Can you oxidize this equation? BrO- +Cr(OH)4*top*1->>>>>Br- + CrO4*top*2-?

Question

I really need help with this equation though

I have been crackin my noggin on this one. Thanks for looking.


Mo2O3 +(Mn2O4)1- ---------->>>>>MoO3 + (Mn2)1+

YES sir. mistake on my part. Sorry. Thanks for the response. Appreciated.

Answer 1

I'm assuming you mean balance it?

Oxidation (presuming we're in basic solution):
Cr(OH)4- + 4 OH- -----> CrO42- + 4 H2O + 3 e-

Reduction:
BrO- + 2 H2O + 2e- ---> Br + 2OH-

Overall reaction:
2 Cr(OH)4- + 2 OH- + 3 BrO- ---> 2 CrO42- + 2 H2O + 3 Br-

For the second one, I'm suspecting that Mn2O4 - should actually be MnO4-, since I've never seen manganese form that ion.

Answer 2

To balance a REDOX equation, you need also to specify the reaction conditions.

I am assuming that the first reaction is in basic solution and the second one is in acidic solution.

Also, do the following steps in balancing redox equations
1. Balance each half-reaction equations ATOMICALLY
2. Balance each half-reaction equations ELECTRONICALLY
3. Combine the half reactions.

For the first one,
Oxidation Half Reaction

Cr(OH)4^-1 + 4 OH^- -----> CrO4^2- + 4 H2O + 3e-

While the reduction is,
BrO^-1 + H2O + 2e- ---> Br^-1 + 2OH-
Multiply to get 6e- on either side,

Thus, the combined reaction is
2Cr(OH)4^- + 2 OH- + 3 BrO^- ---> 2CrO42- + 2 H2O + 3 Br^-
(Charges check at -7)

For the second one, I think there are typos, but I believe your given is
Mo2O3 + MnO4^-1 ---> MoO3 + Mn^+2

Thus, your Oxidation half reaction is
3 H2O + Mo2O3 --> 2MoO3 + 6H^+1 + 6e-

The reduction reaction is
5e- + 8H^+1 + MnO4^-1 --> Mn^+2 + 4H2O

Multiply each side to get 30e to cancel and the final equation is
18H^+1 +5Mo2O3 + 6MnO4^-1 -->10MoO3 + 6Mn^+2 + 9H2O (Charges check at +12)

I hope your givens are correct.