Question

Calculate the electrode potential of \(\text{Mg}^{2+}/\text{Mg}\) electrode having \(0.01\) M concentration of \(\text{Mg}^{2+}\) (\(E^{\circ }\text{Mg}^{2+}/\text{Mg}=-2.36\) V).

• A. \(-2.42\) V
• B. \(-2.30\) V
• C. \(-2.89\) V
• D. \(1.84\) V

Hint:

Apply the Nernst equation for a two electron half cell with the ion concentration at 0.01 M.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For the half cell \(\text{Mg}^{2+}+2e^{-}\rightarrow \text{Mg}\), the Nernst equation at 298 K is \(E=E^{\circ}-\dfrac{0.0591}{n}\log\dfrac{1}{[\text{Mg}^{2+}]}\) with \(n=2\).

Step 2: Calculation
\(E=-2.36-\dfrac{0.0591}{2}\log\dfrac{1}{0.01}\)
\(\log 100=2\), so \(E=-2.36-0.0591\).

Step 3: Compare the options
\(E\approx -2.419\) V, which rounds to \(-2.42\) V, option (A).

Final Answer:
The electrode potential is about \(-2.42\) V. \[ \boxed{\text{(A) }-2.42\text{ V}} \]