Question

For the reduction of $N{O}_3^{-}$ ion in an aqueous solution $E^0$ is + 0.96 V. Values of $E^0$ for some metal ions are given below
$V^{2+}\left(aq\right)+2e^{-}\to V;E^0=-1.19VF{e}^{3+}\left(aq\right)+3e^{-}\to Fe;E^0=-0.04VA{u}^{3+}\left(aq\right)+3e^{-}\to Au;E^0=-1.40VH{g}^{2+}\left(aq\right)+2e^{-}\to Hg;E^0=-0.86V$
The pair (s) of metals that is/ are oxidized by $N{O}_3^{-}$ in aqueous solution is (are)

• A. V and Hg
• B. Hg and Fe
• C. Fe and Au
• D. Fe and V

Answer 1

Answer: (A), (B), (D)

The correct options are
A

V and Hg

B

Hg and Fe

D

Fe and V

Metals with $E^0$ value less than 0.96 V will be able to reduce $N{O}_3^{-}$ in aqueous solution. Therefore, metals V ($E^0$= -1.19 V), Fe ($E^0$= -0.04 V), Hg ($E^0$= -0.86 V) will all reduce $N{O}_3^{-}$ but Au ($E^0$= -1.40 V) cannot reduce $N{O}_3^{-}$ in aqueous solution.