Question

For the reduction of $N{O}_3^{\ominus }$ ion in an aqueous solution, $E^{\ominus }$ is +0.96 V, the values of $E^{\ominus }$ for some metal ions are given below:
$\left(i\right)V^{2+}\left(aq\right)+2e^{-}\to V;E^{\ominus }=-1.19V$
$\left(ii\right)F{e}^{3+}\left(aq\right)+3e^{-}\to Fe;E^{\ominus }=-0.04V$
$\left(iii\right)A{u}^{3+}\left(aq\right)+3e^{-}\to Au;E^{\ominus }=+1.40V$
$\left(iv\right)H{g}^{2+}\left(aq\right)+2e^{-}\to Hg;E^{\ominus }=+0.86V$

The pair(s) of metals that is/are oxidized by $N{O}_3^{\ominus }$ in aqueous solution is/are
(IIT-JEE, 2010)

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

Answer 1

Answer: (B), (C), (D)

The correct options are
B

$Hg$ and $Fe$

C

$V$ and $Hg$

D

$Fe$ and $V$

The basic idea is pretty straightforward here. You need to see if a reaction is spontaneous or not. How do you do this? Calculate the overall $E^0$ of the reaction. If it is +ve, then the reaction would be spontanoues. Easy enough?
You just need to check with every option since this is a multiple answer correct question.

$N{O}_3^{\ominus }+e^{-}\to ?;E_{red}^{\ominus }=0.96V$
Compare the standard reduction potential fo the given metals with that of $N{O}_3^{\ominus }$ reduction.
$E_{reduction}^{\ominus }$ for $N{O}_3^{\ominus }$ is greater than $E_{reduction}^{\ominus }$ of (i), (ii), and (iv).
So, $N{O}_3^{\ominus }$ will be able to oxidize Fe, Hg, and V.
So pairs are as in (b), (c) and (d).