# [GChem] Balancing Redox Reactions

#### neoking77

10+ Year Member

Hey guys,

I need a bit of help with this one. I've been trying to do this question via Chad's videos method of balancing but having a bit of trouble.

So first we need to balance Chromiums so Cr^3+ will become -> 2Cr^3+

Oxidation state of Cr in Cr2O7^2= will be +6

So 3 electrons will need to be transferred for this reduction.

For Br^- -> BrO3^-, 6 electrons are transferred.

So should it not be +22H^+ 2Cr2O7 + 1Br^- -> 4Cr^3+ 1BrO3^- +11H2O?

Is this one of those questions that can't be done with Chad's shortcut method? Where you have to break it down into half reactions? Or is there a way to do it with the shortcut method, and if so where did I go wrong?

D

anyone ?

#### Roy Williams

2+ Year Member
I assume the other guy doesn't need it anymore since it's been quite a while, but here you go:

You have to separate it into half reactions and then add it all back together. But you won't be given a problem this long on the actual DAT because it takes too long. You'll probably be given a half reaction and asked a question about that.

1. Separate into half reactions
For each:
2. Balance everything but oxygen and hydrogen.
3. Add water to one side and then add coefficients to balance the oxygens. Then add hydrogen ion to the other side and add coefficients to balance the hydrogen from the water.
4. Then add electrons (usually same side as hydrogen ion) to balance the charge.
5. Then add both half reactions back together.

D

#### dentalstudent2021

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Last edited by a moderator:

#### Roy Williams

2+ Year Member
I thought 8 was the correct answer? I may have done my electrons wrong (I did this quickly), but that's the same half reaction I have. What's your question? I thought the original question was finding the total number of protons needed to balance the overall reaction. So there was 14 in that one reaction on the left side and then 6 in the other half reaction on the right side, so when you cancel them you have 8 total. I think I'm just not sure what you're trying to ask.

#### Roy Williams

2+ Year Member
Thanks but the answer is the first equation here....

I did chad's way and am somehow missing something.
I mean what is the final answer? Is it not 8? That's what is highlighted in green in the OP, and that's also what I got for my answer.

D

#### dentalstudent2021

I mean what is the final answer? Is it not 8? That's what is highlighted in green in the OP, and that's also what I got for my answer.
sorry! nvm you are correct. sorry for the confusion! and thanks!

rmarder