Complete the following reactions:
(a). `MnO_4^(2-)+H^(o+)toMn^(2+)+?`
(b). `NO_2+H_2Oto?+NO`
(d). `H_2O_2+Sn^(2+)toSn^(4+)+?`

Explanation

`MnO_4^(2-)` is reduced to `Mn^(2+)` (i.e., acts as oxidising agent) and `H^(o+)` is in maximum oxidation state. So, `MnO_4^(2-)` must also be oxidised to `Mn^(7+)` state, i.e., `MnO_4^(ɵ)` will be formed. Hence, a disproportionation reaction occurs.
`MnO_4^(2-)+H^(o+)toMn^(2+)toMn^(2+)+MnO_4^(ɵ)`
The balanced equation is:
`8H^(o+)+cancel(4e^(-))+MnO_4^(2-)toMn+4H_2O`
`underline(MnO_4^(2-)toMnO_4^(ɵ)+cancel(e^(-))]xx4)`
`underline(5MnO_4^(2-)+8H^(o+)toMn^(2+)+4MnO_4^(ɵ)+4H_2O)`
(b). `NO_2(+4` oxidation state) disproportionates [as in (a)] to NO(+2) oxidation state) and `NO_3^(ɵ)` (`+5` oxidation state).
`2H^(o+)+cancel(2e^(-))+NO_2toNO+H_2O`
`underline(H_2O+NO_2toNO_3^(ɵ)+cancel(e^(-))+2H^(o+)]xx2)`
(c). Here `I_2` is reduced to `2I^(ɵ)`, so `H_2O_2` must be oxidised and gives `O_2`.
`cancel(2e^(-))+I_2to2I^(ɵ)`
`underline(H_2O_2toO_2+2H^(o+)+cancel(2e^(-))`
`underline(H_2O_2+I_2toO_2+2I^(ɵ)+2H^(o+))`
(d). Here `Sn^(2+)` is oxidised to `Sn^(4+)`, so `H_2O_2`
`Sn^(2+)toSn^(4+)+2e^(-)`
`underline(2H^(o+)+cancel(2e^(-))+H_2O_2to2H_2O)`
`underline(Sn^(2+)+H_2O_2+2H^(o+)toSn^(4+)+2H_2O)`