Question

If hydrogen electrode is dipped in $2$ solutions of $pH=4$ and $pH=6$ at $1atm$ pressure and salt bridge is connected; the e.m.f. of resulting cell is_______ $V$.

Answer 1

Answer: 0.12

$pH=4,\left[H^{+}\right]=10^{-4}M$
$pH=6,\left[H^{+}\right]=10^{-6}M$
The, electrode which has lower $\left[H^{+}\right]$acts as anode while electrode with higher $\left[H^{+}\right]$acts as cathode.
Hence,
Anode reaction:
$H_2\left(P_1\right)⟶2H^{+}\left(10^{-6}M\right)+2e^{-}$
Cathode reaction:
$2H^{+}\left(10^{-4}M\right)+2e^{-}⟶H_2\left(P_2\right)$
$E_{\text{anode }}=-\frac{RT}{2F}\ln \frac{{\left[H^{+}\right]}^2}{P_1}=-\frac{0.0591}{2}\log \frac{{\left(10^{-6}\right)}^2}{1}$
$E_{\text{cathode }}=-\frac{RT}{2F}\ln \frac{P_2}{{\left[H^{+}\right]}^2}-\frac{0.0591}{2}\log \frac{1}{{\left(10^{-4}\right)}^2}$
$E_{\text{cell }}=E_{\text{anode }}+E_{\text{cathode }}$
$=-\frac{0.0591}{2}\log \frac{{\left(10^{-6}\right)}^2}{1}-\frac{0.0591}{2}\log \frac{1}{{\left(10^{-4}\right)}^2}$
$E_{\text{cell }}=-\frac{0.0591}{2}\log \frac{{\left(10^{-6}\right)}^2}{{\left(10^{-4}\right)}^2}$
$=-0.0591\log \frac{10^{-6}}{10^{-4}}$
$=-0.0591\times \log 10^{-2}$
$=-0.0591\times (-2)=0.12V$