Suppose that the concentrations of NaF and KCl were each 0.20 M in the following

Question

Suppose that the concentrations of NaF and KCl were each 0.20 M in the following cell. (The Ksp's of PbF2 and AgCl are 3.6 10-8 and 1.8 10-10 respectively.)

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

(a) Using the following half-reactions, calculate the cell voltage.

2 AgCl(s) + 2 e- equilibrium reaction arrow 2 Ag(s) + 2 Cl -

PbF2(s) + 2 e- equilibrium reaction arrow Pb(s) + 2 F -  

(b) Now calculate the cell voltage by using the following reactions

. 2 Ag+ + 2 e- equilibrium reaction arrow 2 Ag(s)

Pb2+ + 2 e- equilibrium reaction arrow Pb(s)

For this part, you will need the solubility products for PbF2 and AgCl. . V

Explanation / Answer

a)

AgCl(s) <---->Ag+(aq) +Cl-(aq) Ksp=1.8*10^-10

Ag+ +e ---> Ag(s) , Eo red=+0.80V

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

net rxn: AgCl(s) +e <---> Ag(s) +Cl- (aq)

Eo=RT/nF ln Ksp=0.0591 log Ksp=0.0591 log (1.8*10^-10)=-0.576

Eo(electrode)=Eored+0.0591 log Ksp=0.80-0.576=0.224

Thus electrode potential of half -cell,Cl -(aq) | AgCl(s) | Ag(s)

E=Eo-0.059/n log ([Cl-]

or, E=0.224-0.059*0.2=0.212V

Ered =0.212V

or Eo(red) (cathode)=0.212V

Now, for Pb(s) | PbF2(s) | F -(aq) electrode:

PbF2(s) <--> Pb2+(aq) +2F- (aq) , ksp=3.6*10^-6

Pb2+(aq)+2e ---->Pb(s) .Eored=-0.13V

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

net rxn: PbF2(s)+2e <--->Pb(s) +2F-(aq)

Eo (electrode)=Eo(red)+0.0591 log Ksp=-0.13V+0.0591* log (3.6*10^-8)=-0.57V

Now Eo(red) (anode) =-0.57V

Eo(red) (cathode)=0.212V

For Pb(s) | PbF2(s) | F -(aq) || Cl -(aq) | AgCl(s) | Ag(s) cell:

Ecell=Eo(red) (cathode)-Eo(red) (anode)=0.212-(-0.57)=0.782V

Ecell=0.782V

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

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