JEE Main 2020 — Chemistry Physical Chemistry

2020numerical valuemedium

For an electrochemical cell $\mathrm{Sn}(s)|{\mathrm{Sn}}^{2}+(\mathrm{aq}, 1M)||{\mathrm{Pb}}^{2+}(\mathrm{aq}, 1M)|\mathrm{Pb}(s)$ the ratio $\frac{[{\mathrm{Sn}}^{2+}]}{[{\mathrm{Pb}}^{2+}]}$ when this cell attains equilibrium is _______

(Given: ${E}_{{\mathrm{Sn}}^{2+}|\mathrm{Sn}}^{0}=-0.14 V,{E}_{{\mathrm{Pb}}^{2+}|\mathrm{Pb}}^{0}=-0.13 V, \frac{2.303\mathrm{RT}}{F}=0.06$ )

Official previous-year question

Held on 8 Jan 2020 · Verified 6 Jul 2026.

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Solution

Cell reaction is:

$\text{Sn}(\text{s})+{ \text{Pb}}^{ +2} (\text{aq})\rightarrow { \text{Sn}}^{ +2} (\text{aq})+\text{Pb}(\text{s})$

Apply Nernst equation:

${\text{E}}_{ \text{cell }} ={\text{E}}_{ \text{cell }}^{0} -\frac{ 0.06}{2} \mathrm{log}\frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]} ..... (1)$

${\text{E}}_{ \text{cell }}^{0} =-0.13+0.14=0.01\text{V}$

At equilibrium : ${\text{E}}_{\text{cell }}^{}=0$ Substituting in (1)

$0=0.01-\frac{ 0.06}{2} \mathrm{log}\frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]}$

$\Rightarrow \frac{1}{3} =\mathrm{log}\frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]}$

$\Rightarrow \frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]} =2.15$

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JEE Main 2020 — Chemistry Physical Chemistry

2020numerical valuemedium

(Given: ${E}_{{\mathrm{Sn}}^{2+}|\mathrm{Sn}}^{0}=-0.14 V,{E}_{{\mathrm{Pb}}^{2+}|\mathrm{Pb}}^{0}=-0.13 V, \frac{2.303\mathrm{RT}}{F}=0.06$ )

Official previous-year question

Held on 8 Jan 2020 · Verified 6 Jul 2026.

Did you get this right?

Solution

Cell reaction is:

$\text{Sn}(\text{s})+{ \text{Pb}}^{ +2} (\text{aq})\rightarrow { \text{Sn}}^{ +2} (\text{aq})+\text{Pb}(\text{s})$

Apply Nernst equation:

${\text{E}}_{ \text{cell }} ={\text{E}}_{ \text{cell }}^{0} -\frac{ 0.06}{2} \mathrm{log}\frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]} ..... (1)$

${\text{E}}_{ \text{cell }}^{0} =-0.13+0.14=0.01\text{V}$

At equilibrium : ${\text{E}}_{\text{cell }}^{}=0$ Substituting in (1)

$0=0.01-\frac{ 0.06}{2} \mathrm{log}\frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]}$

$\Rightarrow \frac{1}{3} =\mathrm{log}\frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]}$

$\Rightarrow \frac{ [ { \text{Sn}}^{ +2} ]}{ [ { \text{Pb}}^{ +2} ]} =2.15$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

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JEE Main 2020 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2020 — Chemistry Physical Chemistry

Did you get this right?

Answer

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

Keep practicing Physical Chemistry