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145&146&147. The potential of a silver electrode is measured relative to an Ag-AgCl electrode for the...

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Chemistry

145&146&147. The potential of a silver electrode ismeasured relative to an Ag-AgCl electrode for the titration of100.0 mL of 0.100 M Cl with 0.100 M Ag+. What is the potentialafter 75.00 mL of titrant is added. Eo = 0.799 V for Ag+, E = 0.197V for the Ag-AgCl electrode and Ksp = 1.8 × 10−10.

A) 0.493 V

B) 1.070 V

C) 0.521 V

Answer & Explanation Solved by verified expert
4.0 Ratings (636 Votes)

AgCl(s)+e↔Ag(s) +Cl- Eo=0.197V (anode)

Ag+ +e↔Ag(s) Eo=0.799V (cathode)

Ecell={Eo-0.0591 log1/[Ag+]}-{Eo-0.0591 log1/[Cl-]}

But for reference cell with constant potential,

Ecell ={Eo-0.0591 log1/[Ag+]}                [since 0.0591 log1/[Cl-]}=constant as [Cl-]=constant]

Eocell=0.799V-0.197V=0.602

Ecell=0.602V-0.0591 log1/[Ag+]

Total Cl- in 100 ml=0.100M*100/1000L=0.01 moles

Now, after 75 ml titrant is added, 75% of Cl- is titrated,so only 25% of cl- remain in solution

[Cl-]=fraction in solution*original concentration*initial volume/total volume=0.25*0.1M*100/175=0.0143M

Ag++Cl-→AgCl

Ksp=1.8*10^-10=[Ag+][Cl-]

Or, 1.8*10^-10=[Ag+][Cl-]=[Ag+]*(0.0143M)

[Ag+]=1.8*10^-10/0.0143M=125.87 *10^-10=1.25*10^-8M

Ecell=0.602V-0.0591 log1/[Ag+]

        =0.602V-0.0591 log 1/(1.25*10^-8)

        =0.602V+0.0591 V*log(1.25*10^-8)

      

       =0.602V-0.0591V *(-7.91) =0.602-0.467

Ecell=0.135V
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