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leaf345

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**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!