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Question

For the reduction of NO3 ion in an aqueous solution, E is +0.96 V, the values of E for some metal ions are given below:
(i)V2+(aq)+2eV; E=1.19V
(ii)Fe3+(aq)+3eFe; E=0.04V
(iii)Au3+(aq)+3eAu; E=+1.40V
(iv)Hg2+(aq)+2eHg; E=+0.86V

The pair(s) of metals that is/are oxidized by NO3 in aqueous solution is/are
(IIT-JEE, 2010)


A

Fe and Au

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B

Hg and Fe

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C

V and Hg

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D

Fe and V

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Solution

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 E0 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.

NO3+e?;Ered=0.96V
Compare the standard reduction potential fo the given metals with that of NO3 reduction.
Ereduction for NO3 is greater than Ereduction of (i), (ii), and (iv).
So, NO3 will be able to oxidize Fe, Hg, and V.
So pairs are as in (b), (c) and (d).


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