JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

The equilibrium constant for the reaction $\mathrm{Zn}(s)+{\mathrm{Sn}}^{2+}(\mathrm{aq})\rightleftharpoons {\mathrm{Zn}}^{2+}(\mathrm{aq})+\mathrm{Sn}(s)$ is $1\times {10}^{20}$ at $298K$ . The magnitude of standard electrode potential of $\mathrm{Sn}/{\mathrm{Sn}}^{2+}$ if ${E}_{{\mathrm{Zn}}^{2*}/\mathrm{Zn}}^{0}=-0.76V$ is _____ $\times {10}^{-2}V$ . (Nearest integer)

Given : $\frac{2.303\mathrm{RT}}{F}=0.059V$

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

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Solution

$\mathrm{Zn}(s)+{\mathrm{Sn}}^{2+}(\mathrm{aq})\rightleftharpoons {\mathrm{Zn}}^{2+}(\mathrm{aq})+\mathrm{Sn}(s)$

Given: $\frac{2.303\mathrm{RT}}{F}=0.059V$

${E}_{\mathrm{cell}}^{0}=\frac{2.303\mathrm{RT}}{2F}\mathrm{log}{K}_{c}$

$\Rightarrow {E}_{\mathrm{cell}}^{0}=\frac{0.059}{2}\mathrm{log}({10}^{20})$

$\Rightarrow {E}_{\mathrm{cell}}^{0}=\frac{0.059}{2}\times 20$

$\Rightarrow {E}_{\mathrm{cell}}^{0}=0.59V$

${E}_{\mathrm{cell}}^{0}={E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{0}-{E}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{0}$

$\Rightarrow {E}_{\mathrm{cell}}^{0}={E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{0}-(-0.76)$

$\Rightarrow {E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{0}=-0.76+059=-0.17$

$\Rightarrow {E}_{\mathrm{Sn}/{\mathrm{Sn}}^{2+}}^{0}=0.17=17\times {10}^{-2}$

So, answer is 17.

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JEE Main 2023 — Chemistry Physical Chemistry

2023integerhard

Given : $\frac{2.303\mathrm{RT}}{F}=0.059V$

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\mathrm{Zn}(s)+{\mathrm{Sn}}^{2+}(\mathrm{aq})\rightleftharpoons {\mathrm{Zn}}^{2+}(\mathrm{aq})+\mathrm{Sn}(s)$

Given: $\frac{2.303\mathrm{RT}}{F}=0.059V$

${E}_{\mathrm{cell}}^{0}=\frac{2.303\mathrm{RT}}{2F}\mathrm{log}{K}_{c}$

$\Rightarrow {E}_{\mathrm{cell}}^{0}=\frac{0.059}{2}\mathrm{log}({10}^{20})$

$\Rightarrow {E}_{\mathrm{cell}}^{0}=\frac{0.059}{2}\times 20$

$\Rightarrow {E}_{\mathrm{cell}}^{0}=0.59V$

${E}_{\mathrm{cell}}^{0}={E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{0}-{E}_{{\mathrm{Zn}}^{2+}/\mathrm{Zn}}^{0}$

$\Rightarrow {E}_{\mathrm{cell}}^{0}={E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{0}-(-0.76)$

$\Rightarrow {E}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^{0}=-0.76+059=-0.17$

$\Rightarrow {E}_{\mathrm{Sn}/{\mathrm{Sn}}^{2+}}^{0}=0.17=17\times {10}^{-2}$

So, answer is 17.

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JEE Main 2023 — Chemistry Physical Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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