How Do You Calculate Expected Cell Voltage in Electrochemistry?

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SUMMARY

The discussion focuses on calculating expected cell voltage in electrochemistry using the Nernst equation. The reaction analyzed is 2Ag+ + Cu → 2Ag + Cu2+, with a standard voltage of 0.46V derived from the reduction potentials of silver and copper. The Nernst equation, E = 0.46 - (0.05915/n) * logQ, is applied, where Q is calculated based on the concentrations of the reactants and products. Additionally, a second cell involving Ag and Fe ions is examined, leading to confusion regarding the sign of the calculated voltage compared to the observed value.

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  • Knowledge of standard reduction potentials
  • Basic concepts of concentration and reaction quotient (Q)
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leaf345
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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!
 
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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?
 

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