In: Chemistry
When the following oxidation-reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for H2O, 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
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 4 H2O on product side
Answer: a