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.

Question

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.

Answer 1

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.

Answer 2

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 cathode-Anode) right?

Answer 3

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
0.5-0.0617 = 0.438
Check your work. One of us is punching the wrong keys.

Answer 4

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.