Balancing Redox equation?

Question

My eqaution was

(Mn^+2) + (NaBiO3) -> (Bi^+3) + (MnO4^-1)

if i were to balance this equation where would the Sodium go?
It does not have a Sodium on the other side. how would i balance this?
Please help.
Thank you.

Answer 1

When you balance a redox equation, you only deal with the species that are oxidized or reduced. Actually, this equation looks far-fetched, since MnO4- is a strong oxidizing agent, and it would take a powerful reducing agent to oxidize Mn+2 to Mn+7. But, the routine goes:
The "Red" part: (add H2O to provide enough hydrogen to balance Oxygen and form hydroxyl ion [assume this occurs in base solution] Also add electrons as needed to balance charge.
3H2O + BiO3- + 2e -> Bi+3 + 6 OH-
The "Ox" part (use OH- to supply oxygen
Mn+2 + 8 OH- -> MnO4- + 4H2O +5e
Balance electrons
15H2O + 5 BiO3- + 10e => 5 Bi+3 + 30 OH-
2 Mn+2 + 16 OH- => 2MnO4- + 8 H2O + 10e
Add equations
7H2O+ 5 BiO3 + 2 Mn+2 =>
5 Bi+3 + 2 MnO4- + 14 OH-
7 H2O + 5 NaBiO3 + 2 Mn+2 => 4 Bi(OH)3 + 2NaMnO4 + Bi+3 + 2 Na(OH) + Na+
The final equation show the inbalance of charge, since the cation of Mn+2 was not included.