Question

Calculate the \(E^\circ_{{Mg^{2+}/Mg}}\) potential for the following half-cell at 25°C:
Mg/Mg²⁺(1 × 10⁻⁴ M), E⁰ _Mg²⁺/Mg = +2.36 V

Answer 1

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.