Question

When the following reaction is balanced in BASIC solution, what is the coefficient for S? S2- + MnO4-  S + MnO2 a. 1 b. 2 c. 3 d. 4 e. 5

Answer 1

S in S-2 has oxidation state of -2

S in S has oxidation state of 0

So, S in S-2 is oxidised to S

Mn in MnO4- has oxidation state of +7

Mn in MnO2 has oxidation state of +4

So, Mn in MnO4- is reduced to MnO2

Reduction half cell:

MnO4- + 3e- --> MnO2

Oxidation half cell:

S-2 --> S + 2e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

2 MnO4- + 6e- --> 2 MnO2

Oxidation half cell:

3 S-2 --> 3 S + 6e-

Lets combine both the reactions.

2 MnO4- + 3 S-2 --> 2 MnO2 + 3 S

Balance Oxygen by adding water

2 MnO4- + 3 S-2 --> 2 MnO2 + 3 S + 4 H2O

Balance Hydrogen by adding H+

2 MnO4- + 3 S-2 + 8 H+ --> 2 MnO2 + 3 S + 4 H2O

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

2 MnO4- + 3 S-2 + 8 H+ + 8 OH- --> 2 MnO2 + 3 S + 4 H2O + 8 OH-

Combine H+ and OH- to form water

2 MnO4- + 3 S-2 + 8 H2O --> 2 MnO2 + 3 S + 4 H2O + 8 OH-

Remove common H2O from both sides

Balanced Eqn is

2 MnO4- + 3 S-2 + 4 H2O --> 2 MnO2 + 3 S + 8 OH-

This is balanced chemical equation in basic medium

coefficient of S is 3

Answer: c