Balance the following by ion electron method is basic medium.
`NO_(3)^(ө)+Znrarr Zn^(2+)+NH_(4)^(oplus)`.

Explanation

(i) `Zn+NO_(3)^(-)rarrZn^(2+)+overset(+)(NH_(4))` Skeleton equation
Step I. (a) `overset(0)(Zn)+overset(+5-2)((NO_(3)))rarroverset(+2)((Zn)^(2+))+overset(-3+1)((NH_(4))^(+))`
(b) `overset(0)(Zn)+overset(+5)(NO_(3)^(-))rarr overset(+2)((Zn)^(2+))+overset(-3)((NH_(4))^(+))`
(c)
Step II Oxidation half reaction : `Zn rarr Zn^(+2)`
Since the increase in oxidation number is 2, add 2 electrons to the product side to balance the O.N
`Zn rarr Zn^(2+)+2e^(-)`
Step III. Reduction half reaction : `NO_(3)^(-)rarr NH_(4)^(+)`
Since the decrease in oxidation number is 8, add 8 electrons on the reactant side
`NO_(3)^(-)+8e^(-)rarrNH_(4)^(+)`
To balance O-atoms add `3H_(2)O` to R.H.S
`NO_(3)^(-)+8e^(-)rarrNH_(4)^(+)+3H_(2)O`
To balance H-atoms add `10H_(2)O` to L.H.S and `10 OH^(-)` to R.H.S
`NO_(3)^(-)+8e^(-)+10H_(2)OrarrNH_(4)^(+)+3H_(2)O+10OH^(-)`
or `NO_(3)^(-)+8e^(-)+7H_(2)O rarr NH_(4)^(+)+10OH^(-)`
Step IV. Adding the two half reactions
`{:(" "ZnrarrZn^(2+)+2e^(-)"]"xx4),(NO_(3)^(-)+7H_(2)O+8e^(-)rarrNH_(4)^(+)+10OH^(-)),(bar(4Zn+NO_(3)^(-)+7H_(2)Orarr4Zn^(2+)+NH_(4)^(+)+10OH^(-))):}`
[The reduction half reaction is multiplied by four in order to equate the number of electrons]