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Electrochemistry question

2023 · 6 Apr · Shift 2 · Q17
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Electrochemistry question

2023 · 6 Apr · Shift 2 · Q17

JEE MainChemistryElectrochemistryNumerical+4 / −1
The standard reduction potentials at 298 K298 \mathrm{~K}298 K for the following half cells are given below: NO3−+4H++3e−→NO(g)+2H2OEθ=0.97 VV2+(aq)+2e−→VEθ=−1.19 VFe3+(aq)+3e−→FeEθ=−0.04 VAg+(aq)+e−→Ag(s)Eθ=0.80 VAu3+(aq)+3e−→Au(s)Eθ=1.40 V\mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O} \quad \mathrm{E}^{\theta}=0.97 \mathrm{~V}\mathrm{V}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{V} \quad\quad\quad \mathrm{E}^{\theta}=-1.19 \mathrm{~V}\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad\quad\quad \mathrm{E}^{\theta}=-0.04 \mathrm{~V}\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \quad\quad\quad \mathrm{E}^{\theta}=0.80 \mathrm{~V}\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}(\mathrm{s}) \quad\quad\quad \mathrm{E}^{\theta}=1.40 \mathrm{~V}NO3−​+4H++3e−→NO(g)+2H2​OEθ=0.97 VV2+(aq)+2e−→VEθ=−1.19 VFe3+(aq)+3e−→FeEθ=−0.04 VAg+(aq)+e−→Ag(s)Eθ=0.80 VAu3+(aq)+3e−→Au(s)Eθ=1.40 V The number of metal(s) which will be oxidized by NO3−\mathrm{NO}_{3}^{-}NO3−​ in aqueous solution is ‾\underline{\hspace{2cm}}​.
Numerical answer
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