Question

A voltaic cell represented by the following cell diagram has Ecell= 1.409 V . Zn(s)|Zn2+(1.00M)||Ag+(saturated Ag2X)|Ag(s) What must be the Ksp of Ag2X? Use the following standard electrode potentials: Zn2+(aq) + 2e- →Zn(s) E° = -0.764 V Ag+(aq) + e- →Ag(s) E° = 0.799 V To enter your answer, multiply your Ksp by 1x109 then enter it to 2 decimal places.

Answer 1

Let , [Ag⁺ ] = x M  

Zn (s) + 2 Ag⁺ (x M) Zn²⁺ (1.00 M) + 2 Ag (s)

cathode reaction is Ag⁺ + e Ag

anode reaction  is Zn - 2e   Zn²⁺  

hence, E⁰_cell = E⁰_{cathode -} E⁰_anode = 0.799 - ( - 0.764)

= 1.563 V

Nernst equation is

E_cell = E⁰_cell -  log

1.409 = 1.563 - log

or, 1.409 - 1.563 = - log

or, 0.154 = log

or, log   = 5.211

or, = 10^5.211 = 162744

or, x² = (1.00/ 162744)

or x ² = 6.144 *10^-6

or, x = 0.00248 M  

so. [Ag⁺] = 0.00248 M

Ag₂X 2Ag²⁺ + X^2-

so, [X^2-] =  [Ag⁺]/2 = (0.00248/2) = 0.00124 M

now

Ksp of Ag₂X

= [Ag⁺]²[X^2-]

= (0.00248)² * (0.00124)

= 7.62 *10^-9

so final answer is

or Ksp = 7.62 *10^-9 *10⁹ = 7.62