Calculate the \( E_{Mg^{2+}/Mg} \) potential for the following half cell at 25°C :
Mg/Mg2+(1 × 10−4 M); E°Mg2+/Mg = +2.36 V
[Given : log 10 = 1]
Calculate the \( E_{Mg^{2+}/Mg} \) potential for the following half cell at 25°C :
Mg/Mg2+(1 × 10−4 M); E°Mg2+/Mg = +2.36 V
[Given : log 10 = 1]
Solution and Explanation
Calculation of Electrode Potential using the Nernst Equation
We use the Nernst equation:
\[ E = E^\circ - \frac{0.0592}{n} \log Q \]
Given:
- \(E^\circ = 2.36\ \text{V}\)
- \(n = 2\) (since 2 electrons are involved in the half-cell reaction)
- \([{Mg^{2+}}] = 1 \times 10^{-4}\ \text{M} \Rightarrow Q = \frac{1}{[{Mg^{2+}}]} = 10^4\)
Substitute values into the Nernst equation:
\[ E = 2.36 - \frac{0.0592}{2} \log (10^4) \]
\[ E = 2.36 - \frac{0.0592}{2} \times 4 \]
\[ E = 2.36 - 0.1184 = 2.2416\ \text{V} \]
Final Answer: The electrode potential is 2.2416 V.
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