1 Answer

we must first learn about electrolytes.

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

Use the Nernst equation:

ξ = ξo - (RT/nF) ln Q

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.... hoping if you can double check this for me. thanks!