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**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!

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