# Question #056c1

Dec 23, 2014

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?

Jan 1, 2015

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:

$K M n {O}_{4} + N {a}_{2} S {O}_{3} + {H}_{2} S {O}_{4} \to M n S {O}_{4} + {K}_{2} S {O}_{4} + N {a}_{2} S {O}_{4} + {H}_{2} O$

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

$M {n}^{+ 7} + 5 {e}^{-} = M {n}^{+ 2}$ and
${S}^{+ 4} - 2 {e}^{-} = {S}^{+ 6}$

Multiply the first equation by 2 and the second by 5 for a total of $10 {e}^{-}$ transferred, and get the balanced equation

$2 K M n {O}_{4} + 5 N {a}_{2} S {O}_{3} + 3 {H}_{2} S {O}_{4} \to 2 M n S {O}_{4} + 5 N {a}_{2} S {O}_{4} + 3 {H}_{2} O + {K}_{2} S {O}_{4}$