When the reaction below is balanced under basic conditions, what is the coefficient in front of the hydroxide ion?

C2H6O2 + Cr2O7 2- == C2O4 2- + Cr(OH)4 -

Answer is 6, but what are the steps in solving this problem? I tried adding OH- on either side and balance the other side with H2O but that didn't seem right.

Question

When the reaction below is balanced under basic conditions, what is the coefficient in front of the hydroxide ion?

C2H6O2 + Cr2O7 2-  ==> C2O4 2- + Cr(OH)4 -


Answer is 6, but what are the steps in solving this problem? I tried adding OH- on either side and balance the other side with H2O but that didn't seem right.

anonymous · Answer

first u shud balance charge

anonymous · Answer

Please elaborate, do I use e- or OH-. And to which side should I add OH-.

anonymous · Answer

first use e- then balance H & O by OH- or H+ acc. to medium

anonymous · Answer

I don't get it. Can you show steps?

anonymous · Answer

Why do I need any e- o balance charge if my O and H arent can't be balanced by using 6OH-. I tried both sides and they won;t balance

anonymous · Answer

wait i'm doing

anonymous · Answer

sorry i tried a lot but i was unable to balance H. I'm still working

anonymous · Answer

It's alright, thanks for trying. I'll go ask my teachr about this later.

anonymous · Answer

yw