Calculate Eo potential for voltaic cell?

Question

Hello, i am preparing for an exam, and i would like help with this question.

Calculate the cell potential, Eo, for a voltaic cell composed of one compartment containing a Pb(s) electrode immersed in 1.0M Pb2+(aq), and a second compartment containing a Pt(s) electrode immersed in water that is saturated with H2(g)

please show your steps, as i have the answer, but i don't know how to get it. (study guide)

Answer 1

I'll assume the H2 half-cell is also 1 M in H+. (Standard H2 half cell).

Standard reduction potentials are:

Pb2+ + 2e- ---->  Pb   Eo = -0.13 V

2H+ + 2e- ----> H2  Eo = 0.00 V

Reverse the Pb2+ half reaction and add to the 2H+ half reaction.

Pb ---->  Pb2+ + 2e-  Eo = +0.13 V

2H+ + 2e- ----> H2  Eo = 0.00 V

Pb + 2H+ ---->  Pb2+ + H2  Ecell = +0.13 V

Answer 2

The Pt eletrode is in water so you will have to use the Nernst equation to work out the actual half cell potential as the hydrogen electode is in 1M [H+] and water is 10^-7 [H+] also the Pb2+ may be 1M but it will not be even close to unit activity so an adjustment will have to be made for that also.

Depends on how deep you are expected to go with your answer.