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

(b) `overset(+7)MnO_4^(-) (aq) +overset(+4)(SO_2)(g) rarr overset(+2)(M)n^(2+)(aq)+overset(+5)(HSO_4^(-))(aq)`
In this reaction , manganeses undergoes reduction because oxidation number of Mn decrease from +7 (in `MnO_4^(-)` ) to +2 (in `Mn^(2+)` ) . `SO_2` gets oxidised because oxidation number of sulphur increases from +4 (in `SO_2`) to 6 in `(HSO_4^(-))` .
Oxidation half reaction : `SO_2(g) rarr HSO_4^(-)(aq)`
Reduction half reaction : `MnO_4^(-) (aq) rarr Mn^(2+) (aq)`
Balancing each half reaction separately as :
`SO_2(g) rarr HSO_4^(-)(aq)`
(i) Balance all atoms other than H and O (already done).
(ii) Sulphur changes its oxidation number from +4 to +6 and there is a difference of 2 electrons To balance the electrons,
`SO_2(g) rarr HSO_4^(-)(aq) +2e^(-)`
(iii) Since the reaction takes place acidic medium add `3H^+` on the right hand side and two `H_2O` molecules on the left hand side.
`SO_2(g) +2H_2O(l) rarr HSO_4^(-) (aq) +3H^(+)+3H^(+)+2e^(-)`
To balance reduction half reaction.
`MnO_4^(-)(aq) rarr Mn^(2+) (aq)`
(i) Balance all atoms other than H and O (already done ).
(ii) Manganese change its oxidaiton number from +7 to -2 and there is difference of 5 electrons . To balance the electrons .
`MnO_4^(-)(aq) +5e^(-) rarr Mn^(2+)(aq)`
(iii) Since the reaction takes place in acidic medium add `8H^+` on the left to the right to balance oxygen atoms .
`MnO_4^(-) +8 H^(+) +5e^(-)rarr Mn ^(2+)`
`MnO_4^(-) +8H^(+) +5e^(-) rarr Mn^(2+)+4H_2O`
Adding two half reaction by equation the number of electrons.
`{:(MnO_4^(-)(aq)+8H^(+)+5e^(-)rarrMn^(2+)(aq)+4H_2O(l)xx2),(SO_2(g)+2H_2OrarrHSO_4^(-)(aq)+3H^(+)(aq)+2e^(-)"]"xx5),(bar(2MnO_4^(-)(aq)+5SO_2(g)+2H_2O(l)+H^(+)(aq)rarr2Mn^(2+)(aq)+5HSO_4^(-)(aq))):}`
(c) `H_2overset(-1)O_2(aq)+overset(+2)(Fe^(2+))(aq)rarroverset(+3)(Fe^(3+))(aq)+H_2overset(-2)O(l)`
In this reaction , the oxidation of Fe increases from +2 (in `Fe^(2+)` ) to 3 (in `Fe^(3+)` ) and therefore , `Fe^(2+)` has been oxidised The oxidation number of oxygen has decreased from -1 (in `H_2O_2`) to -2 in `H_2O` and therefore , `H_2O_2` has been reduced .
Oxidation half reaction :
`Fe^(2+)(aq) rarr Fe^(3+)(aq)`
`Fe^(2+) (aq) rarr Fe^(3+)(aq) +e^(-) " "...(i)`
Reduction half reaction
`H_2O_2(aq) rarr H_2O(l)`
`H_2O_2(aq) +e^(-) rarr H_2O(l)`
`H_2O_2(aq) +2H^(+)(aq) +2e^(-) rarr 2H_2O(l)" " ....(ii)`
Multiply eq (i) by 2 and it to eq. (ii) to balance electrons.
Reduction half reaction :
`Cr_2O_7^(2-) (aq) rarr Cr^(3+)(aq)`
`Cr_2O_7^(2-) rarr 2Cr^(3+)(aq)`
`Cr_2O_7^(2-) (aq) +6e^(-) rarr 2Cr^(3+)(aq)`
`Cr_2O_7^(2-) (aq) +6e^(-) +14 H^(+) (aq) rarr 2Cr^(3+)(aq)`
`Cr_2O_7^(2-) (aq) +6e^(-) +14 H^(+) (aq) rarr2Cr^(3+)(aq) +7 H_2O....(ii)`
Multiply eq. (i) by 3 and add it to eq (ii) to balance electrons :
`Cr_2O_7^(2-) (aq) +3SO_2(g) +2H^(+) (aq) rarr 2Cr^(3+) (aq) +3SO_4^(2-) (aq) +H_2O`