Question

For the reduction of $N{O}_3^{-}$ ion in an aqueous solution $E^{\circ }$ is
$+0.96V$. Values of $E^{\circ }$ for some metal ions are given below
$V^{2+}(aq)+2e^{-}⟶V;E^{\circ }=-1.19Vtt$
$F{e}^{3+}(aq)+3e^{-}⟶Fe;E^{\circ }=-0.04V$
$A{u}^{3+}(aq)+3e^{-}⟶Au;E^{\circ }=+1.40V$
$H{g}^{2+}(aq)+2e^{-}⟶Hg;E^{\circ }=+0.86V$
The pair(s) of metals that is/are oxidised 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)

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