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MHTCET 25 Apr 2025 Shift 1 Paper

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Question : 94 of 150

Marks: +1, -0

If

E^{\circ}\left(\mathrm{Zn}_{(\mathrm{aq})}^{+2} \mid \mathrm{Zn}_{(\mathrm{s})}\right)=-0.76 \mathrm{~V}

Calculate potential for

\mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{+2}(0.01 \mathrm{M})+2 e^{-}

at 298 K .

+0.8192 V
-0.8192 V
+0.7008 V
-0.7008 V

Solution:

Concept:

Use the Nernst equation for oxidation potential at

298\ \mathrm{K}

.

Explanation:

For the oxidation reaction

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

, the standard oxidation potential is

E^{\circ}_{\text{ox}} = +0.76\ \mathrm{V}

.

The Nernst equation for this process is:

E = E^{\circ}_{\text{ox}} - \frac{0.0592}{n}\log[\mathrm{Zn}^{2+}]

Here,

n = 2

and

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

.

E = 0.76 - \frac{0.0592}{2}\log(0.01)

Since

\log(0.01) = -2

,

E = 0.76 - 0.0296(-2) = 0.76 + 0.0592 = +0.8192\ \mathrm{V}

Answer:

+0.8192\ \mathrm{V}

Correct option: A.

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