Question

There are two hydrogen electrodes in a cell. The anode is in contact with the hydrogen ion concentration of 10⁻⁶ M. The emf of the cell at 25°C is 0.118 volt. Calculate the hydrogen ion concentration at cathode.

Answer 1

The given cell can be represented as:

\[\ce{Pt,H2 (1 atm) | H+ (10^{-6} M) || H+ (Unknown conc.) | H2 (1 atm), Pt}\]

The half cell reactions and the overall cell reaction are as follows.

At anode: \[\ce{\frac{1}{2} H2_{(g)} -> H+ (10^{-6} M) + e-}\]

At cathode: \[\ce{H+ (Unknown conc.) + e- -> \frac{1}{2} H2_{(g)}}\]

Overall cell reaction: \[\ce{H+ (Unknown conc.) -> H+ (10^{-6} M)}\]

Obviously, for the cell reaction, n = 1; 

According to the Nernst equation, at 25°C

\[\ce{E = E{^{\circ}_{cell}} - \frac{0.0591}{n} log_10 \frac{[H+] known}{[H+] unknown}}\]

Given, E_cell = 0.118 V, [H⁺]_known = 10⁻⁶ M

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{H^{+}/\frac{1}{2} H_2}} - E{^{\circ}_{H^{+}/\frac{1}{2} H_2}}}\]

= 0.00 − 0.00

= 0.00

Substituting the values, we have

\[\ce{0.118 = 0.00 - \frac{0.059}{1} log_10 \frac{10^{-6}}{[H+]_{unknown}}}\]

or, \[\ce{log_10 \frac{10^{-6}}{[H+]_{unknown}} = -\frac{0.118}{0.059}}\] = −2

or, \[\ce{log_{10} 10^{-6} - log_{10} [H+]_{unknown} = -2}\]

or, \[\ce{-log_{10} [H+]_{unknown} = -2 + 6}\] = 4

or, \[\ce{[H+]_{unknown} = antilog_10(-4) = 1 × 10^{-4} M}\]

Hence, the hydrogen ion concentration at the cathode is 1 × 10⁻⁴ M.