In the resction `2NH_(4)^(+)+6NO_(3)^(-)(aq)+4H^(+)(aq)rarr 6NO_(2)(g)+N_(2)(g)+6H_(2)O` the reducing agent is

Options

A. `NH_(4)^(+)`
B. `NO_(3)^(-)`
C. `H^(+)(aq)`
D. `NO_(2)`

Answer

1

Explanation

`2NH_(4)+6NO_(3)^(-)(aq)+4H^(+)(aq)rarr 6NO_(2)(g)+6H_(2)O`
Oxidation No. of `N` in `NH_(4)^(+)=-3`
Oxidation No. of `N` in `NO_(3)^(-)=+4`
Oxidation No. of `N` in `N_(2)=0`
So, reducing agent ( which oxidese ) `=NH_(4)^(+)` from `-3` to `0 (N_(2))` and `+4(NO_(2))`.