## Winter 2013 Midterm Q7

$\Delta G^{\circ} = -nFE_{cell}^{\circ}$

Christopher Reed 1H
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### Winter 2013 Midterm Q7

Hi!

The question asks to calculate K and dG0 at 298 K for the following redox reaction:

Mn2+ + Br2 --> MnO4- +Br-

I was able to solve for dG0 by using dG0 = -nFE0 and obtained the same answer as the answer key. When I try to solve for K using dG0 = -RTln(K) I find the K is equal to 1.35x10^-72. The course reader finds that using this method yields a K of 9.2x10^-72.

The values I used were dG0 = 4.1x10^5 , R=8.314 , T=298K.

Am I doing something wrong, or is this an error in the reader?

Thanks.

Mana_Sheykhsoltan_1A
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### Re: Winter 2013 Midterm Q7

I found K by using lnK=(nE(cell))/(0.025693 V) and got the same answer as the course reader did, 1.0 x 10^-71. When I used delta(G)=-RTlnK, I got the same answer as you did, K=1.35 x 10^-72. I used the values delta(G)=4.1 x 10^5 J/mol, R=8.314 J/(K mol), T=298 K. I think the course reader may be wrong. Hope this helps!

ntruong2H
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### Re: Winter 2013 Midterm Q7

Christopher Reed 1H wrote:Hi!

The question asks to calculate K and dG0 at 298 K for the following redox reaction:

Mn2+ + Br2 --> MnO4- +Br-

I was able to solve for dG0 by using dG0 = -nFE0 and obtained the same answer as the answer key. When I try to solve for K using dG0 = -RTln(K) I find the K is equal to 1.35x10^-72. The course reader finds that using this method yields a K of 9.2x10^-72.

The values I used were dG0 = 4.1x10^5 , R=8.314 , T=298K.

Am I doing something wrong, or is this an error in the reader?

Thanks.

I used the second equation as well; however, I think you have to put in the non-rounded answer. I used the value 405,237 J/mol for dG and got the answer listed in the course reader.