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**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! **

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