Question

If the standard reduction potential is \( E^\circ \left( \text{Mg}^{2+}_{\text{(aq)}} \mid \text{Mg}_{\text{(s)}} \right) = -2.37 \, \text{V} \), then what is the electrode potential for the following reaction at \( 298 \, \text{K} \)? 
\[ \text{Mg}_{\text{(s)}} \longrightarrow \text{Mg}^{2+}_{\text{(aq)}}(0.01\,\text{M}) + 2e^- \]

• A. +2.3108 V
• B. -2.3108 V
• C. +2.4292 V
• D. -2.4292 V

Hint:

The reaction is an oxidation, so use the oxidation potential and apply the Nernst equation with the given ion concentration.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For the oxidation \(\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-\), the standard oxidation potential is \(+2.37\) V (the reverse of the reduction potential).

Step 2: Nernst Equation
\(E = E^{\circ} - \dfrac{0.0591}{n}\log\left[\text{Mg}^{2+}\right]\) with \(n = 2\).

Step 3: Calculation
\(E = 2.37 - \dfrac{0.0591}{2}\log(10^{-2}) = 2.37 + \dfrac{0.0591}{2}\times 2 = 2.37 + 0.0591 \approx 2.4291\) V.

Step 4: Compare the options
The value is about \(+2.4292\) V, the third option.

Final Answer:
The electrode potential for the oxidation is about plus 2.4292 V. \[ \boxed{\text{(C) }+2.4292\ \text{V}} \]