A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C .

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 5.00×102 M and 1.40 M , respectively.

a. What is the initial cell potential? (Express your answer using two significant figures)

b. What is the cell potential when the concentration of Cu2+ has fallen to 0.230 M ? (Express your answer using two significant figures)

c. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V ? (Enter your answers numerically separated by a comma. Express your answer using two significant figures)

Explanation / Answer

Pb---> Pb+2 +2e- E0= 0.13V

Cu+2+2e---> Cu Eo=0.34V

Pb+Cu+2----> Pb+2+Cu E0=0.34+0.13=0.47V

E = E0 - (0.5916/z)*log10([Pb^2+]/[Cu^2+]

z=2 and [Pb+2] =5*10-2 and Cu+2= 1.4M

E= 0.47- (0.5916/2)log (5*10-2/1.4)=0.898V

b) when Cu+2 drops to 0.23M

E= 0.47- 0.2958*log (5*10-2/0.23)=0.66V

C) 0.37= 0.47- 0.2958*log(Pb+2/Cu+2)

-0.1= -0.2958 log (Pb+2/Cu+2)

pb+2/Cu+2= 2.18

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