Suppose that the concentrations of NaF and KCI were each 0.14 M in the cell Pb(s

Question

Suppose that the concentrations of NaF and KCI were each 0.14 M in the cell Pb(s) | PbF_2(s) | F^-(aq) || Cl^-(aq) | AgCI(s) | Ag(s). (a) Using the half-reactions 2 AgCI(s) + 2 e^- 2 Ag(s) + 2 Cl^-(aq) and PbF_2(s) +2 e^- Pb(s) + 2 F^-(aq), calculate the cell voltage. (Assume that the Nernst potential is 0.05916 V.) V (b) In which direction do electrons flow? Electrons will flow clockwise through the wire from the left half-cell to the right half-cell. Electrons will flow counterclockwise through the wire from the right half-cell to the left half-cell. (c) Now calculate the cell voltage by using the reactions 2 Ag^+(aq) + 2 e^- 2 Ag(s) and Pb^2+(aq) + Pb(s). (Assume K_sp of PbF_2 = 3.6 times 10^-8, K_sp) of AgCl = 1.8 times 10^-10.) V

Explanation / Answer

Suppose that the concentrations of NaF and KCl were each 0.14 M in the cell:

Pb(s)|PbF2(s)|F- (aq)||Cl- (aq)|AgCl(s)|Ag(s).

a) Using the half-reactions

2AgCl(s) + 2e- 2Ag(s) + 2Cl- and

PbF2(s) + 2e- Pb(s) + 2F- , calculate the cell voltage.

E = Ered – Eox (note both quantities are calculated as reduction potentials) with

Ered = 0.222 V – (0.05916/2)log[Cl- ] 2

= 0.222 V + 0.05916 V = 0.281 V

Eox = -0.350 V – (0.05916/2)log[F- ] 2 = -0.350 V - (0.05916/2)log(0.10)2 = -0.350 + 0.05916

or Eox = -0.291 V

E = Ered – Eox = 0.281 V – (-0.291 V) = 0.572 V

======================================================

b) in which direction do electrons flow?

OPTOPN A

The electrons are supplied by the oxidation reaction or

from the Pb/PbF2 electrode and go to the Ag/AgCl electrode.

====================================================

c) Now calculate the cell voltage by using the reactions

2Ag+ + 2e- 2Ag(s)

Pb2+ + 2e- Pb(s).

[Ag+ ] and [Pb2+]. [Ag+ ] = Ksp/[Cl- ] = 1.8 x 10-10/0.10

M = 1.8 x 10-9 M

[Pb2+] = 3.6 x 10-8/(0.1)2 = 3.6 x 10-6 M

E = Ered – Eox with Ered =0.799 V – (0.05916/2)log(1/[Ag+ ] 2 ) =0.799 – 0.517 V = 0.282 V

Eox = -0.126 V – (0.05916/2)log(1/[Pb2+]) = -0.287 V

E = Ered – Eox = 0.282 V – (-0.287 V) = 0.569 V

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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