Question

The observed emf of the cell,
$pt|H_2\left(1atm\right)|H^{+}(3\times {10}^{-4}M)\parallel H^{+}\left(M_1\right)|H_2\left(1atm\right)|Pt$ is $0.154V$. Calculate the value of $M_1$ and $pH$ of cathodic solution.

Answer 1

$E_{cell}=0.0591\log \frac{M_1}{3\times {10}^{-4}}$
or $\log \frac{M_1}{3\times {10}^{-4}}=\frac{0.154}{0.0591}=2.6058$
$\frac{M_1}{3\times {10}^{-4}}=4.034\times {10}^2$
$M_1=4.034\times {10}^2\times 3\times {10}^{-4}M$
$=0.121M$
$pH=-\log \left[H^{+}\right]=-\log 0.121=0.917$.