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MHTCET 22 Apr 2025 Shift 2 Paper

Question : 63 of 150

Marks: +1, -0

E^{o}\left(\mathrm{AI}_{(\mathrm{eq})}^{+3} \mid \mathrm{AI}_{(\mathrm{s})}\right)=-1.66 \mathrm{~V}

\mathrm{Al}_{(\mathrm{s})} \longrightarrow \mathrm{Al}^{+3}(0 \cdot 1 \mathrm{M})+3 e^{-}

at 298 K ?

Solution:

Concept:

For a half-cell, the Nernst equation relates electrode potential to ion concentration at a given temperature.

Explanation:

The standard oxidation potential for

\mathrm{Al} \to \mathrm{Al}^{3+} + 3e^-

+1.66\ \mathrm{V}

At 298 K, the Nernst equation for this oxidation is:

E = E^\circ_{\text{ox}} - \frac{0.0592}{n} \log_{10}[\mathrm{Al}^{3+}]

Here,

n = 3

and

[\mathrm{Al}^{3+}] = 0.1\ \mathrm{M}

Substituting the values:

E = 1.66 - \frac{0.0592}{3} \log_{10}(0.1)

Since

\log_{10}(0.1) = -1

E = 1.66 - 0.0197 \times (-1)

E = 1.66 + 0.0197 = 1.679\ \mathrm{V}

Answer:

The potential is

+1.679\ \mathrm{V}

, which matches option C.

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