Question

Consider the following cell at 298 K:
\[ {Mg(s)} | {Mg}^{2+}(x M) || {Zn}^{2+}(y M) | {Zn(s)} \] The cell reaction reached the equilibrium state. What is the value of \(\log \frac{[{Mg}^{2+}]}{[{Zn}^{2+}]}\)? Given: \[ E^\circ_{{Mg}^{2+}/{Mg}} = -2.36\, V; E^\circ_{{Zn}^{2+}/{Zn}} = -0.76\, V; \frac{2.303 RT}{F} = 0.06\, V \]

• A. 53.33
• B. 5.333
• C. 26.67
• D. 2.667

Hint:

At equilibrium, cell potential is zero; use Nernst equation to find concentration ratios.

Answer 1

The Correct Option is A

At equilibrium, \(E_{{cell}} = 0\). Using Nernst equation: \[ E_{{cell}} = E^\circ_{{cell}} - \frac{0.06}{n} \log Q = 0 \] Where \(n = 2\). Calculate \(E^\circ_{{cell}} = E^\circ_{{cathode}} - E^\circ_{{anode}} = (-0.76) - (-2.36) = 1.60\, V\). Thus, \[ 0 = 1.60 - \frac{0.06}{2} \log \frac{[{Mg}^{2+}]}{[{Zn}^{2+}]} \] \[ \frac{0.06}{2} \log \frac{[{Mg}^{2+}]}{[{Zn}^{2+}]} = 1.60 \] \[ \log \frac{[{Mg}^{2+}]}{[{Zn}^{2+}]} = \frac{1.60 \times 2}{0.06} = \frac{3.2}{0.06} = 53.33 \]