Question

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

What is the initial cell potential, what is the cell potential when the concentration of Cu2+ has fallen to 0.200 V, and what are the concentrations of Pb2+ and Cu2+when the cell potential falls to 0.370 V?

I honestly have no idea what to do, any help would be appreciated! Thank you.

Answer 1

Given data Concentration, [Pb-J-5.1x10-2 M Concentration, Cu1.6 M Write the reduction half reaction at cathode Cu2+ +2e-- Cu (s) E thode-0.3 Write the oxidation half reaction at anode Pb(s)Pb2 2 , 0.13 V Calculate the standard cell potential Substitute the values in the above equation. F0at 0.34-(-0.13) = 0.47 V Write the Nemst equation to calculate the cell potential. E. 00592 b* Cu2- 10 cell- "cell Here. Ecell is the cell potential and n-2 is the number of moles of the transferred electrons Substitute the values in the above expression. 0.0592, (5.1x102 Ect 0.47- 1.6 0.47 +0.044 =0.514 V Hence, the initial cell potential is 0.514 V

Concentration. 「Cu2+-0.240 M Make the ICE table Cu(s) Pb I 1.6 C -x E 0.24 0.051 0.051+x From the ICE table 1.6-x 0.24 x= 1.36 Therefore, Concertation, Pb0.051+x 0.051 +1.36 Write the Nemst equation to calculate the cell potential cetl-lo Cu Substitute the values in the above expression. 0.0592, 1.411 0.240 cell =0.47- = 0.47-0.023 0.447 V Hence, cell potential when the concentration of Cu2 has fallen to 0.240 M is 0.447 V Given data, Cell potential, 0.360 v Make the ICE table

Cu Pb(s) Cu(s)Pb I 0.24 + Pb(s)Cu(s) Pb 1.411 E 0.24-x 1.411+x Write the Nemst equation to calculate the cell potential E 0.0592,Pb e-lo Substitute the values in the above expression 0.36 0.47-0.0592 , (1.411+x lo 0.24-х (1.411+-3.716 1.411+ 5199.96 5200.96x 1246.579 x = 0.24 Calculate the concentration of Cu [Cu*) 0.24-x Substitute the value of x in the above expression. Cu20.24-0.24 Calculate the concentration of Pb Pb-1.411+x Substitute the value of x in the above expression. Pb"--1.411+0.24 -1.651 M Hence, the concentration [Pb ]is 0.61M and the concentration [Cu ] is OM