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: Identify the anode and cathode.
In cell notation the electrode written on the left is the anode (oxidation) and the one on the right is the cathode (reduction). So magnesium is oxidised and copper ion is reduced.
Anode: \( \text{Mg} \rightarrow \text{Mg}^{2+} + 2e^- \)
Cathode: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
Overall: \( \text{Mg} + \text{Cu}^{2+} \rightarrow \text{Mg}^{2+} + \text{Cu} \), so the number of electrons transferred \( n = 2 \).

Step 2: Standard cell potential.
\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-2.37) = 2.71\,V \).

Step 3: Write the Nernst equation at 298 K.
For the cell reaction, only ionic concentrations appear (solids taken as 1):
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\,\log\frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \]

Step 4: Substitute the values.
\[ E_{cell} = 2.71 - \frac{0.0591}{2}\,\log\frac{0.001}{0.0001} \]
\[ E_{cell} = 2.71 - 0.02955 \times \log(10) \]

Step 5: Simplify.
Since \( \log 10 = 1 \):
\[ E_{cell} = 2.71 - 0.02955 = 2.6805\,V \]

Result:
\[ \boxed{E_{cell} \approx 2.68\,V} \]
The positive value confirms the reaction is spontaneous.