqrt{1\times 1.8\times {10}^{-5}}$
$=4.2426\times {10}^{-3}M$
$E_{cell}=0.0591\log \frac{[H^{+}{]}_{RHS}}{[H^{+}{]}_{LHS}}=0.0591\log \frac{4.2426\times {10}^{-3}}{0.9407\times {10}^{-2}}$
$=-0.0204volt$.