How Do You Calculate Expected Cell Voltage in Electrochemistry?

  • Thread starter Thread starter leaf345
  • Start date Start date
  • Tags Tags
    Electrochemistry
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 5K views
leaf345
Messages
9
Reaction score
0
Hey guys, I just did a lab on electrochem but we haven't started learning this in our lectures so I'm not too confident on it. One of the lab question asks us to calculate the expected cell voltage. From reading the textbook, this is what I did:
The reaction was 2Ag+ + Cu -> 2Ag + Cu2+
I added reduction potentials of Ag and Cu to get 0.46V for the standard voltage. Then I used the nersnt eq'n
E= 0.46 -(0.05915/n)*logQ
What I'm not sure about is...
We used 0.1 M AgNO3 and Cu(NO3)2, so would n be 2 or 0.2?
And would Q= [0.1 Cu2+][1]/[0.1 Ag+]^2*[1]=10?
Help would be appreciated!
 
Physics news on Phys.org
Also, we did another cell that is giving me trouble too.
One 1/2 cell was:
Ag electrode
0.1 M AgNO3 solution

The other was:
Pt electrode
0.1 M Fe2+ and 0.1 M Fe3+

I think the overall redox rx'n was Ag+ + Fe2+ --> Fe3+ + Ag
so I calculated the expected cell voltage. But I get a value that is opposite to the sign of the voltage we observed.
What I did was
E=Eo -(0.05915/n)*logQ
where Eo= 0.8-0.77 because that is the difference between the redox potentials of Ag and Fe.
n=1 electron
Q= [Fe3+]/[Fe2+]*[Ag+] = 0.1/(0.1*0.1)=10
so E= negative something

Am I doing this one correct?