Electrochemistry - Result Question 78
####26. For the reduction of $NO _3^{-}$ion in an aqueous solution $E^{\circ}$ is $+0.96 V$. Values of $E^{\circ}$ for some metal ions are given below
$$ \begin{aligned} V^{2+}(a q)+2 e^{-} \longrightarrow V ; & & E^{\circ}=-1.19 V \ Fe^{3+}(a q)+3 e^{-} \longrightarrow Fe ; & & E^{\circ}=-0.04 V \ Au^{3+}(a q)+3 e^{-} \longrightarrow Au ; & & E^{\circ}=+1.40 V \ Hg^{2+}(a q)+2 e^{-} \longrightarrow Hg ; & & E^{\circ}=+0.86 V \end{aligned} $$
The pair(s) of metals that is/are oxidised by $NO _3^{-}$in aqueous solution is (are)
(2009)
(a) $V$ and $Hg$
(b) $Hg$ and $Fe$
(c) $Fe$ and $Au$
(d) Fe and V
Numerical Value Based Question
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Answer:
Correct Answer: 26. $\left(8 \times 10^{-5} M\right) 3$
Solution:
- Metals with $E^{\circ}$ value less than $0.96 V$ will be able to reduce $NO _3^{-}$ in aqueous solution. Therefore, metals $V\left(E^{\circ}=-1.19 V\right)$, $Fe\left(E^{\circ}=-0.04 V\right), Hg\left(E^{\circ}=0.86 V\right)$ will all reduce $NO _3^{-}$but $Au$ $\left(E^{\circ}=1.40 V\right)$ cannot reduce $NO _3^{-}$in aqueous solution.
