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Redox Reactions and Electrochemistry: Chemistry | JEE Main

The standard reduction potential at

298\mathrm{~K}

for the following half cells are given below :-

\mathrm{NO}_{3}^{-}+4\mathrm{H}^{+}+3\mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{g})+2\mathrm{H}_{2}\mathrm{O} \quad \mathrm{E}^{0}=0.97\mathrm{V}

\mathrm{V}^{2+}(\mathrm{aq})+2\mathrm{e}^{-} \rightarrow \mathrm{V} \quad \mathrm{E}^{0}=-1.19\mathrm{V}

\mathrm{Fe}^{3+}(\mathrm{aq})+3\mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathrm{E}^{0}=-0.04\mathrm{V}

\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \quad \mathrm{E}^{0}=0.80\mathrm{V}

\mathrm{Au}^{3+}(\mathrm{aq})+3\mathrm{e}^{-} \rightarrow \mathrm{Au}(\mathrm{s}) \quad \mathrm{E}^{0}=1.40\mathrm{V}

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

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

in aqueous solution is

\dots\dots\dots\dots

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Step-by-step solutionView

Correct answer

Three metals (V, Fe, and Ag) have a standard reduction potential lower than that of the nitrate half-cell ( $0.97\text{ V}$ ) and will be oxidized by $\mathrm{NO}_3^-$ , so the answer is 3.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Reduction potentials are: $E^\circ(\mathrm{NO}_3^-/\mathrm{NO}) = +0.97\text{ V}$ $E^\circ(\mathrm{V}^{2+}/\mathrm{V}) = -1.19\text{ V}$ $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}) = -0.04\text{ V}$ $E^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\text{ V}$ $E^\circ(\mathrm{Au}^{3+}/\mathrm{Au}) = +1.40\text{ V}$

02find

Identify the number of metals (among $\mathrm{V}, \mathrm{Fe}, \mathrm{Ag}, \mathrm{Au}$ ) that can be oxidized by $\mathrm{NO}_3^-$ in aqueous medium.

03strategise

For $\mathrm{NO}_3^-$ to oxidize a metal $\mathrm{M}$ to its cation, the cell potential $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ(\mathrm{NO}_3^-/\mathrm{NO}) - E^\circ(\mathrm{M}^{n+}/\mathrm{M})$ must be positive ( $E^\circ_{\text{cell}} > 0$ ). Therefore, $E^\circ(\mathrm{M}^{n+}/\mathrm{M}) < E^\circ(\mathrm{NO}_3^-/\mathrm{NO}) = +0.97\text{ V}$ .

04execute

Compare each metal's reduction potential with $0.97\text{ V}$ : - For $\mathrm{V}$ : $-1.19\text{ V} < 0.97\text{ V}$ (Oxidized, $E^\circ_{\text{cell}} = 0.97 - (-1.19) = +2.16\text{ V} > 0$ ) - For $\mathrm{Fe}$ : $-0.04\text{ V} < 0.97\text{ V}$ (Oxidized, $E^\circ_{\text{cell}} = 0.97 - (-0.04) = +1.01\text{ V} > 0$ ) - For $\mathrm{Ag}$ : $+0.80\text{ V} < 0.97\text{ V}$ (Oxidized, $E^\circ_{\text{cell}} = 0.97 - 0.80 = +0.17\text{ V} > 0$ ) - For $\mathrm{Au}$ : $+1.40\text{ V} > 0.97\text{ V}$ (Not oxidized, $E^\circ_{\text{cell}} = 0.97 - 1.40 = -0.43\text{ V} < 0$ ) Total count = 3 ( $\mathrm{V}, \mathrm{Fe}, \mathrm{Ag}$ ).

✓verify

Only Au has a higher reduction potential than nitrate, which matches the chemical intuition that gold requires aqua regia to dissolve.

