Question

The cell potential for the following cell:

\[\ce{Pt | H2_{(g)} | H{^+_{(aq)}} || Cu^+2 (0.01 M) | Cu}\] is 0.576 at 298 K.

The pH of the solution is ______. (Nearest integer)

Answer 1

The pH of the solution is 5.

Explanation:

The given cell is \[\ce{Pt | H2_{(g)} | H{^+_{(aq)}} || Cu^+2 (0.01 M) | Cu}\]

E_cell = 0.576 V

T = 298 K

Where \[\ce{E^{\circ}_{cell}}\] = 0.34 V

Cu²⁺ = 0.01 

n = 2

Use the Nernst equation:

\[\ce{E = E{^{\circ}_{cell}} - \frac{0.0591}{n} log \frac{[H^{+}]^2}{[Cu^{2+}]}}\]

\[\ce{0.576 = 0.34 - \frac{0.0591}{2} log \frac{[H^{+}]^2}{0.01}}\]

\[\ce{0.576 - 0.34 = -0.02955 log \frac{[H^{+}]^2}{0.01}}\]

\[\ce{0.236 = -0.02955 log \frac{[H^{+}]^2}{0.01}}\]

Divide both sides

\[\ce{log \frac{[H^{+}]^2}{0.01} = -7.986}\]

\[\ce{log \frac{[H^{+}]^2}{0.01}}\] ≈ −8.0

So

\[\ce{log \frac{[H^{+}]^2}{0.01} = 10^8}\]

[H⁺]² = 10⁻¹⁰

[H⁺] = 10⁻⁵

pH = −log[H⁺]

= −log(10⁻⁵)

= 5​