Question

A cell containing two H electrodes. The negative electrode is in contact with a solution of ${10}^{-6}M{H}^{+}$ ion. The e.m.f. of the cell is 0.118 volt at 25${}^{0}C$. $\left[H^{+}\right]$ at the positive electrode is :

• A. ${10}^{-4}M$
• B. ${10}^{-5}M$
• C. ${10}^{-3}M$
• D. None of these

Answer 1

The correct option is A ${10}^{-4}M$

Anode: $H_2\to 2H^{+}+2e;\left[H^{+}\right]={10}^{-6}M$
(Negative polarity)

Cathode: $2H^{+}+2e\to H_2;say\left[H^{+}\right]=aM$
(positive polarity)

Now for the above redox changes:

$E_{cell}=E_{O{P}_H}+E_{R{P}_H}=E_{O{P}_H}^o-\frac{0.059}{2}log[H^{+}{]}_{anode}^2+E_{R{P}_H}^o+\frac{0.059}{2}log[H^{+}{]}_{cathode}^2$

$E_{cell}=\frac{0.059}{2}log\frac{[H^{+}{]}_{cathode}^2}{[H^{+}{]}_{anode}^2}$

$=\frac{0.059}{1}log\frac{[H^{+}{]}_{cathode}}{[H^{+}{]}_{anode}}$

or $0.118=\frac{0.059}{1}log\frac{[H^{+}{]}_{cathode}}{{10}^{-6}}$

$\therefore [H^{+}{]}_{cathode}={10}^{-4}M$

Hence, option A is correct.