Question

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

\[\ce{Pt, H2 (1 atm) | H2SO; E_{H/H^+}}\] = 0.3 V

Answer 1

The electrode reaction is:

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

According to the Nernst equation, at 25°C:

\[\ce{E_{H^+/\frac{1}{2} H_2} = E^{\circ}_{H^+/\frac{1}{2} H_2} - \frac{0.059}{n} log_10 \frac{1}{[H^+_{ (aq)}]}}\] 

\[\ce{E_{H^+/\frac{1}{2}H_2} = -E_{\frac{1}{2}H_2/H^+}}\]

= −0.3V, n = l

Substituting. the values; we have,

−0.3 = \[\ce{0.00 - \frac{0.059}{1} log_10  \frac{1}{[H^+_{ (aq)}]}}\]

−0.3 = 0.00 − 0.059 pH

or, pH = `0.3/0.059`

= 5.08