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MHTCET 23 Apr 2025 Shift 1 Paper

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Question : 90 of 150

Marks: +1, -0

What is value of

E_{\text {cell }}

at 298 K for the reaction,

\mathrm{Zn}_{(\mathrm{s})}+\mathrm{Cu}^{+2}(0.1 \mathrm{M}) \longrightarrow \mathrm{Zn}^{+2}(0.1 \mathrm{M})+\mathrm{Cu}_{(\mathrm{s})}

E_{\text {cell }}^{\circ}=1.1 \mathrm{~V}

?

1.1 V
0.11 V
1.0408 V
0.0296 V

Solution:

Concept:

Use the Nernst equation to relate

E_{\mathrm{cell}}

to concentration and

E_{\mathrm{cell}}^{\circ}

.

Explanation:

For the cell reaction,

\mathrm{Zn}_{(\mathrm{s})} + \mathrm{Cu}^{2+}_{(0.1\mathrm{M})} \rightarrow \mathrm{Zn}^{2+}_{(0.1\mathrm{M})} + \mathrm{Cu}_{(\mathrm{s})}

, two electrons are transferred, so

n = 2

.

The Nernst equation at 298 K is:

E_{\mathrm{cell}} = E_{\mathrm{cell}}^{\circ} - \frac{0.0592}{n}\log_{10}\frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]}

Substitute

E_{\mathrm{cell}}^{\circ} = 1.1\mathrm{V}

,

[\mathrm{Zn}^{2+}] = 0.1\mathrm{M}

,

[\mathrm{Cu}^{2+}] = 0.1\mathrm{M}

, and

n = 2

:

E_{\mathrm{cell}} = 1.1 - \frac{0.0592}{2}\log_{10}\frac{0.1}{0.1} = 1.1 - 0.0296\log_{10}(1)

Since

\log_{10}(1) = 0

, we get:

E_{\mathrm{cell}} = 1.1 - 0 = 1.1\mathrm{V}

Answer:

E_{\mathrm{cell}} = 1.1\mathrm{V}

, which corresponds to option A.

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