Question

When the following equation is balanced properly under acidic conditions, what are the coefficients of the species shown?

Mn2+ + Sb2O5 = MnO4- + SbO+

Water appears in the balanced equation as a (reactant, product, neither) with a coefficient of . (Enter 0 for neither.)

How many electrons are transferred in this reaction?

Answer 1

Mn in Mn+2 has oxidation state of +2
Mn in MnO4- has oxidation state of +7
So, Mn in Mn+2 is oxidised to MnO4-

Sb in Sb2O5 has oxidation state of +5
Sb in SbO+ has oxidation state of +3
So, Sb in Sb2O5 is reduced to SbO+





Reduction half cell:
Sb2O5 + 4e- --> 2 SbO+
Oxidation half cell:
Mn+2 --> MnO4- + 5e-
Balance number of electrons to be same in both half reactions

Reduction half cell:
5 Sb2O5 + 20e- --> 10 SbO+
Oxidation half cell:
4 Mn+2 --> 4 MnO4- + 20e-

Lets combine both the reactions.
5 Sb2O5 + 4 Mn+2 --> 10 SbO+ + 4 MnO4-
Balance Oxygen by adding water

5 Sb2O5 + 4 Mn+2 + H2O --> 10 SbO+ + 4 MnO4-

Balance Hydrogen by adding H+
5 Sb2O5 + 4 Mn+2 + H2O --> 10 SbO+ + 4 MnO4- + 2 H+
This is balanced chemical equation in acidic medium

Answer:
H2O appear as a reactant with coefficient 1
number of electrons transferred = 20