Posted by Mima on Sunday, April 29, 2012 at 1:20pm.
Calculate the cell potential for the following reaction as written at 79C given that [Zn^+2] = 0.819 M and [Ni^+2] = 0.014 M.
Zn + Ni^+2 = Zn^+2 + Ni
I know: E=E RT/nF ln Q
reduction potential for Ni= 0.26
reduction potential for Zn = 0.76
so E standard potential = 0.5
E= 0.5  (8.315)(352)/(2)(96480) ln 0.819/0.014.
I'm getting E= 1.989 and my answer is wrong.

Chemistry  DrBob222, Sunday, April 29, 2012 at 1:55pm
The set up looks ok but I would use 8.314 for R and 96,485 for F. If I run through that I obtain 0.438.

Chemistry  Mima, Sunday, April 29, 2012 at 2:16pm
E= 0.438?
I got the same answer (1.987) when I used the numbers you mentioned. Is the E standard potential = 0.26 (0.76)?
Am I doing that right?(its cathodeAnode) right?

Chemistry  DrBob222, Sunday, April 29, 2012 at 2:55pm
Ecell = 0.5v. You are correct. I plugged in your numbers and obtained 0.438 with them, also. This is like looking for a needle in haystack. The BEST think to have done was type in your work and let someone look at it. As it is both are floundering. Here is your work.
E= 0.5  (8.315)(352)/(2)(96480) ln 0.819/0.014.
In pieces:
8.314*352/(2*96480) = 0.01517
0.819/0.014 = 58.5
ln 58.5 = 4.060
0.01517*4.069 = 0.0617 and
0.50.0617 = 0.438
Check your work. One of us is punching the wrong keys.

Chemistry  Mima, Sunday, April 29, 2012 at 6:46pm
Got it, you are right, it's 0.439. Thank you so much, I was doing an easy math mistake (substracting before multiplying). Thanks again.
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