Question

When the following oxidation–reduction reaction in a basic solution is balanced, what is the lowest whole-number coefficient for OH−and on which side of the balanced equation should it appear?

S2O82–(aq)+NO(g)→SO42–(aq)+NO3–(aq)

a. 4, product side

b. 8, reactant side

c. 12, reactant side

d. 8, product side

e. 4, reactant side

Answer 1

S in S2O8-2 has oxidation state of +7

S in SO4-2 has oxidation state of +6

So, S in S2O8-2 is reduced to SO4-2

N in NO has oxidation state of +2

N in NO3- has oxidation state of +5

So, N in NO is oxidised to NO3-

Reduction half cell:

S2O8-2 + 2e- --> 2 SO4-2

Oxidation half cell:

NO --> NO3- + 3e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

3 S2O8-2 + 6e- --> 6 SO4-2

Oxidation half cell:

2 NO --> 2 NO3- + 6e-

Lets combine both the reactions.

3 S2O8-2 + 2 NO --> 6 SO4-2 + 2 NO3-

Balance Oxygen by adding water

3 S2O8-2 + 2 NO + 4 H2O --> 6 SO4-2 + 2 NO3-

Balance Hydrogen by adding H+

3 S2O8-2 + 2 NO + 4 H2O --> 6 SO4-2 + 2 NO3- + 8 H+

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

3 S2O8-2 + 2 NO + 4 H2O + 8 OH- --> 6 SO4-2 + 2 NO3- + 8 H+ + 8 OH-

Combine H+ and OH- to form water

3 S2O8-2 + 2 NO + 4 H2O + 8 OH- --> 6 SO4-2 + 2 NO3- + 8 H2O

Remove common H2O from both sides

Balanced Eqn is

3 S2O8-2 + 2 NO + 8 OH- --> 6 SO4-2 + 2 NO3- + 4 H2O

This is balanced chemical equation in basic medium

There are 8 OH- on reactant side

Answer: b