Respuesta :
Answer: The value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
[tex]AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)[/tex]
Its solubility product will be as follows.
[tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]
Cell reaction for this equation is as follows.
[tex]Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)[/tex]
Reduction half-reaction: [tex]Ag^{+} + 1e^{-} \rightarrow Ag(s)[/tex], [tex]E^{o}_{Ag^{+}/Ag} = 0.799 V[/tex]
Oxidation half-reaction: [tex]Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}[/tex], [tex]E^{o}_{AgCl/Ag}[/tex] = ?
Cell reaction: [tex]Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)[/tex]
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product, [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]
= [tex]1.77 \times 10^{-10}[/tex]
Therefore, according to the Nernst equation
[tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]
At equilibrium, [tex]E_{cell}[/tex] = 0.00 V
Putting the given values into the above formula as follows.
[tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]
[tex]0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}[/tex]
[tex]E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}[/tex]
= [tex]0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}[/tex]
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
[tex]E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}[/tex]
[tex]0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}[/tex]
[tex]0.577 V = 0.799 V - E^{o}_{AgCl/Ag}[/tex]
[tex]E^{o}_{AgCl/Ag}[/tex] = 0.222 V
Thus, we can conclude that value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.
