Complete Question
The complete question is shown on the first uploaded image
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]
d
[tex]\Delta G ^o = 70.478 \ K J /mol[/tex]
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
[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
[tex]0.73 = \frac{0.0591}{n_e} log \frac{1}{k_{sp} }[/tex]
=> [tex]log \frac{1}{K_{sp}} = \frac{0.73}{0.0591}[/tex]
=> [tex]K_{sp} = 4.45 *10^{-13}[/tex]
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]
