Complete Question
The complete question is shown on the first uploaded image Â
Answer:
a
 From  the saturated side to the '0.10 M ' side
b
 The diluted [tex]Ag^+[/tex] is  [tex][Ag^+] = 4.45*10^{-12} \ M[/tex]
c
 The experimental values of [tex]k_{sp}[/tex] is  [tex]K_{sp} = 4.45 *10^{-13}[/tex] Â
d
  [tex]\Delta G ^o = 70.478 \ K J /mol[/tex]
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Explanation:
  From the question we are told that
    The concentration of [tex]Ag^+ _{(aq)}[/tex] is  [tex][Ag_{(aq)}] = 0.10 M[/tex]
    The experiment value is [tex]E_e = 0.603 V[/tex]
     The value of the  Nernst plot is [tex]E_n = - 0.0591 \ V[/tex]
     The concentration of the halide is [tex][Br] = 0.10 M[/tex]
From the question we are told that one side has a saturated [tex]AgBr[/tex] and the other side has a ion 0.10M  Ag+(aq) this will cause a difference in potential causing the electron to flow from the the saturated side to the side with Ag+(aq) in other to reduce  it
 The equation for the reaction is given as.
    [tex]AgBr_{aq} + e^- \to Ag_{s} + Br^-_{aq}[/tex]   [tex]E^o = 0.07 V[/tex]
    [tex]AgBr_{aq} + e^- \to Ag_{s} + Br^-_{aq}[/tex]   [tex]E^o = 0.80 V[/tex]
This values are constant potential  data for standard reduction of AgBr
   The overall equation of reaction in the cell is Â
    [tex]Ag ^+ + Br^- \to AgBr _{(aq)}[/tex]
[tex]E_{cell} = 0.80 - 0.07[/tex]
   [tex]E_{cell} = 0.73 V[/tex]
This potential for the cell can also be represented as
    [tex]E_{cell} = \frac{0.0591}{n_e} log \frac{1}{k_{sp} }[/tex]
Where  is the number of moles of electron which 1 from the chemical equation
[tex]k_{sp}[/tex] is the solubility product which is mathematically represented as
     [tex]k_{sp} = [Ag^+][Br^-][/tex]
So Â
  [tex]0.73 = \frac{0.0591}{n_e} log \frac{1}{k_{sp} }[/tex]
=> Â [tex]log \frac{1}{K_{sp}} = \frac{0.73}{0.0591}[/tex]
=> Â [tex]K_{sp} = 4.45 *10^{-13}[/tex] Â
 Now
   [tex]k_{sp} = [Ag^+][Br^-][/tex]
So
   [tex][Ag^+] = \frac{k_{sp} }{[Br^-]}[/tex]
   [tex][Ag^+] = \frac{4.45*10^{-13} }{0.10}[/tex]
    [tex][Ag^+] = 4.45*10^{-12} \ M[/tex]
The Free energy ΔG° is mathematically represented as
   [tex]\Delta G ^o = RTln k_{sp}[/tex]
Where R is the gas constant with value  [tex]R = 8.314 J/K \cdot mol[/tex]
    So
        [tex]\Delta G ^o = -8.314 * 298 * ln (4.45*10^{-13})[/tex]
            [tex]\Delta G ^o = 70478.3 J /mol[/tex]
Converting to KJ
            [tex]\Delta G ^o = 70.478 \ K J /mol[/tex]
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