Balance each of the following redox reactions occurring in acidic solution

You need to know and understand how to do these yourself. So rather than give you the balanced equations, I will tell you that

a)S changes oxidation state from +4 on the left to +6 on the right. Mn changes its oxidation state from +7 on the elft to +2 on the right.

b)S changes oxidation state from 2 (for each S) on the left to +6 on the right.
Cl changes from zero on the left to -1 on the right.

If this doesn't help all that much, tell me what you don't understand about how to balance redox equation.

I'm mainly having trouble with adding H20/H+ and then balancing it!

Post your work on the first one as far as you can go and I'll help you through it.

A)

OX
H20+SO32- --> SO42- +2H+ 2e-

RED

16H+ +2MnO4- ---> 2Mn2+ 8H20

Both half reactions look balanced to me; the second one is twice what is necessary. Notice you can reduce each coefficient by 1/2 to make them 8,1,1,4.

Looks like you did a good job to me.

Here is a method I use.

SO3^- ==> SO4^-2

1. S changes from +4 to 6. Add electrons to the appropriate side (right for this one) to balance the change in oxidation state.
2. SO3^-2 ==> SO4^-2 + 2e
3. Now count the charge on each side. The left is -2 and the right is -4; therefore, add
(a) H^+ to balance the charge if it is acid solution or
(b)OH^- to balance the charge if it is basic or neutral solution.
This is acid so we add 2H^+ to balance the charge.
4. SO3^-2 ==> SO4^-2 + 2e + 2H^+
5. Now add water (usually to the opposide side) to balance the H atoms. Oxygen SHOULD balance at that point.
H2O + SO3^-2 ==> SO4^-2 + 2e + 2H^+

For the Mn, without all the explaining, but I'll follow the same format.

MnO4^- ==> Mn^+2
Mn goes from 7 to 2; therefore, add 5e to the left side.
MnO4^- + 5e ==> Mn^+2

Charge is -6 on the left, +2 on the right, add 8H^+ on the left to balance the charge.
8H^+ + 5e + MnO4^- ==>Mn^+2

Now add 4H2O to the right to balance the H atoms
8H^+ + 5e + MnO4^- ==> Mn^+2 + 4H2O

Thank you