Question

Write the Nernst equation of the following cell at 298 K and calculate its electromotive force.
\( \text{Mg(s)} \mid \text{Mg}^{2+}\,(0.001\,M) \parallel \text{Cu}^{2+}\,(0.0001\,M) \mid \text{Cu(s)} \)
Given: \( E^\circ_{Mg^{2+}/Mg} = -2.37\,V \) and \( E^\circ_{Cu^{2+}/Cu} = 0.34\,V \). (4)

Hint:

Mg is the anode, Cu the cathode; use \( E_{cell}=E^\circ_{cell}-\frac{0.0591}{n}\log\frac{[Mg^{2+}]}{[Cu^{2+}]} \) with \( E^\circ_{cell}=2.71\,V \) and \( n=2 \).

Answer 1

Step 1: Fix the direction of the reaction.
The half-cell with the lower (more negative) reduction potential drives oxidation. Here \(E^\circ_{Mg^{2+}/Mg} = -2.37\,V\) is far below \(E^\circ_{Cu^{2+}/Cu} = 0.34\,V\), so Mg is oxidised (anode) and \(Cu^{2+}\) is reduced (cathode). Net: \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\) with \(n = 2\).

Step 2: Emf under standard conditions.
\(E^\circ_{cell} = 0.34 + 2.37 = 2.71\,V\).

Step 3: Set up the reaction quotient.
\(Q = \dfrac{[Mg^{2+}]}{[Cu^{2+}]} = \dfrac{0.001}{0.0001} = 10\).

Step 4: Apply the Nernst form using \(2.303RT/F\).
At 298 K, \(\dfrac{2.303RT}{F} = 0.0591\,V\), hence
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q = 2.71 - \frac{0.0591}{2}\log 10 \]

Step 5: Evaluate.
\(\dfrac{0.0591}{2} = 0.02955\) and \(\log 10 = 1\), so the correction is only \(0.03\,V\).
\[ E_{cell} = 2.71 - 0.030 = 2.68\,V \]
\[ \boxed{E_{cell} \approx 2.68\,V} \]
Because the ion concentrations are dilute and their ratio is small, the emf stays very close to \(E^\circ_{cell}\).