[vnveteran]

Br- + Cr2O7^2- + H+ = Cr^3- + H2O + 3Br2



ok this is what i did im just confused wether or not to bring down the Br2 with the 3 or not

Br- ---> 3Br2 or Br2? i don't know which one to do?

oh and one more thing how would i find the equilibrium constant from that equation when there is 3 on one side and 3 on the other?? usually there is only 2 on each side. aA +bB <->cC +dD

[DemonicAcid]

Look at the half reactions and you should be able to figure it out.

[Borek]

ok this is what i did im just confused wether or not to bring down the Br2 with the 3 or not

Do you know what it means "balanced reaction equation"?

[vnveteran]

thats what i am trying to do. i am trying towrite a balanced overall equation. but i when i do the half reactions for it do i use             Br-  ---> 3Br2 or Br-  ---->Br2

this is what i got but i am not sure it is right please help me

my overall all was this 6Br- + Cr2O7^2- + 14H+  ==> 2Cr^3+ + 7H2O + 3Br2

[Borek]

thats what i am trying to do. i am trying towrite a balanced overall equation. but i when i do the half reactions for it do i use             Br-  ---> 3Br2 or Br-  ---->Br2

Half reaction must be balanced just like every other reaction, neither of those two you listed is balanced.

6Br- + Cr2O7^2- + 14H+  ==> 2Cr^3+ + 7H2O + 3Br2

This is easy to check - balanced reaction must have identical total charge and identical number of atoms on both sides.

[vnveteran]

i dont think anyone is understanding my question. nvm.

[AWK]

6Br- + Cr2O7^2- + 14H+  ==> 2Cr^3+ + 7H2O + 3Br2

Check by calculation numbers of each atom type and sum of charges on both side of equation.
This gives you an answer about correctness of your work.

By the way its OK

[Borek]

i dont think anyone is understanding my question. nvm.

I am afraid that's because you are misunderstanding something and your question doesn't make much sense.

do i use             Br-  ---> 3Br2 or Br-  ---->Br2

Short answer - neither, as both are wrong. You should start with

2Br^- -> Br₂ + 2e

and multiply it by 3 (to balance elecrons with the other half reaction), which yields

6Br^- -> 3Br₂ + 6e

That's what you should use - it differs from both reactions you were asking about, so it was impossible to answer your question in a simple way.