Question

Balance the following reactions by oxidation number method :

a. Cr₂O₇^2Θ_(aq) + SO₃^2Θ_(aq) → Cr^3⊕_(aq) + SO₄^2Θ_(aq) (acidic)

b. MnO₄^Θ_(aq) + Br^Θ_(aq) → MnO_2(s) + BrO₃^Θ_(aq)(basic)

Answer 1

a. Cr₂O₇^2Θ_(aq) + SO₃^2Θ_(aq) → Cr^3⊕_(aq) + SO₄^2Θ_(aq) (acidic)

Cr₂O₇^2-_(aq) + SO₃^2-_aq) → Cr³⁺_(aq) + SO₄^2-_(aq) (acidic)

Step 1: 

Write skeletal equation and balance the elements other than O and H.

Cr₂O₇^2-_(aq) + SO₃^2-_(aq) → Cr³⁺_(aq) + SO₄^2-_(aq)

Step 2 :

Assign oxidation number to Cr and S. 

Calculate the increase and decrease in the oxidation number and make them equal.

Increase in oxidation number :

(Increase per atom = 2)

Decrease in oxidation number :

(Decrease per atom = 3)

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. 

(There are already 2 Cr atoms.)

Step 3 :

Balance 'O' atoms by adding 4H₂O to the right-hand side.

Cr₂O₇^2-_(aq) + 3SO₃^2-_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) +4H₂O_(l)

Step 4 :

The medium is acidic. 

To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.

Cr₂O₇^2-_(aq) + 3SO₃^2-_(aq) + 8H⁺_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) +4H₂O_(l)

Step 5 : 

Check two sides for balance of atoms and charges. 

Hence, balanced equation :

Cr₂O₇^2-_(aq) + 3SO₃^2-_(aq) + 8H⁺_(aq) → 2Cr³⁺_(aq) + 3SO₄^2-_(aq) +4H₂O_(l)

b. MnO₄^Θ_(aq) + Br^Θ_(aq) → MnO_2(s) + BrO₃^Θ_(aq)(basic)

MnO₄^-_(aq) + Br^-_(aq) → MnO_2(s) + BrO₃^-_(aq) (basic)

Step 1 : 

Write skeletal equation and balance the elements other than O and H.

MnO₄^-_(aq) + Br^-_(aq) → MnO_2(s) + BrO₃^-_(aq)

Step 2 : 

Assign oxidation number to Mn and Br. 

Calculate the increase and decrease in the oxidation number and make them equal.

Increase in oxidation number :

(Increase per atom = 6)

Decrease in oxidation number :

(Decrease per atom = 3)

To make the net increase and decrease equal, we must take 2 atoms of Mn.

2MnO₄^-_(aq) + Br^-_(aq) → MnO_2(s) + BrO₃^-_(aq)

Step 3 : 

Balance 'O' atoms by adding H₂O to the right-hand side.

2MnO₄^-_(aq) + Br^-_(aq) → 2MnO_2(s) + BrO₃^-_(aq) + H₂O_(l)

Step 4 : 

The medium is basic. 

To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

2MnO₄^-_(aq) + Br^-_(aq) + 2H⁺_(aq) → 2MnO_2(s) + BrO₃^-_(aq) + H₂O_(l)

Add OH^- ions equal to the number of H⁺ ions on both sides of the equation.

2MnO₄^-_(aq) + Br^-_(aq)+ 2H⁺_(aq)+ 2OH^-_(aq) → 2MnO_2(s) + BrO₃^-_(aq) + H₂O_(l) + 2OH^-_(aq)

The H+ and OH- ions appearing on the same side of the reaction are combined to give H₂O molecules.

2MnO₄^-_(aq) + Br^-_(aq)+ 2H₂O_(l)→ 2MnO_2(s) + BrO₃^-_(aq) + H₂O_(l) + 2OH^-_(aq)

2MnO₄^-_(aq) + Br^-_(aq)+ H₂O_(l)→ 2MnO_2(s) + BrO₃^-_(aq) + H₂O_(l) + 2OH^-_(aq)

Step 5 :

Check two sides for balance of atoms and charges.

Hence,

Balanced eqution :

2MnO₄^-_(aq) + Br^-_(aq)+ H₂O_(l)→ 2MnO_2(s) + BrO₃^-_(aq) + H₂O_(l) + 2OH^-_(aq)