I need help balancing this redox equation:

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.

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.

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

Your thinking and work are correct. Here's the step-by-step process for balancing the redox equation in basic solution:

1. Identify the oxidation states of each element:
H2O2 (H: +1, O: -1)
Ni+2 (Ni: +2)
H2O (H: +1, O: -2)
Ni+3 (Ni: +3)

2. Write the two half-reactions, balancing the atoms and charges on each side:
Oxidation (loss of electrons):
Ni+2 --> Ni+3 + 1e-

Reduction (gain of electrons):
H2O2 + 2OH- --> 2H2O + 2e-

3. Balance the number of electrons in both half-reactions by multiplying each half-reaction by the appropriate coefficient:
2Ni+2 --> 2Ni+3 + 2e-
H2O2 + 2OH- --> 2H2O + 2e-

4. Multiply the entire equation to make the number of electrons equal in both half-reactions:
2H2O2 + 4OH- + 2Ni+2 --> 4H2O + 2Ni+3 + 2e-

5. Combine the half-reactions, cancelling out the electrons:
2Ni+2 + 2H2O2 + 4OH- --> 2Ni+3 + 4H2O

6. Simplify the equation by removing any spectator ions or redundant species:
Ni+2 + H2O2 + 2OH- --> Ni+3 + 2H2O

Therefore, the balanced redox equation in basic solution is:
H2O2 + Ni+2 + OH- --> Ni+3 + H2O

Your thinking and work are correct.

To balance the second half-reaction, we start with the reaction:

H2O2 ==> H2O

First, we need to balance the oxygens. Since there are 2 oxygens in H2O2 and 1 oxygen in H2O, we add an additional H2O to the right side:

H2O2 ==> 2H2O

Now, we need to balance the hydrogens. Since there are 2 hydrogens in H2O2 and 4 hydrogens in 2H2O, the hydrogens are already balanced.

Next, we need to balance the charges. The total charge on the left side is 0 (since H2O2 is neutral), and the total charge on the right side is also 0 (since there are no ions or charges in H2O). Therefore, the charges are already balanced.

Finally, to balance the atoms, we need to add OH- ions to the right side because this is a basic solution. Since we have 2 hydrogens in H2O2, we need to add 2 OH- ions to balance the hydrogens:

H2O2 + 2OH- ==> 2H2O + O2- + 2e

Now the half-reaction is balanced.

Overall, the balanced redox equation in basic solution is:

H2O2 + Ni2+ ==> H2O + Ni3+ + 2OH-