I need help balancing this redox equation:

Question

H2O2 + Ni+2 --> H2O + Ni+3 in basic solution.

Here's what I have so far:
Ni+2 --> Ni+3 + 1e-
H2O2 --> H2O

I can't seem to make the second half-reaction balance.

If you follow the rules for balancing basic redox solutions, this should work as follows:
H2O2 ==> H2O
First, add 2 to right to make two O atoms (so we are comparing the same number).
H2O2 ==> 2H2O.
Total charge on left O is -2. Total on right O is -4
Add 2 electrons to left to balance charge.
2e + H2O2 ==> 2H2O
Charge on left is -2. Charge on right is 0. Add OH^- to balance charge. So add 2 OH^- to right (this is a basic solution).
2e + H2O2==>2H2O + 2 OH^-
Then add water to other side to balance H atoms.
2e + 2H2O + H2O2==>2H2O + 2OH^-
Now cancel the 2H2O on each side to arrive at the final half equation.
2e + H2O2 ==> 2OH^- OR
we could have done it easier by simply going to OH^- initially.

H2O2 ==> 2 OH^-
Charge on left O is -2. right O is -4
Add 2e to left.
2e + H2O2 ==> 2 OH^-

Check my thinking and my work.

Answer 1

I can do it two ways. I'm not sure which way is right though. Here is way 1.

(If you don't make the oxygen into O2.)

Also I think you just confusing yourself. You don't need to add any OH.

This is how I would do it.

Equation: H2O2 ==> H2O +O
Charge: -1 ==> -2
Solution: add +1e- to the left hand side so the charges equal.

Therefore:

H2O2 +1e- ==> H2O + O

Way 2.

Equation:2H2O2 ==> 2H2O +O2
Charge: -2 0
Solution: add 2 electrons (+ 2e-) to the right and side.

Solution: 2H2O2 ==> H2O + O2 +2e-
-2 -2

I think the first way is right.

Answer 2

He's right. When you reduce H2O2 in basic solutions you have to add OH- to balance the equation.