Question

If \(\mathrm{E}^*\left(\mathrm{Mg}^{+2}{ }_{(\mathrm{aq})} \mid \mathrm{Mg}_{(\mathrm{s})}\right)=-2.37 \mathrm{~V}\).

What is potential for \(\mathrm{Mg}_{(\mathrm{s})} \rightarrow \mathrm{Mg}^{+2}(0.1 \mathrm{M})+2 \mathrm{e^-}\) at 298 K ?

• A. +2.3996 V
• B. -2.3996 V
• C. +2.3404 V
• D. -2.3404 V

Hint:

The oxidation potential is the negative of the reduction potential, then apply the Nernst equation for the given concentration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The reaction \(\text{Mg}\rightarrow\text{Mg}^{2+}+2e^-\) is an oxidation, so its standard potential is the reverse of the reduction value: \(E^0_{\text{ox}}=+2.37\text{ V}\).

Step 2: Nernst Equation
\(E=E^0-\dfrac{0.0591}{n}\log[\text{Mg}^{2+}]\) with \(n=2\).

Step 3: Calculation
\(E=2.37-\dfrac{0.0591}{2}\log(0.1)=2.37+0.02955=2.39955\approx+2.3996\text{ V}\)

Final Answer:
The potential is +2.3996 V. \[ \boxed{\text{(A) }+2.3996\text{ V}} \]