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For the reduction of NO(3)^(c-) ion in a...

For the reduction of `NO_(3)^(c-)` ion in an aqueous solution, `E^(c-)` is `+0.96V`, the values of `E^(c-)` for some metal ions are given below `:`
`i.V^(2+)(aq)+2e^(-)rarr V, " "E^(c-)=-1.19V`
`ii. Fe^(3+)(aq)+3e^(-) rarr Fe, " "E^(c-)=-0.04V`
`iii. Au^(3+)(aq)+3e^(-) rarr Au, " "E^(c-)=+140V`
`iv. Hg^(2+)(aq)+2e^(-) rarr Hg, " "E^(c-)=+0.86V`
The pair`(s)` of metals that is `//` are oxidized by `NO_(3)^(c-)` in aqueous solution is `//` are

A

`Fe` and `Au`

B

`Hg` and `Fe`

C

`V` and `Hg`

D

`Fe` and `V`

Text Solution

Verified by Experts

The correct Answer is:
b,c,d
`NO_(3)^(c-)+e^(-) rarr ?, E^(c-)._(red)=0.96V`
Compare the standard reduction potential of the given metals with that of `NO_(3)^(c-)` reduction.
`E^(c-)._(reduction)` for `NO_(3)^(c-)` is greater than `E^(c-)._(reduction)` of `(i),(ii),` and `(iv).`
So, `NO_(3)^(c-)` will be able to oxidize `Fe, Hg` and `V.`
So pairs are as in `(b),(c),` and `(d)`.
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Knowledge Check

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