Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Expected Cell Voltage (Galvanic Cell)?
What is the expected cell voltage for:
Zn | 0.10M ZnSO4 || 0.10M AgNO3 | Ag
E = -0.76V for Zn -> 2e + Zn2+
E = 0.80V for Ag+ + e -> Ag
How do I do this? Do I just add up the two numbers?
3 Answers
- A.S.Lv 79 years agoFavorite Answer
you have not mentioned the temperature so i am assuming it as 298 K (25 degree C)
the cell representation shows that Ag is getting reduced at cathode (right compartment)
and Zn is getting oxidised at anode (left compartment)
at cathode : Ag+ + e- ------> Ag ....Eo = 0.80 V.....(1)
Zn ------> Zn2+ + 2e- .....Eo = -0.76 V.....(2)
multiplying reaction (1) by 2 and adding in (2)
Zn + 2Ag+ + 2e- ------> Zn2+ + 2e- + 2Ag
2e- will cancel out....
Zn + 2Ag+ ------> Zn2+ + 2Ag
Eo(cell) = Eocathode - Eoanode = 0.8 - -0.76 = 1.56 V
now this would have been the e.m.f. of cell if every species were at 1 M.....but since their conc. is different so we will have to use the nernst equation...
E = Eo(cell) - RT/nF ln [Zn2+] / [Ag+]^2
where R = 8.314 j/K/mole
T = 298 K
n = no. of electrons invoved in electrode reaction = 2
F = 96500 coulombs
[Zn2+] = 0.1
[Ag+] = 0.1
Eo = 1.56
putting the values..
E(cell) = 1.56 - (8.314 X 298)/ (2X96500) ln 0.1/0.1^2
E(cell) = 1.56 - 0.0128 X ln 0.1/0.01
E(cell) = 1.56 - 0.0128 X ln 10
E(cell) = 1.56 - 0.0295
E(cell) = 1.5305 V
please chck the maths..
see the examples given in this link---
http://www.science.uwaterloo.ca/~cchieh/cact/c123/...
feel free to ask any questions...
