Ncert Exercises & Solutions

24. Balance the following equations by the oxidation number method.

(a) ${Fe}^{2 +} + H^{+} + {Cr}_{2} O_{7}^{- 2} \to {Cr}^{2 +} + {Fe}^{3 +} + H_{2} O$

(b) $I_{2} + {NO}_{3}^{-} \to {NO}_{2} + {IO}_{3}^{-}$

(d) ${MnO}_{2} + C_{2} O_{4}^{2 -} \to {Mn}^{2 +} + {CO}_{2}$

Oxidation number method

(a)

(Multiply Cr³⁺ by 2 because there are 2Cr atoms in

{Cr}_{2} O_{7}^{2 -}

ion.)

Balance increase and decrease in oxidation number.

6 {Fe}^{2 +} + H^{+} + {Cr}_{2} O_{7}^{- 2} \to 2 {Cr}^{2 +} + 6 {Fe}^{3 +} + H_{2} O

Balance charge by multiplying H⁺ by 14.

6 {Fe}^{2 +} + 14 H^{+} + {Cr}_{2} O_{7}^{- 2} \to 2 {Cr}^{2 +} + 6 {Fe}^{3 +} + H_{2} O

Balance H and O-atoms by multiplying H₂O by 7.

6 {Fe}^{2 +} + 14 H^{+} + {Cr}_{2} O_{7}^{- 2} \to 2 {Cr}^{2 +} + 6 {Fe}^{3 +} + 7 H_{2} O

This represents a balanced redox reaction.

(b)

(Multiply

S_{2} O_{3}^{2 -}

by 2 because there are 4 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.

(c)

Increase and decrease in oxidation number is already balanced.

Add 4H⁺ towards LHS of the equation to balance charge.

{MnO}_{2} + C_{2} O_{4}^{2 -} + 4 H^{+} \to {Mn}^{2 +} + 2 {CO}_{2}

Add 2H₂O towards RHS of the equation to balance H-atoms.

{MnO}_{2} + C_{2} O_{4}^{2 -} + 4 H^{+} \to {Mn}^{2 +} + 2 {CO}_{2} + 2 H_{2} O

This represents a balanced redox reaction.