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) following the genal procedure for ion electron method detailed on the balanced half reaction are
oxidation half equation `SO_(2)(g)+2H_(2)O(l)rarrHSO_(4)^(-)(aq)+3H^(+)(aq)+2e^(-)`
reductin half equation : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)rarrMn^(2+)(aq)+4 H_(2)O(l)`
(c ) oxidation half equation `Fe^(2+)(aq)rarrFe^(3+)(aq)+e^(-)` Multiply eq (i) by 2 and add it to eq (ii) we have
`H_(2)O_(2)(aq)+2Fe^(2+)(aq)+2h^(+)(aq)rarr2Fe^(3+)(aq)+2H_(2)O(l)`
(d) following the general procedure for ion electron detailed on the balanced half reaction equation are
Oxidation half equation `SO_(2)(g)+2H_(2)O(l)rarrSO_(4)^(2-)(aq)+4H^(+)(aq)+2e^(-)`
multiply eq (i) by 3 and add it to eq (ii) we have
`Cr_(2)O_(7)^(2)(aq)+3 SO_(2)(q)+2H^(+)(aq)+2H^(+)(aq)rarr2Cr^(3+)(aq)+3SO_(4)^(2-)(aq)+H_(2)O(l)`