Question

Calculate the pH of the following half cell solutions (at 25°C):

\[\ce{Pt, H2 (1 atm) | HCl; E_{H/H^+}}\] = 0.25 V

Answer 1

The electrode reaction is

\[\ce{H{^+_{(aq)}} + e- -> \frac{1}{2} H2_{(g)}}\]

According to the Nernst equation, at 25°C

\[\ce{E_{{H^{+}/{\frac{1}{2}H_{2}}}} = E^{\circ}_{{H^{+}/{\frac{1}{2}H_{2}}}} - \frac{0.059}{n} log_{10} \frac{1}{[H{^{+}_{(aq)}}]}}\]

For the given reaction, n = 1,

\[\ce{E_{{H^{+}/{\frac{1}{2}H_{2}}}} = -E_{{\frac{1}{2}H_{2}}/{H^{+}}}}\]

= −0.25 V and

\[\ce{E^{\circ}_{{H^{+}/{\frac{1}{2}H_{2}}}} = 0.00}\]

Substituting the values, we have

\[\ce{-0.25 = 0.00 - \frac{0.059}{1} log_{10} \frac{1}{[H{^{+}_{(aq)}}]}}\]

or −0.25 = 0.00 − 0.059 pH    ...(∵ pH = −log₁₀ [H+(aq)])

∴ \[\ce{{pH = \frac{0.25}{0.059}}}\]

= 4.237