Balance the following equation in basic medium by ion-electron method and oxidation number methods & Identify the oxidising agent of & the reducing agent.
`N_(2)H_(4)(l)+CIO_(3)^(-)(aq)toNO(g)+CI^(-)(g)`

Explanation

Ion -electron :
Oxidation reaction
`N_(2)H_(4)(l)to2NO(g)`
Balancing N- atoms on both sides ,we have
`N_(2)H_(4)(l)to2NO(g)`
To balance H and O -atoms we ,add `8OH^(-)` to the left hand side and `6H_(2)O` to the right hand side
`N_(2)H_(4)(l)+8OH^(-)(aq)to2NO(g)+6H_(2)O(l)`
To balance the change ,we add 8 electron to the right side
`N_(2)H_(4)(l)+8OH^(-)(aq)to2NO(g)+6H_(2)O(l)+8e`
Reduction reaction : `CIO_(3)^(-)(aq)toCI^(-)(aq)`
Balancing O-atoms and charge on both sides we have
`3N_(2)H_(4)(l)+4CIO_(3)^(-)(aq)to6NO(g)+4CI^(-)(aq)+6H_(2)O(l)`
This is the balanced equation for the reaction.
Oxidation number method :
`overset(-2)N_(2)H_(4)(l)+overset(=5)CIO_(3)^(-)(aq)overset(+2)NO(g)+overset(-1)CI^(-)(aq)`
In the reacation ,total oxidation number for two N- atoms increases by 2xx4=8 unit and total oxidation number of CI -atom decreases by 6 unit
So ,the ratio of increase to decrease in oxidation number =8:6=4
To balance the increase and decrease in oxidation number ,we multiply `N_(2)H_(4)by2and CIO_(3)^(-)by4.`
`3N_(2)H_(4)(l)+4CIO_(3)^(-)(aq)to6NO(g)+4CI^(-)(aq)`
To balance H and O-atoms on both sides ,we add `6H_(2)O`to the right hand side.
`3N_(2)H_(4)(l)+4CIO_(3)^(-)(aq)to6NO(g)+4CI^(-)(Aq)+6H_(2)O(l)`
This is the balanced equation for the reaction.