Question

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.00×10−2 M and 1.50 M , respectively

What is the initial cell potential?

0.51V

What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ?

0.45V

Part C What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.360 V ?

I need help with Part C

Answer 1

The formula to be used here ( for a,b,c) is:

E_cell = E^o_cell – (RT/nF)lnQ

        = E^o_cell – 0.0296 V· logQ

        = (0.337+0.126)V - 0.0296 V· log ([Pb²⁺]/[Cu²⁺])

        = 0.463V - 0.0296 V· log ([Pb²⁺]/[Cu²⁺])

1. E_cell = 0.463V - 0.0296 V· log (0.0500/1.50) = 0.507 V
2. The concentration variation for Cu²⁺ was (1.50 – 0.200) = 1.30 M

             The new Pb²⁺ conc. is (0.050 + 1.300 = 1.35 M

           E_cell = 0.463V - 0.0296 V· log (1.35/0.200) = 0.438 V

      c.

            E_cell = 0.463 V- 0.0296 V· log ([Pb²⁺]/[Cu²⁺])

          0.360 V = 0.463V - 0.0296 V· log ([Pb²⁺]/[Cu²⁺])

          log ([Pb²⁺]/[Cu²⁺]) = 3.480

          [Pb²⁺]/[Cu²⁺] = 3.020 x10³ = 3020

          If x is the concentration variation for each ion, then:

           ( 0.05 + x) / (1.50 –x) = 3020

            0.05 + x = 4530– 3020x

             3021 x = 4530

                      x = 1.49933

            [Pb²⁺] = 0.0500 M+ 1.499 M = 2.00 M

            [Cu²⁺] = 1.500 - 1.49933 = 0.00066 = 6.7x10^-4M    

An approximate, but simple alternative solution:

For a ratio [Pb²⁺]/[Cu²⁺] = 3.020 x10³, suppose Cu²⁺ is quite consumed, thus

[Pb²⁺] = 0.0500 M+ 1.50 M = 2.00 M

and [Cu²⁺] = 2.00 M / 3020 = 0.00066 M