The standard reduction potentials at $295K$ for the following half cells are given below: ${{\mathrm{NO}}_{3}}^{-}+4{H}^{+}+3{e}^{-}\rightarrow \mathrm{NO}(g)+2{H}_{2}O$ ${E}^{o}=0.97V$ ${V}^{2+}(\mathrm{aq})+2{e}^{-}\rightarrow V(s)$ ${E}^{o}=-1.19V$ ${\mathrm{Fe}}^{3+}(\mathrm{aq})+3{e}^{-}\rightarrow \mathrm{Fe}(s)$ ${E}^{o}=-0.04V$ ${\mathrm{Ag}}^{+}(\mathrm{aq})+{e}^{-}\rightarrow \mathrm{Ag}(s)$ ${E}^{o}=0.80V$ ${\mathrm{Au}}^{3+}(\mathrm{aq})+3{e}^{-}\rightarrow \mathrm{Au}(s)$ ${E}^{o}=1.40V$ The number of metal(s) which will be oxidised by ${{\mathrm{NO}}_{3}}^{-}$ in aqueous solution is _____.

JEE Main 2023 — Chemistry Physical Chemistry

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The standard reduction potentials at $295K$ for the following half cells are given below:

${{\mathrm{NO}}_{3}}^{-}+4{H}^{+}+3{e}^{-}\rightarrow \mathrm{NO}(g)+2{H}_{2}O$	${E}^{o}=0.97V$
${V}^{2+}(\mathrm{aq})+2{e}^{-}\rightarrow V(s)$	${E}^{o}=-1.19V$
${\mathrm{Fe}}^{3+}(\mathrm{aq})+3{e}^{-}\rightarrow \mathrm{Fe}(s)$	${E}^{o}=-0.04V$
${\mathrm{Ag}}^{+}(\mathrm{aq})+{e}^{-}\rightarrow \mathrm{Ag}(s)$	${E}^{o}=0.80V$
${\mathrm{Au}}^{3+}(\mathrm{aq})+3{e}^{-}\rightarrow \mathrm{Au}(s)$	${E}^{o}=1.40V$

The number of metal(s) which will be oxidised by

{{\mathrm{NO}}_{3}}^{-}

in aqueous solution is _____.

Official previous-year question

Held on 6 Apr 2023 · Verified 6 Jul 2026.

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Solution

For feasibility, check , ${{E}^{o}}_{\mathrm{cell}}={E}_{\mathrm{cathode}(\mathrm{reduction})}^{o}-{E}_{\mathrm{anode}(\mathrm{oxidation})}^{o}>0$

The ${E}^{o}$ values when metal acts as anode and ${\mathrm{NO}}_{3}^{-}$ reaction is cathodic reaction.

For Vanadium metal, ${E}_{\mathrm{cell}}^{o}=0.97+1.19=2.16V$

For Iron metal, ${E}_{\mathrm{cell}}^{o}=0.97+0.04=1.01V$

For silver metal, ${E}_{\mathrm{cell}}^{o}=0.97-0.80=0.17V$

For Gold metal, ${E}_{\mathrm{cell}}^{o}=0.97-1.140=-0.17V$

For electrodes having oxidation potential greater than $-0.97V$ , ${{E}^{o}}_{\mathrm{cell}}>0$

$\therefore \mathrm{Ag},\mathrm{Fe}$ and $V$ can be oxidised

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Solution

For feasibility, check , ${{E}^{o}}_{\mathrm{cell}}={E}_{\mathrm{cathode}(\mathrm{reduction})}^{o}-{E}_{\mathrm{anode}(\mathrm{oxidation})}^{o}>0$

The ${E}^{o}$ values when metal acts as anode and ${\mathrm{NO}}_{3}^{-}$ reaction is cathodic reaction.

For Vanadium metal, ${E}_{\mathrm{cell}}^{o}=0.97+1.19=2.16V$

For Iron metal, ${E}_{\mathrm{cell}}^{o}=0.97+0.04=1.01V$

For silver metal, ${E}_{\mathrm{cell}}^{o}=0.97-0.80=0.17V$

For Gold metal, ${E}_{\mathrm{cell}}^{o}=0.97-1.140=-0.17V$

For electrodes having oxidation potential greater than $-0.97V$ , ${{E}^{o}}_{\mathrm{cell}}>0$

$\therefore \mathrm{Ag},\mathrm{Fe}$ and $V$ can be oxidised

JEE Main 2023 — Chemistry Physical Chemistry

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JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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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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