Question

Calculate at 25°C the electrode potential of Mg²⁺/Mg electrode in which the concentration of Mg²⁺ ions is 0.1 M. Given \[\ce{E^{\circ}_{Mg^{2+}/Mg}}\] = −2.36 V, R = 8.314 JK⁻¹, F = 96500 coulombs mol⁻¹.

Answer 1

Given: \[\ce{E^{\circ}_{Mg^{2+}/Mg}}\] = −2.36 V,

T = 25°C = 298 K,

R = 8.314 JK⁻¹,

F = 96500 coulombs mol⁻¹,

n = 2 (Since Mg²⁺ involves 2 electrons)

Mg²⁺ = 0.1 M

Apply values in the Nernst equation:

\[\text{E = E}^{\circ} - \frac{\text{RT}}{\text{nF}} \text{ln} \frac{1}{[\text{Mg}^{2+}]}\]

\[\text{E} = -2.36 - \frac{(8.314)(298)}{(2)(96500)} \text{ln} \left(\frac{1}{0.1}\right)\] \[\text{E} = -2.36 - \frac{2477.6}{193000} \text{ln}(10)\]

E = −2.36 − (0.01283)(2.3026)

E = −2.36 − 0.02956

E = −2.3896 V