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) The skeleton equation is:
`MnO_(4)^(-)+I^(-)(aq)rarrMnO_(2)(s)+I_(2)(S)`
step I sepparation of the equation in two half reactiopns
(i) Write the O.N of the atoms involved in the equaiton
`(overset(+7)(Mn)overset(-2)O_(4))^(-)+ (overset(-1)I)^(-)rarr(overset(+4)Mn overset(-2)O_(2))+overset(0)I_(2)`
(ii)Identify the atoms which undergo changei in O.N
`overset(+7)(MnO_(4))^(-)+overset(-1)(I)^(-)rarroverset(+4)MnO_(2)+overset(0)I_(2)`
(iii) Find out the speices involved in the oxidation and reduction half reations:

Step II Balancing the oxidation half reation
`2I^(-)rarrI_(2)+2e^(-)`
stepIII Balaning the reduction half reation
The reduction half reaction is :
`MnO_(4)^(-)rarrMnO_(2)`
stepIV Adding the two half reaction
In order to equate the electrons , multiply eqn (i) by 3 and eqn (ii) by 2. Add the two eqns.
`2I^(-)rarrI_(2)+2e^(-)xx3`
`[MnO_(4)^(-)+3e^(-)+2H_(2)OrarrMnO_(2)+4OH^(-)]xx23`
`2MnO_(4)^(-)+6I^(-)+4H_(2)Orarr3I_(2)2MnO_(2)+8OH^(-)`
or `2MnO_(4)^(-)(aq)+6I^(-)4H_(2)Orarr3I_(2)+2MnO_(2)+8OH^(-)`
(b) The skeletion equation is :
`MnO_(4)^(-)(aq)+SO_(2)(g)rarrMn^(2)rarrMn^(2+)(aq)+HSO_(4)^(-)(aq)`
Step I. Separation of the equation in two half reactions
(i) write the O.N of the satoms involved in the equation
(ii) Identify the atoms which undergo change in O.N
(iii) Find out the species involved in the oxidation and reduction half reaction

step II Balancing the oxidation half reaction
The oxidation half reaction is:
`SO_(2)rarrHSO_(4)^(-) +3H^(+)+2e^(-)`
step III Balancing the reductionn half rection
The reduction half rection is :
`MnO_(4)^(-)rarrMn^(2+)`
(i) As the decrease in O.N is 5, therefore , add `5e^(-)` on the rectant side,
(ii) In order to balance the no of , oxygen atoms add four `H_(2)O` mileucles on the product side and then to balance H atoms, add `8H^(+)` on the reactant side.
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)` step IV Adding the two half reactions
In order to equate the elecrons multiply eqn (i) by 3 and eqn (ii) by 2 the two eqns
Step I. Separation of the equation in two half reacins
(i) Write the O.N of the atoms involved in the equation
`overset(+7)Moverset(-2)O_(4)^(-)+overset(+4)Soverset(-2)O_(2)rarr(overset(+2)Mn)^(2+)+(overset(+1)Hoverset(+6)Soverset(-2)0_(4))^(-)`
(ii)Identify the (iii) Find out the species involved in the oxidation and reduction half rections
`{:(SO_(2)+2H_(0)OrarrHSO_(4)^(-)+3H^(+)2e^(-)xx5),(MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+Mn^(2)+4H+_(2)Oxx2)/(2Mno_(4)^(-)(aq)+5SO_(2)(g)+2h_(2)O(l)+H^(+)(aq)rarr5HSO_(4)^(-)(aq)+2Mn^(2)+(aq)):}`
(c) solve on similar lines
(d) Solve on similar lines.