Question

Question #056c1

Answer 1

You can't.

This is an impossible equation.

The oxidation number of Mn changes from +7 to +2. But nothing else changes oxidation number.

A minor problem is that you have Na₂SO₄ on each side of the equation.

Could you have mis-typed your question?

Answer 2

Like Ernest said, it's very possible the question was mistyped. The closest reaction I could think of that resembles what the posted reaction was is this one:

`KMnO_4 + Na_2SO_3 + H_2SO_4 -> MnSO_4 + K_2SO_4 + Na_2SO_4 + H_2O`

The oxidation number of `Mn` would change from +7 to +2, while the oxidation number of `S` would go from +4 to +6. You'd have

`Mn^(+7) + 5e^(-) = Mn^(+2)` and
`S^(+4) - 2e^(-) = S^(+6)`

Multiply the first equation by 2 and the second by 5 for a total of `10e^(-)` transferred, and get the balanced equation

`2KMnO_4 + 5Na_2SO_3 + 3H_2SO_4 -> 2MnSO_4 + 5Na_2SO_4 + 3H_2O + K_2SO_4`