Electrochemistry 1 Question 26
26. For the reduction of $\mathrm{NO} _{3}^{-}$ion in an aqueous solution $E^{\circ}$ is $+0.96 \mathrm{V}$. Values of $E^{\circ}$ for some metal ions are given below
$ \begin{aligned} \mathrm{V}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{V} ; & & E^{\circ}=-1.19 \mathrm{V} \\ \mathrm{Fe}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Fe} ; & & E^{\circ}=-0.04 \mathrm{V} \\ \mathrm{Au}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Au} ; & & E^{\circ}=+1.40 \mathrm{V} \\ \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg} ; & & E^{\circ}=+0.86 \mathrm{V} \end{aligned} $
The pair(s) of metals that is/are oxidised by $\mathrm{NO} _{3}^{-}$in aqueous solution is (are)
(2009)
(a) $\mathrm{V}$ and $\mathrm{Hg}$
(b) $\mathrm{Hg}$ and $\mathrm{Fe}$
(c) $\mathrm{Fe}$ and $\mathrm{Au}$
(d) Fe and V
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Answer:
Correct Answer: 26. (a,b,d)
Solution:
- Metals with $E^{\circ}$ value less than $0.96 \mathrm{V}$ will be able to reduce $\mathrm{NO} _{3}^{-}$ in aqueous solution. Therefore, metals $\mathrm{V}\left(E^{\circ}=-1.19 \mathrm{V}\right)$, $\mathrm{Fe}\left(E^{\circ}=-0.04 \mathrm{V}\right), \mathrm{Hg}\left(E^{\circ}=0.86 \mathrm{V}\right)$ will all reduce $\mathrm{NO} _{3}^{-}$but $\mathrm{Au}$ $\left(E^{\circ}=1.40 \mathrm{V}\right)$ cannot reduce $\mathrm{NO} _{3}^{-}$in aqueous solution.
