Question

Calculate the emf and ΔG for the cell reaction at 298 K:

Mg_(s) | Mg²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu_(s)

Given: \[\ce{E^{\circ}_{cell}}\] = 2.71 V

1 F = 96500 C

Answer 1

Given: The given cell reaction is:

Mg_(s) | Mg²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu_(s)

Using the Nernst equation:

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

Where:

\[\ce{E^{\circ}_{cell}}\] = +2.71 V

n = 2 (number of electrons transferred)

Mg²⁺ = 0.1 M

Cu²⁺ = 0.01 M

E_cell = \[\ce{2.71 - \frac{0.0591}{2} log \frac{0.1}{0.01}}\]

= 2.71 − 0.02955 log (10)

= 2.71 − 0.02955 × 1

= 2.71 − 0.02955

= 2.680 V