Question

Balance the reaction using half reactions under BASIC conditions:

Al + CrO4^-2 ---> Al(OH)3 + Cr(OH)3

I tried to solve this question and ended up with this:

Here are the two half equations
Al ---> Al(OH)3
CrO4^-2 ---> Cr(OH)3

For Al ---> Al(OH)3, I did
3OH^- + 3H2O + Al ---> Al(OH)3 + 3H^+ + 3OH^-
3OH^- + 3H2O + Al ---> Al(OH)3 + 3H2O
3OH^- + Al ---> Al(OH)3
3e^- + 3OH^- + Al ---> Al(OH)3

For CrO4^-2 ---> Cr(OH)3 I did:
5OH^- + 5H^+ CrO4^-2 ---> Cr(OH)3 + H2O + 5OH^-
5H2O + CrO4^-2 ---> Cr(OH)3 + H2O + 5OH^-
4H2O + CrO4^-2 ---> Cr(OH)3 + 5OH^-
3e^- + 4H2O + CrO4^-2 ---> Cr(OH)3 + 5OH^-

After combining the reactions I got:
3e^- + 3OH^- + Al + Cr(OH)3 + 5OH^- ---> Al(OH)3 + 3e^- + 4H2O + CrO4^-2

simplified it and got:
8OH^- + Al + Cr(OH)3 ---> Al(OH)3 + 4H2O + CrO4^-2

Is this correct?

Answer 1

The Al is not right.
Al atoms balance.
H atoms balance
O atoms balance.
Charge does not balance. You have 6- on the left and zero on the right. You can fix that by placing the 3e on the right.

The CrO4^2- half reaction is balanced.

The total equation is not balanced.
Al is ok
Cr is ok
O is ok
H is ok
charge does not balance. You have 8- on the left and 2- on the right. I think you just didn't add the two equations right. You somehow reversed the 5OH and 4H2O. Each belongs on the other side.

If you add the two equations you get
4H2O + CrO4^2- + 3OH^- + Al ==> Al(OH)3 + Cr(OH)3 + 5OH^- , then you adjust the OH^- on each side by removing the 3 OH^- on the left and reducing the 5OH^- on the right to 2 OH^-