I'm having trouble balancing this equation in a basic solution.

Question

Cr(OH)3 + Br2 -----> CrO4{2-} + Br{-}

Do I do start this with half reactions

Thanks

Answer 1

I always revert to half reactions when I have trouble. Most students forget to compare apples with applies; i.e., you should make that 2Br^ on the right to sstart. But let's do try this way.
Cr is +3 on the left and +6 on the right so a gain of 3e.
Br is zero on the left and 2Br^- is -2 on the right or a loss of 2. Cr x 2 and Br x 3 gives
2Cr(OH)3 + 3Br2 ==> 2CrO4^2- + 6Br^-

The charge on the right is -10 and zero on the left so add 10 OH to the left.
10 OH^- + 2Cr(OH)3 + 3Br2 ==> 2CrO4^2- + 6Br^-.

On the left there are 16H so on the right place 8 H2O
10 OH^- + 2Cr(OH)3 + 3Br2 ==> 2CrO4^2- + 6Br^- + 8H2O

See if that checks. It looks ok to me.