[murrayt6]

I did a search first, but I'm still having trouble with this and any help would be greatly appreciated:

In the redox equation: 2MnO4,- + 5H2SO3  -->  2Mn,2+ + 3H2O + 5SO4,2- + 4H+ what is the change in the oxidation number of the element that is reduced? Enter your answer as a number with the correct sign, for example, -3 or +2.

On the left side, 2Mn04-: Mn is +7 to balance O(-2x4) -1.
                       5H2SO3: (H2 is +2), (O is -6), so (S must be +4)

On the products side
(I think this is where I run : should 2Mn, 2+ be 0 since it is by itself or should it be +2?
into trouble)                    3H2O: (H2 is a +2), (O is a -2)
                                    5S04 2-: (0 is -8), then with 2-, S must be +6
                                    4H+:H is a +1 but here is it a +2 because of the + after it?

I'm just getting really frustrated and I know I'm probably making it more difficult than it really is, but I need to know the oxidation # of the element that is reduced.  Thanks for taking the time to help out.

[Borek]

Read how to assign oxidation numbers, especially in the case of charged mono atomic ions.

[qwerty44]

Mn would go from +7 to +2 (because Mn2+ has a charge). So I think the reduction would be -5.