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

Answer

b,c,d

Explanation

`(b,c,d)`
`NO_(3)^(c-)+e^(-)rarr" "E_(reduction )^(c-)=0.96V`
Compare the standard reduction potential of the given metals with that of `NO_(3)^(c-)` reduction.
`E_(reduction)^(c-) ` for `NO_(3)^(c-)` is greater than `E_(reduction)^(c-)` of `(i),(ii)` and `(iii).`
So, `NO_(3)^(c-)` will be able to oxidize, `Fe, Hg,` and `V`. So correct options are `(b),(c)` and `(d).`