In acidic medium .for the reduction of each `NO_(3)^(-)` ion in the given reaction ,how many electrons will be required `NO_(3)^(-)toNH_(2)OH`

Explanation

`NO_(3)^(-)toNH_(2)OH`: For equalising th e number of O-atoms on both sides ,two `H_(2)O` molecules are added to the right side (heaving lesser number of O -atoms ) and two `H^(+)1` ions are added to the left side for each molecule of `H_(2)O` added
`NO_(3)^(-)+4H^(+)toNH_(2)OH+2H_(2)O`
For equalising the number of H- atoms on both sides , three additonal `H^(+)` ions are required on the left side .
So we get `NO_(3)^(-) +7H^(+)toNH_(2)OH+2H_(2)O`
To balance the change on both sides ,6 electrons are added to the left side of the equation .
`NO_(3) ^(-) 7H^(+)+6eto NH_(2)OH+2H_(2)O`
Hence for the reduction of each `NO_(3)^(-)` ion into `NH_(2)OH` molecule 6 electrons are required .