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 5.30×102 M and 1.50 M , respectively. A-What is the cell potential when the concentration of Cu2+ has fallen to 0.210 M ? Express your answer using two significant figures.

B- 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. the unit is  (M )

Explanation / Answer

A) Pb2+ + 2e- -------------------> Pb Eo = - 0.13 V

Cu2+ + 2e- --------------------> Cu Eo = + 0.34 V

-----------------------------------------------------------------------------

Pb + Cu2+ ------------> Pb2+ + Cu

Eocell = + 0.34 V - (-0.13 V) = + 0.47 V

no of electrons transferred n = 2

Ecell = Eocell - (0.059/n) log [Pb2+]/[Cu2+]

= 0.47 - (0.059/2) log [5.30×102 M]/[ 0.210 M]

= + 0.487 V

Ecell = + 0.487 V

Therefore,

cell potential = + 0.487 V

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.