Question

Balance the following redox reaction equation first in acidic solution and then in a basic solution:

Cr (s) + MnO₂ (s) --> Mn⁺² (aq) + Cr⁺³ (aq)

Answer 1

1)

Cr in Cr has oxidation state of 0

Cr in Cr+3 has oxidation state of +3

So, Cr in Cr is oxidised to Cr+3

Mn in MnO2 has oxidation state of +4

Mn in Mn+2 has oxidation state of +2

So, Mn in MnO2 is reduced to Mn+2

Reduction half cell:

MnO2 + 2e- --> Mn+2

Oxidation half cell:

Cr --> Cr+3 + 3e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

3 MnO2 + 6e- --> 3 Mn+2

Oxidation half cell:

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

Lets combine both the reactions.

3 MnO2 + 2 Cr --> 3 Mn+2 + 2 Cr+3

Balance Oxygen by adding water

3 MnO2 + 2 Cr --> 3 Mn+2 + 2 Cr+3 + 6 H2O

Balance Hydrogen by adding H+

3 MnO2 + 2 Cr + 12 H+ --> 3 Mn+2 + 2 Cr+3 + 6 H2O

This is balanced equation in acidic medium

Answer:

3 MnO2(s) + 2 Cr(s) + 12 H+(aq) -> 3 Mn+2(aq) + 2 Cr+3(aq) + 6 H2O(l)

b)

We have:

3 MnO2 + 2 Cr + 12 H+ --> 3 Mn+2 + 2 Cr+3 + 6 H2O

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

3 MnO2 + 2 Cr + 12 H+ + 12 OH- --> 3 Mn+2 + 2 Cr+3 + 6 H2O + 12 OH-

Combine H+ and OH- to form water

3 MnO2 + 2 Cr + 12 H2O --> 3 Mn+2 + 2 Cr+3 + 6 H2O + 12 OH-

Remove common H2O from both sides

Balanced Eqn is

3 MnO2 + 2 Cr + 6 H2O --> 3 Mn+2 + 2 Cr+3 + 12 OH-

This is balanced chemical equation in basic medium

Answer:

3 MnO2(s) + 2 Cr(s) + 6 H2O(l) -> 3 Mn+2(aq) + 2 Cr+3(aq) + 12 OH-(aq)