Balancing (in acidic medium)

[Jules18]

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 canceled 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.

 

[Nino MMP]

Your reaction is balanced correctly. The coefficient of H2O in this equation is 6. The answer key is incorrect.

 

[Blueshift5]

I can confirm that your attempt to balance the above reaction using the half-reaction method is correct. However, I believe the discrepancy in the coefficient for H2O may be due to the fact that there are multiple ways to balance a chemical equation and still arrive at the same overall reaction. In this case, both 4 and 6 are valid coefficients for H2O. It is possible that the answer key used a different method or approach to balance the equation, leading to a different coefficient for H2O. Therefore, I would suggest double-checking your work and discussing with your instructor or peers to understand the differences in approaches and coefficients. Overall, your attempt is scientifically accurate and valid.