Question: The reaction of Cr{2+} with Cr2O7{2-} in acid solution to formCr{3+}.

Question

Question:

The reaction of Cr{2+} with Cr2O7{2-} in acid solution to formCr{3+}. Calculate Delta Grxn. Then calculate K.

I started by dividing into half reactions, then balancing with H+and H2O. Final equation is:

6Cr{2+} + Cr2O7{2-} + 14H{+} ---> 8Cr{3+} + 7H2O

Then, I obviously need to calculate Ecell by using the ReductionValues of the two half-reactions and the formula:

Ecell = Eox + Ered.

Reduction:
Cr2O7{2-} + 14H{+} + 6e- ---> 2Cr{3+} + 7 H2O Ered= 1.33V

Cr{3+} + e- ---> Cr{2+} Ered = -0.50V

Since in my equation we're actually oxidizing Cr{2+} to makeCr{3+}, it's reduction value needs to be flipped, making ittherefore 0.50V.

So my formula is:

Ecell= 0.50V + 1.33V
Ecell= 1.83V

Correct?

This is where I'm like, huh? How?

Obviously I need to use the formula:

DeltaG = -nFEcell

n = 6
F = in my book uses 96,500, but I want the answer in kJ, so96.5
Ecell = 1.83V

When I do the math, I get -1059.6 kJ. But, the answer in the backof the book states it as 1059.6 kJ. How are they getting it aspositive? I don't see how - (6*96.5*1.83) comes out positive...

Nonetheless, I assume they calculate (how I don't know) Ecell asbeing -1.83 so I use that in next part.

log K = Ecell * n/0.0592

n = 6
K comes out to be some RIDICULOUS number like 3.2 x 10^-186 orsomething like that. Can't remember. Obviously not correct. What doI do here?

Anyone able to tell me why Grxn is positive and not negative like Icalculated? And then how to get K?

Thank you. Sorry for such a long post. I figure, the more detail,the better someone can help me.

Explanation / Answer

I have been told by others that my calculations are correct, andthat it should be -1059.6 kJ. Are they correct? If so, then K isstill not possible because it becomes logK = 185.5, K = 10^185.5.Which is just not correct. I'm stumped. Somebody PLEASE help. Willrate lifesaver for sure.

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