JEE Main 2019 — Chemistry Physical Chemistry

2019mcqhard

In the cell, $\mathrm{Pt}(s)|{H}_{2}(g, 1\mathrm{bar})| \mathrm{HCl} (\mathrm{aq})|\mathrm{AgCl}(s)|\mathrm{Ag}(s)|\mathrm{Pt}(s),$ the cell potential is $0.92 V$ when a ${10}^{-6}$ molar $\mathrm{HCl}$ solution is used. The standard electrode potential of $\mathrm{Ag}|\mathrm{AgCl}|{\mathrm{Cl}}^{-}$ electrode is:

(Given, $\frac{2.303\mathrm{RT}}{F}=0.06 V$ at $298 K$ )

Official previous-year question

Held on 10 Jan 2019 · Verified 6 Jul 2026.

Options

A
$0.76 V$
B
$0.20 V$
C
$0.40 V$
D
$0.94 V$

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Solution

$A:\frac{1}{2}{H}_{2}(g)\rightarrow {H}^{+} (\mathrm{aq}.)+{e}^{-};{E}^{o}=0.0 V$

$C :\mathrm{AgCl} (s)+{e}^{-}\rightarrow \mathrm{Ag} (s)+{\mathrm{Cl}}^{-};{E}^{o}=\mathrm{xV}$

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{o}-\frac{0.06}{1}\mathrm{log}⁡{[{H}^{+}][{\mathrm{Cl}}^{-}]}$

$0.92=x-\frac{0.06}{1}\mathrm{log}⁡({10}^{-12})$

$0.92=x+0.72$

$x=0.92-0.72$

$=0.2V$

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Solution

$A:\frac{1}{2}{H}_{2}(g)\rightarrow {H}^{+} (\mathrm{aq}.)+{e}^{-};{E}^{o}=0.0 V$

$C :\mathrm{AgCl} (s)+{e}^{-}\rightarrow \mathrm{Ag} (s)+{\mathrm{Cl}}^{-};{E}^{o}=\mathrm{xV}$

${E}_{\mathrm{cell}}={E}_{\mathrm{cell}}^{o}-\frac{0.06}{1}\mathrm{log}⁡{[{H}^{+}][{\mathrm{Cl}}^{-}]}$

$0.92=x-\frac{0.06}{1}\mathrm{log}⁡({10}^{-12})$

$0.92=x+0.72$

$x=0.92-0.72$

$=0.2V$

JEE Main 2019 — Chemistry Physical Chemistry

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JEE Main 2019 — Chemistry Physical Chemistry

Options

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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