A cell is containing two H electrodes. The negative electrode is in contact with a solution of 10^{-6} M H+ ions. The emf of the cell is 0.118 V at $25^{\circ} {C}$. What is the $\left[{H}^{+}\right]$at positive electrode?

Question

A cell is containing two ${H}$ electrodes. The negative electrode is in contact with a solution of $10^{-6} {M}$ ${H}^{+}$ions. The e.m.f. of the cell is $0.118 \mathrm{~V}$ at $25^{\circ} {C}$. What is the $\left[{H}^{+}\right]$at positive electrode?

(a) $10^{-4} {M}$

(b) $10^{-6} {M}$

(c) $10^{-2} {M}$

(d) $10^{-8} {M}$

Answer 1

Correct Option : (A)

Explanation—

At anode $: {H} \longrightarrow {H}^{+}+e^{-}$

At cathode : ${H}^{+}+e^{-} \longrightarrow {H}$

Given that, $\quad E_{\text {cell }}=0.118 {~V} ;\left[{H}^{+}\right]_{\text {anode }}=10^{-6} {M}$ $E_{\text {cell }\left({H}^{+} / {H}\right)}^{\circ}=0 ; n=1$

We know, $\quad E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.059}{n} \log \frac{\left[{H}^{+}\right]_{\text {Anode }}}{\left[{H}^{+}\right]_{\text {Cathode }}}$

$\therefore \quad 0.118=0-\frac{0.059}{1} \log \frac{10^{-6}}{\left[{H}^{+}\right]_{\text {Cathode }}}$

or $\log \frac{10^{-6}}{\left[{H}^{+}\right]_{\text {Cathode }}}=-2 \Rightarrow \frac{10^{-6}}{\left[{H}^{+}\right]_{\text {Cathode }}}=10^{-2}$

or $\left[{H}^{+}\right]_{\text {Cathode }}=\frac{10^{-6}}{10^{-2}}=10^{-4} {M}$