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Write the balanced ionic equation which ...

Write the balanced ionic equation which represents the oxidation of iodine `(I^(-))` ion by per-manganate ion in basic medium to give iodine (I) and manganese dioxide `(MnO_(2))`.

Text Solution

Verified by Experts

Basic equation is
`MnO_(4)^(-)+I^(-)MnO_(2)+I_(2)` (Basic medium)
(a) `MnO_(4(aq))^(-)+I_((aq))^(-) to MnO_(2(s))+I_(2(s))` (Basic medium)
Reduction half cell
`MnO_(4)^(-)to MnO_(2)`
`MnO_(4)^(-)to MnO_(2)+2H_(2)O` (Oxygens balanced)
`MnO_(4)^(-)+4H_(2)O to MnO_(2)+2H_(2)O+4OH^(-)`
(Hydrogents balanced)
`MnO_(4)^(-)+2H_(2)O to MnO_(2)+4 OH^(-)`
`MnO_(4)^(-)+2H_(2)O+3e^(-)to MnO_(2)+4OH^(-)` (charge balanced)
Oxidation half cell
`I^(-)to I_(2)`
`2I^(-)to I_(2)` (Iodines balanced)
`2 I^(-)to I_(2)+2e^(-)` (Charge balanced)
`overset(2[MnO_(4)+2HO_(2)+3e^(-)to MnO_(2)+4OH])(3xx[2I^(-)to I_(2)+2e^(-)])/(2MnO_(4)^(-)+4 H_(2)O+6e^(-)to 2MnO_(2)+8OH^(-))`
`6I^(-)to 3I_(2)+6e^(-)`
`2MnO_(4)^(-)+4 H_(2)O+6I^(-)to 2MnO_(2)+3I_(2)+8OH^(-)`
The above equation is balanced equation .
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