Test Index

MHTCET 22 Apr 2025 Shift 1 Paper

Question : 100 of 150

Marks: +1, -0

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

\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}^{+2}(0.01 \mathrm{M})+2 e^{-}

at 298 K ?

Solution:

Concept:

Use the Nernst equation for the oxidation half-cell after reversing the given reduction potential.

Explanation:

Given reduction potential:

E^\circ(\mathrm{Mg}^{2+}_{(aq)}|\mathrm{Mg}_{(s)}) = -2.37\ \mathrm{V}

For oxidation,

\mathrm{Mg}_{(s)} \rightarrow \mathrm{Mg}^{2+}_{(aq)} + 2e^-

, the standard oxidation potential is the reverse sign:

E^\circ_{\text{ox}} = +2.37\ \mathrm{V}

Apply the Nernst equation for oxidation at 298 K:

E = E^\circ_{\text{ox}} - \frac{0.0592}{n}\log[\mathrm{Mg}^{2+}]

Here,

n = 2

and

[\mathrm{Mg}^{2+}] = 0.01\ \mathrm{M}

E = 2.37 - \frac{0.0592}{2}\log(0.01)

E = 2.37 - \frac{0.0592}{2}(-2)

E = 2.37 + 0.0592 = 2.4292\ \mathrm{V}

Thus, the potential is positive and equals

+2.4292\ \mathrm{V}

Answer:

+2.4292\ \mathrm{V}

(Option C).

Go to Question:

Prev Question

Next Question