Question

Balance the following redox reaction that occurs in base ClO3-1(aq) + Cr+3(aq) ↔ Cr2O7-2(aq) + ClO-1(aq)

Answer 1

Cl in ClO3- has oxidation state of +5

Cl in ClO- has oxidation state of +1

So, Cl in ClO3- is reduced to ClO-

Cr in Cr+3 has oxidation state of +3

Cr in Cr2O7-2 has oxidation state of +6

So, Cr in Cr+3 is oxidised to Cr2O7-2

Reduction half cell:

ClO3- + 4e- --> ClO-

Oxidation half cell:

2 Cr+3 --> Cr2O7-2 + 6e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

3 ClO3- + 12e- --> 3 ClO-

Oxidation half cell:

4 Cr+3 --> 2 Cr2O7-2 + 12e-

Lets combine both the reactions.

3 ClO3- + 4 Cr+3 --> 3 ClO- + 2 Cr2O7-2

Balance Oxygen by adding water

3 ClO3- + 4 Cr+3 + 8 H2O --> 3 ClO- + 2 Cr2O7-2

Balance Hydrogen by adding H+

3 ClO3- + 4 Cr+3 + 8 H2O --> 3 ClO- + 2 Cr2O7-2 + 16 H+

Add equal number of OH- on both sides as the number of H+

3 ClO3- + 4 Cr+3 + 8 H2O + 16 OH- --> 3 ClO- + 2 Cr2O7-2 + 16 H+ + 16 OH-

Combine H+ and OH- to form water

3 ClO3- + 4 Cr+3 + 8 H2O + 16 OH- --> 3 ClO- + 2 Cr2O7-2 + 16 H2O

Remove common H2O from both sides

Balanced Eqn is

3 ClO3- + 4 Cr+3 + 16 OH- --> 3 ClO- + 2 Cr2O7-2 + 8 H2O

This is balanced chemical equation in basic medium

Answer:

Cl in ClO3- has oxidation state of +5

Cl in ClO- has oxidation state of +1

So, Cl in ClO3- is reduced to ClO-

Cr in Cr+3 has oxidation state of +3

Cr in Cr2O7-2 has oxidation state of +6

So, Cr in Cr+3 is oxidised to Cr2O7-2

Reduction half cell:

ClO3- + 4e- --> ClO-

Oxidation half cell:

2 Cr+3 --> Cr2O7-2 + 6e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

3 ClO3- + 12e- --> 3 ClO-

Oxidation half cell:

4 Cr+3 --> 2 Cr2O7-2 + 12e-

Lets combine both the reactions.

3 ClO3- + 4 Cr+3 --> 3 ClO- + 2 Cr2O7-2

Balance Oxygen by adding water

3 ClO3- + 4 Cr+3 + 8 H2O --> 3 ClO- + 2 Cr2O7-2

Balance Hydrogen by adding H+

3 ClO3- + 4 Cr+3 + 8 H2O --> 3 ClO- + 2 Cr2O7-2 + 16 H+

Add equal number of OH- on both sides as the number of H+

3 ClO3- + 4 Cr+3 + 8 H2O + 16 OH- --> 3 ClO- + 2 Cr2O7-2 + 16 H+ + 16 OH-

Combine H+ and OH- to form water

3 ClO3- + 4 Cr+3 + 8 H2O + 16 OH- --> 3 ClO- + 2 Cr2O7-2 + 16 H2O

Remove common H2O from both sides

Balanced Eqn is

3 ClO3- + 4 Cr+3 + 16 OH- --> 3 ClO- + 2 Cr2O7-2 + 8 H2O

This is balanced chemical equation in basic medium

Answer:

3 ClO₃^- + 4 Cr⁺³ + 16 OH^- -> 3 ClO^- + 2 Cr₂O₇^-2 + 8 H₂O