Balance the following redox reactions by ion electron method:
a. `MnO_(4)^(Θ)(aq)+I^(Θ)(aq) rarr MnO_(2)(s)+I_(2)(s)` (in basic medium)
b. `MnO_(4)^(Θ)(aq)+SO_(2)(g) rarr Mn^(2+)(aq)+HSO_(4)^(Θ)(aq)` (in acidic solution)
c. `H_(2)O_(2)(aq)+Fe^(2+)(aq) rarr Fe^(3+)(aq)+H_(2)O(l)` (in acidic solution)
d. `Cr_(2)O_(7)^(2-)+SO_(2)(g) rarr Cr^(3+)(aq)+SO_(4)^(2-)(aq)` (in acidic solution)

Explanation

(a) Step 1 : The two half reactions involved in the given reaction are :
Oxidation half reaction `overset(-1)(I)_((aq)) to overset(0)(I)_(2 (s))`
Reduction half `overset(+7)(M)n O_(4 (aq))^(-) to overset(+4)(M) n O_(2 (aq))`
Step 2 :
Balancing l in the oxidation half reaction , we have:
`2I_((aq))^(-) to I_(2 (s))`
Now , to balance the charge , we add `2 e^(-)` to the RHS of the reaction .
`2 I_((aq))^(-) to I_(2 (s)) + 2 e^(-)`
Step 3 :
In the reduction half reaction , the oxidation state of Mn has reduced from `+7` to +4 .
Thus, 3 electrons are added to the LHS of the reaction.
`MnO_(4 (aq))^(-) + 3 e^(-) to MnO_(2 (aq))`
Now , to balance the charge , we add 4 `OH^(-)` ions to the RHS of the reaction as the reaction is taking place in a basic medium.
`MnO_(4(aq))^(-) + 3e^(-) to MnO_(2 (aq)) + 4 OH^(-)`
Step 4 : In this equation , there are 6 O atoms on the RHS and 4 O atoms on the LHS .
Therefore , two water molecules are added to the LHS .
`MnO_(4 (aq))^(-) + 2 H_(2)O + 3 e^(-) to Mn O_(2 (aq)) + 4 OH^(-)`
Step 5 :
Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
`6I_(aq)^(-) to 3I_(2 (s)) + 6 e^(-)`
`2MnO_(4 (aq))^(-) + 4H_(2)O + 6 e^(-) to MnO_(2 (s)) + 8 OH_((aq))^(-)`
Step 6 :
Adding the two half reactions, we have the net balanced redox reaction as:
`6 I_((aq))^(-) + 2 MnO_(4 (aq))^(-) + 4 H_(2)O_((l)) to 3 I_(2 (s)) + 2 Mn O_(2 (s)) + 8 OH_((aq))^(-)`
(b)Following the steps as in part (a), we have the oxidation half reaction as:
`SO_(2 (g)) + 2 H_(2)O_((l)) to HSO_(4 (aq))^(-) + 3 H_((aq))^(+) + 2 e_((aq))^(-)`
And the reduction half reaction as :
`MnO_(4 (aq))^(-) + 8 H_((aq))^(+) + 5 e^(-) to Mn_((aq))^(2+) + 4 H_(2)O_((l))`
Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:
`2MnO_(4 (aq))^(-) + 5 SO_(2 (g)) + 2H_(2)O_((l)) + H_((aq))^(+) to 2 Mn_((aq))^(2+) + 5 HSO_(4 (aq))^(-)`
(c) Following the steps as in part (a) , we have the oxidation half reaction as :
`Fe_((aq))^(2+) to Fe_((aq))^(3+) + e^(-)`
And the reduction half reaction as :
`H_(2)O_(2 (aq)) + 2 H_((aq))^(+) + 2 e^(-) to 2H_(2)O_((l))`
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
`H_(2)O_(2 (aq)) + 2 Fe_((aq))^(2+) + 2 H_((aq))^(+) to 2 Fe_((aq))^(3+) + 2 H_(2)O_((l))`
(d) Following the steps as in part (a) , we have the oxidation half reaction as :
`SO_(2 (g)) + 2H_(2)O_((l)) to SO_(4 (aq))^(2-) + 4 H_((aq))^(+) + 2e^(-)`
And the reduction half reaction as :
`Cr_(2)O_(7 (aq))^(2-) + 14 H_((aq))^(+) + 6 e^(-) to 2 Cr_((aq))^(3+) + 7 H_(2)O_((l))`
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
`Cr_(2)O_(7 (aq))^(2-) + 3 SO_(2 (g)) + H_((aq))^(+) to 2 Cr_((aq))^(3+) + 3 SO_(4 (aq))^(2-) + H_(2)O_((l))`