Hints that build this answer step by step

Under standard conditions, what criterion must a metal half-cell $\mathrm{M}^{n+}/\mathrm{M}$ satisfy to be spontaneously oxidized by the nitrate half-cell?
$E^0(\mathrm{M}^{n+}/\mathrm{M}) < E^0(\mathrm{NO}_3^-/\mathrm{NO}) = 0.97\text{ V}$
Comparing the given reduction potentials to $0.97\text{ V}$ , which metals satisfy $E^0 < 0.97\text{ V}$ ?
V ( $-1.19\text{ V}$ ), Fe ( $-0.04\text{ V}$ ), and Ag ( $0.80\text{ V}$ )
What is the total number of metals that will be oxidized by $\mathrm{NO}_3^-$ ?
3

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why must the SRP of the metal be less than that of nitrate?

Nitrate acts as the oxidizing agent and is itself reduced (acting as cathode). The metal is oxidized (acting as anode). For a spontaneous reaction, $E^\circ_{\text{cell}} = E^\circ_{\text{red}}(\text{cathode}) - E^\circ_{\text{red}}(\text{anode}) > 0$ , requiring $E^\circ_{\text{red}}(\text{metal}) < E^\circ_{\text{red}}(\text{nitrate})$ .

Answer

Three metals (V, Fe, and Ag) have a standard reduction potential lower than that of the nitrate half-cell ( $0.97\text{ V}$ ) and will be oxidized by $\mathrm{NO}_3^-$ , so the answer is 3.

Why each option works or fails

Step-by-step solution

given: Reduction potentials are: $E^\circ(\mathrm{NO}_3^-/\mathrm{NO}) = +0.97\text{ V}$ $E^\circ(\mathrm{V}^{2+}/\mathrm{V}) = -1.19\text{ V}$ $E^\circ(\mathrm{Fe}^{3+}/\mathrm{Fe}) = -0.04\text{ V}$ $E^\circ(\mathrm{Ag}^+/\mathrm{Ag}) = +0.80\text{ V}$ $E^\circ(\mathrm{Au}^{3+}/\mathrm{Au}) = +1.40\text{ V}$
find: Identify the number of metals (among $\mathrm{V}, \mathrm{Fe}, \mathrm{Ag}, \mathrm{Au}$ ) that can be oxidized by $\mathrm{NO}_3^-$ in aqueous medium.
strategise: For $\mathrm{NO}_3^-$ to oxidize a metal $\mathrm{M}$ to its cation, the cell potential $E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = E^\circ(\mathrm{NO}_3^-/\mathrm{NO}) - E^\circ(\mathrm{M}^{n+}/\mathrm{M})$ must be positive ( $E^\circ_{\text{cell}} > 0$ ). Therefore, $E^\circ(\mathrm{M}^{n+}/\mathrm{M}) < E^\circ(\mathrm{NO}_3^-/\mathrm{NO}) = +0.97\text{ V}$ .
execute: Compare each metal's reduction potential with $0.97\text{ V}$ : - For $\mathrm{V}$ : $-1.19\text{ V} < 0.97\text{ V}$ (Oxidized, $E^\circ_{\text{cell}} = 0.97 - (-1.19) = +2.16\text{ V} > 0$ ) - For $\mathrm{Fe}$ : $-0.04\text{ V} < 0.97\text{ V}$ (Oxidized, $E^\circ_{\text{cell}} = 0.97 - (-0.04) = +1.01\text{ V} > 0$ ) - For $\mathrm{Ag}$ : $+0.80\text{ V} < 0.97\text{ V}$ (Oxidized, $E^\circ_{\text{cell}} = 0.97 - 0.80 = +0.17\text{ V} > 0$ ) - For $\mathrm{Au}$ : $+1.40\text{ V} > 0.97\text{ V}$ (Not oxidized, $E^\circ_{\text{cell}} = 0.97 - 1.40 = -0.43\text{ V} < 0$ ) Total count = 3 ( $\mathrm{V}, \mathrm{Fe}, \mathrm{Ag}$ ).
verify: Only Au has a higher reduction potential than nitrate, which matches the chemical intuition that gold requires aqua regia to dissolve.