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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Jules18 on September 01, 2009, 05:48:28 PM

Title: Balancing (in acidic medium)
Post by: Jules18 on September 01, 2009, 05:48:28 PM

Cr2O7(-2) + HNO2 + H(+) --> Cr(+3) + NO3(-) + H2O

The question asks me to balance the above rxn, and then say what the coefficient in front of H2O is.
The answer key says it's 4, but I've tried three times and I keep getting 6.
I've always had trouble with these, so chances are I'm doing it wrong. Here was my attempt, using the half-reaction method.

Cr2O7(-2) --> Cr(+3)

HNO2 --> NO3(-)

1. I balanced Cr and N
2. Added H2O to balance O
3. Added H(+) to balance H
4. Added e- to balance charge
5. Combined both half-rxns into one and cancelled terms that were on both sides.

So I got this result:

4e- + 11H(+) + Cr2O7(-2) + HNO2--> 2Cr(+3) + NO3(-) + 6H2O

If anyone could help me on this I'd really appreciate it! Thanks ahead of time for any responses.

Title: Re: Balancing (in acidic medium)
Post by: renge ishyo on September 01, 2009, 07:03:10 PM

The problem is bolded:

4e- + 11H(+) + Cr2O7(-2) + HNO2--> 2Cr(+3) + NO3(-) + 6H2O

You are not allowed to have electrons in your answer like this, they have to cancel completely as you combine your half reactions. Your first half reaction had 6 e's and your second half reaction had 2e's. What can you multiply the entire second equation by so that the electrons will be 6 so you can cancel? I checked and you will end up with 4H₂0 if you do it right.

Title: Re: Balancing (in acidic medium)
Post by: Borek on September 02, 2009, 02:40:15 AM

http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method

Title: Re: Balancing (in acidic medium)
Post by: Jules18 on September 03, 2009, 10:52:06 AM

Thank you so much!!

That is the most helpful thing ever.