How Do You Calculate Expected Cell Voltage in Electrochemistry?

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To calculate the expected cell voltage in electrochemistry, the standard cell potential is determined by adding the reduction potentials of the half-reactions. For the reaction involving silver and copper, the standard voltage calculated was 0.46V. The Nernst equation is then applied, but clarification is needed on the value of n and the reaction quotient Q. In a second experiment with silver and iron, the calculated voltage was found to have an opposite sign compared to the observed voltage, raising questions about the accuracy of the calculations. Overall, understanding the correct application of the Nernst equation and the values for n and Q is crucial for accurate voltage predictions.
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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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