How do you balance this?

Question

KMnO4 + Na2SO3 + H2O---> MnO2 + Na2SO4 + KOH

Could you please show me how to balance this by adding OH-
Also what are the 1/2 equations?

Answer 1

To balance this equation in basic ambient we divide the equation in two half-reacrions without the spectator ions :
(4 H+ + 3e- + MnO4- >> MnO2 + 2 H2O ) x 2
( H2O + SO32- >> SO42- +2 H+ + 2e-) x 3

8H+ + 6e- + 2MnO4- >> 2 MnO2 + 4H2O
3H2O + 3 SO32- >> 3SO42- + 6H+ + 6e-

we add the two half-reactions cancelling any duplications

2H+ + 2 MnO4- + 3 SO32- >> 2 MnO2 + 3 SO42- + H2O

Now we add 2 OH- on the left and on the right side
On the left side 2 H+ + 2 OH- >> 2 H2O

H2O + 2 MnO4- + 3 SO32- >> 2 MnO2 + 3 SO42- + 2 OH-

we add the spectator ions

H2O + 2 KMnO4 + 3 Na2SO3 >> 2 MnO2 + 3Na2SO4 + 2 KOH

Answer 2

Try (MnO4)- + 3e- + 2H2O -----> MnO2 + 4OH-

and

(SO3)2- + 2OH- ------> (SO4)2- + 2e- + H2O

Now multiply the first equation by 2, and the second by 3, and you're away!