Balance the following equations :
a. `Fe^(3+) + Sn^(+2) rarr Sn^(4+) + Fe^(2+)`
b. `MnO_(4)^(Ө) + H_(2)S rarr S+Mn^(2+)`
c. `Cr_(2)O_(7)^(2-) + 2I^(Ө) rarr 2Cr^(3+) + I_(2)`
d. `Zn + NO_(3)^(Ө) rarr Zn^(2+) + NH_(4)^(o+)`
e. `MnO_(4)^(Ө)+SO_(3)^(2-) rarr SO_(4)^(2-)+MnO_(2)`
f. `Cl_(2)+IO_(3)^(Ө) rarr IO_(4)^(Ө)` (in basic medium)

Explanation

Oxidation number method
(a)

(Multiple `Cr^(3+)` by 2 because there are 2Cr atoms in `Cr_(2)O_(7)^(2-)` ion.).
Balance increase and decreases in oxidation number
`6Fe^(2+)+H^(+)+Cr_(2)O_(7)^(2-)to2Cr^(3+)+6Fe^(3+)+H_(2)O`
Balance charge by multiplying `H^(+)` by 14
`6Fe^(2+)+14H^(+)+Cr_(2)O_(7)^(2-)to2Cr^(3+)+6Fe^(3+)+H_(2)O`
Balance H and O-atom by multiplying `H_(2)O` by 7.
`6Fe^(2+)+14H^(+)+Cr_(2)O_(7)^(2-)to2Cr^(3+)+6Fe^(3+)+7H_(2)O`
This represents a balanced redox reaction.
(b)
Balance increase and decreases in oxidation number
`I_(2)+10NO_(3)^(-)to10NO_(2)+2IO_(3)^(-)`
Balance charge by writing `8H^(+)` in LHS of the equation.
`I_(2)+10NO_(3)^(-)+8H^(+)to10NO_(2)+2IO_(3)^(-)`
Balance H-atoms by writing `4H_(2)O` in RHS of the equation
`I_(2)+10NO_(3)^(-)+8H^(+)to10NO_(2)+2IO_(3)^(+)+4H_(2)O`
Oxygen atoms are automatically balanced.
This represents a balanced redox reaction.
(c )
(Multiply `S_(2)O_(3)^(2-))` by 2 because there are S-atoms in `S_(4)O_(6)^(2-)` ion.)
Increase and decrease in oxidation number is already balanced. Charge and oxygen atoms are also balanced.
This represents a balanced redox reaction.
(d)
Increase and decreases in oxidation number is already balanced
Add `4H^(+)` towards LHS of the equation to balance charge.
`MnO_(2)+C_(2)O_(4)^(2-)+4H^(+)toMn^(2+)+2CO_(2)`
Add `2H_(2)O` towards RHS of the equation to balance H-atoms
`MnO_(2)+C_(2)O_(4)^(2-)+4H^(+)toMn^(2+)+2CO_(2)+2H_(2)O`
This represents a balanced redox reaction.