Permangante (VII) ion, in basic solution oxidizes iodide ion I- to produce molecular iodine `(I_2)` and manganes (IV) oxide `(MnO_2)`. Write a balanced ionic equation to represent this redox reaction.

Explanation

Step -1 write the skeletal equation for the given reaction
`MnO_4^(-)(aq)+1^(-)(aq)toMnO_2(s)+l_2(s)....(i)`
Step-2: Write the O.N. of all the elements above their respective symbols.

`Mn_n^(+7)O_4^(2-)toM_n^(+4)O_2` (ON of Mn from +7t +4 i.e., 3 unit charge)....(ii)
`I^(-)toI_2^(0)` (ON of iodine from -1 to 0)......(iii)
Oxidation half equation `I^(-) (aq) to I_2 (s)`..........(iv)
Reduction half equation `MnO_4^(-)(aq)toMnO_2(s)`......(v)
Step 4: To balance oxidation half Eq.
(a) Balance all atoms `2l^(-) (aq)toI_2(s)`
(b) Balance O.N. by adding electrons
`2l^(-) (aq)toI_2(s)+2e^(-)` .....(vi)
Charge on either side of Eq (v) is balanced. THus eq (vI) represents the balanced oxidation half equation.
Step 5: Balance the reduction half equation (v)
Balance O.N. by adding electrons.
`MnO_4^(-)(aq)+2H_2O(l)+2e^(-)toMnO_2(s)+4OH^(-) (aq)`...(vii)
Step 6: To balance the electrons lost in Eq. (vi) and gained in Eq. (vii) Equation (vi) `times 3+` Equation (vii) `times 2` we have,
`6I^(-) (aq)to3I_2(s)+6e^(-)`
`2MnO_4^(-) (aq) +4H_2O(l)+6e^(-) to2MnO_2(s)+8OH^(-) (aq)`
`overline(underline(2MnO_4^(-) (aq) +6I^(-) (aq)+4H_2O(I)to2MnO_2(s)+3I_2(s)+8OH^(-) (aq)))`