Consider the follwing reaction in basic medium :
`NH_3 + O Cl^(-) rarr N_2 H_4 + Cl^(-)`
The coefficient fo `N_2H_4` in the balanced equation will be .

Options

A. `4`
B. `3`
C. `1`
D. `2`

Answer

C

Explanation

`overset(-3)(NH_(3))+overset(+1)(OCl^(-))rarr overset(-2)(N_(2))H_(4)+overset(-1)(Cl^(-))`
Oxidation half-reaction
`2overset(-3)(NH_(3))rarroverset(-2)(N_(2))H_(4)+2e^(-)`
Reduction half-reaction
`2e^(-)+overset(+1)(OCl^(-)) rarr overset(-1)(Cl^(-))`
Add `OH^(-)` to balance the ionic charges
`2NH_(3)+2OH^(-) rarr N_(2)H_(4)+2e^(-)`
`OCl^(-)+2e^(-) rarr Cl^(-)+2OH^(-)`
Add `H_(2)O` molecules to balance H and O atoms:
`2NH_(3)+2OH^(-) rarr N_(2)H_(4)+2H_(2)O+2e^(-)`
`OCl^(-)+H_(2)O+2e^(-) rarr Cl^(-)+2OH^(-)`
Finally add them
`2NH_(3)+OCl^(-)rarr N_(2)H_(4)+H_(2)O+Cl^(-)`