Suppose that the concentrations of NaF and KCI were each 0.30 M in the following

Question

Suppose that the concentrations of NaF and KCI were each 0.30 M in the following cell. (The K_sp's of PbF_2 and AgCI are 3.6 Times 10^-8 and 1.8 Times 10^-10 respectively.) Pb(s) | PbF_2(s) | F^-(aq) || Cl^- (aq) | AgCl(s) | Ag(s) Using the following half-reactions, calculate the cell voltage. 2 AgCI(s) + 2 e^- 2 Ag(s) + 2 Cl^- PbF_2(s) + 2e6- Pb(s) + 2F^- By the reasoning in this figure, in which direction do electrons flow? Now calculate the cell voltage by using the following reactions. 2 Ag^+ 2 e^- 2 Ag(s) Pb^2+ + 2e^- Pb(s) For this part, you will need the solubility products for PbF_2 and AgCI.

Explanation / Answer

Cathode reaction: 2AgCl + 2e = 2Ag + 2Cl-

Anode reaction : Pb + 2F- = PbF2 + 2e

Overall reaction : 2AgCl + Pb + 2F- = 2Ag + 2Cl- + PbF2

Ecell =E0cell - (0.059/2) ln [Cl-]2/[F-]2 .....[Solid reactant and products are not inserted in the equation]

= 1.242 - (0.059/2) ln (0.3)2/(0.3)2

= 1.242 V

Electrons always flow from the anode to the cathode, i.e., left to right.

Cathode reaction: 2Ag+ +2e = 2Ag

Anode reaction : Pb = Pb2+ + 2e

Overall reaction : 2Ag+ + Pb = 2Ag + Pb2+

Ecell =E0cell - (0.059/2) ln [Pb2+]/[Ag+]2

Now for PbF2, Ksp = S.(2S)2 = 3.6*10-8 ......[S= solubility]

So, SPb2+ = 2.08*10-3

Now for AgCl, Ksp = S.S = 1.8*10-10

SAg+ = 1.34*10-5

Ecell =1.242- (0.059/2) ln [2.08*10-3]/[ 1.34*10-5]2

= 0.762 V

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