Chemistry
posted by Mima .
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.

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.

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? 
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. 
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.
Respond to this Question
Similar Questions

chemistry
Calculate the cell potential for the following reaction as written at 53 °C, given that [Zn2 ] = 0.862 M and [Ni2 ] = 0.0110 M. Standard reduction and oxidation potentials can be found here. Zn(s) + Ni^2+(aq) > Zn^2+(aq) + Ni(s) 
Chemistry
Consider the following cell: PtH2(g, 0.460 atm)H (aq, ? 
CHEMISTRY
Consider the following cell: PtH2(g, 0.130 atm)H (aq, ? 
chemistry
Calculate the cell potential for the following reaction as written at 51 degrees C, given that [Zn2+] = 0.837 M and [Sn2+] = 0.0180 M. Zn(s) + Sn^2+(aq) = Zn^2+(aq)+Sn(s) E = __ V Reduction HalfReaction Standard Potential Ered° (V) … 
Chemistry
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.842 M and [Ni2 ] = 0.0100 M. Standard reduction potentials can be found here. reaction: Zn(s)+Ni^2+(aq)>Zn^2+(aq)+Ni(s) standard … 
Chemistry
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.859 M and [Ni2 ] = 0.0190 M. Standard reduction potentials can be found here. Cr+Ni(2+) >Cr(2+) + Ni 
Chemistry
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.859 M and [Ni2 ] = 0.0190 M. Standard reduction potentials can be found here. 
Chemistry
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.859 M and [Ni2 ] = 0.0190 M. Standard reduction potentials can be found here. 
Chemistry (Electrochemistry)
The standard potential for the following galvanic cell is 1.72V : Be(s)Be2+(aq)Pb2+(aq)Pb(s) The standard reduction potential for the Pb2+/Pb halfcell: E = .13V Calculate the standard reduction potential for the Be2+/Be halfcell. … 
Chemistry DR BOB
Under standard conditions of 1 atm and 298.15K, the halfcell reduction potential E,zero for the anode in a voltaic cell is 0.37V. The halfcell reduction potential E,zero for the cathode of the cell is 0.67V. The number of moles, …