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Electrochemistry - Nernst Equation

1. Homework Statement
Electrochemistry - Nernst Equation

Pb2+ + 2 e- → Pb (s) ξo = -0.13 V
Ag+ + 1 e- → Ag (s) ξo = 0.80 V

What is the voltage, at 298 K, of this voltaic cell starting with the
following non-standard concentrations:

[Pb2+] (aq) = 0.08 M
[Ag+] (aq) = 0.5 M

2. Homework Equations
Use the Nernst equation:

ξ = ξo - (RT/nF) ln Q

3. The Attempt at a Solution
there are 2 parts to this question one is to find Q and then use that to
find E.
ξo=0.80-(-0.13)=0.93V

first I balanced the equation:

2(Ag+ + 1 e- → Ag (s))
Pb(s)→ Pb2+ + 2 e-
--------------------
2Ag+ + Pb(s) --> Pb2+ + 2Ag(s)

Q = [products]^p/[reactants]^r
so Q = 0.08/0.5^2 = 0.32

i used ξ = ξo - (RT/nF) ln Q
ξ = 0.93V - ((8.314)(298K)/(2)(96500)) ln0.32 = 0.9446V

I also used E=Eo-(0.05916/n)logQ
E = 0.93V - (0.05916/2)log0.32= 0.9446

I am down to my last try and hoping if anyone can double check this for me. thanks!
 
Last edited:

Answers and Replies

458
0
How can your answer be negative when the ln of a fraction is a negative, and subtracting a negative number from a positive number should make your answer positive?
 
o thanks i edited it, i just typed in the - sign for some reason XD, but does the process seem correct?
 
458
0
Yes.
 

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