Permanganate (VII) ion, in basic solution oxidize iodide ion `I^(-)` to produce molecular iodine `I_(2)` and manganese (IV) oxide `MnO_(2)`. Write a balanced ionic equation to represent this redox reaction.

Explanation

Step 1 write the skeletal equatoin for the given reaction
`MnO_(4)^(-)(aq)+I^(-)(aq)rarrMnO_(2)(s)+I_(2)(s)`
step 2 write the O.N of all the elements above their respective symbol

step 3 find oiut the species which hve been oxidised and reduced and split the given skeletal equation in to two rection
step 4 To balance oxidation half eq(ii)
(a) balance all atoms other than O and H atoms since there are two I atoms on the R.H.S of Eq (ii) but only one on ethe L.H.S therefore multiply `I^(-)` ion by 2 we have
(b) balance O.N by adding electrons
the O.N of I in `I^(-)` ion is -1 while that in `I_(2)` is 0 thus each `I^(-)` ions loses one electron since there are two `I^(-)` ions on the L.H.S therefore add `2e^(-)` to the R.H.S of Eq (iv) we have
`2I^(-)(aq)rarrI_(2)(s)+2e^(-)`
(c ) balance charge
not needed since charge on either siede of equ(v) is balaned thus eq (v) represent the balanced oxidation half equation
step 5 balance teh reduction half equation (iii)
(a) balance all atoms other tha O and H
not needed because Mn is already balanced ltvbrgt (b) balanced O.N by adding electrons
the O.N of Mn in `Mn_(4)^(-)` on the L.H.S of equ (iii) is +7 while the O.N of Mn in `MnO_(2)` is +4 on the R.H.S therefore and `3e^(-)` to R.H.S of eq (iii) we have
`MnO_(4)^(-)(aq)+3e^(-)rarrMnO_(2)(s)`
(c ) balance charge by adding `OH^(-)` ions sicne the reaction occurs in the basic medium
Therefore add add 4 `OH^(-)` to the R.H.S of Eq (vi) we have
(d) balance O atoms
The R.H.S of eq (vii) has six o atom s while the L.H.S has only four therefore add `2H_(2)O` to the L.H.S of E(vii) we
`MnO_(4)^(-)(aq)+2H_(2)O(aq)+2H_(2)O(l)+3e^(-)rarrMnO_(2)(s)+4OH^(-)(aq)`
by doing so H atoms are automatically balanced therefore eq (viii) represents the balance reduction half equatoin
step 6 to balance the electrons lost in Qe (v) and gained eq (viii) multiply eq (v) by 3 and eq (viii) by 2 and add we have
`6I^(-)(aq)rarr3I_(2)(s)+6e^(-)`
`2MnO_(4)^(-)(aq)+4H_(2)O(l)+6e^(-)rarr2MnO_(2)(s)+8OH^(-)(aq)`
`2MnO_(4)^(-)(aq)+6I^(-)(aq)+4H_(2)Orarr2MnO_(2)(s)+3I_(2)(s)+8OH^(-)(aq)`
this represent the final balanced redox equation
step 7 verification total charge on L.H.S of eq (ix) =2(-1) +6(-1)=-8
total charge of R.H.S of Eq (ix) =8 since the magnitude of charge on either side of Eq (ix) is equal therefore eq (ix) represent the correct balance redox equation