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JEE Advanced 2025 - Mathematics
Numerical Type
Question
The standard reduction potential at 298 K for the following half cells are given below: $$ \mathrm{NO}_3^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O} \quad \mathrm{E}^0=0.97 \mathrm{~V} $$ $\begin{array}{ll}\mathrm{V}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{V} & \mathrm{E}^0=-1.19 \mathrm{~V} \\ \mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathrm{E}^0=-0.04 \mathrm{~V} \\ \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) & \mathrm{E}^0=0.80 \mathrm{~V} \\ \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au}(\mathrm{s}) & \mathrm{E}^0=1.40 \mathrm{~V}\end{array}$ The number of metal(s) which will be oxidized by $\mathrm{NO}_3^{-}$in aqueous solution is $\_\_\_\_$ .
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Question Tags
JEE Main
Chemistry
Medium
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