Explanation:
Expression for [tex]K_{sp}[/tex] of the given reaction is as follows.
[tex]K_{sp} = [Cu^{2+}]^{2}[Fe(CN)_{6}][/tex]
Let us assume that the concentration of given species is "s". As the value of [tex]K_{sp}[/tex] is given as [tex]1.3 \times 10^{-16}[/tex].
[tex]K_{sp} = [Cu^{2+}]^{2}[Fe(CN)_{6}][/tex]
[tex]1.3 \times 10^{-16} = s^{2} \times s[/tex]
[tex]s^{3} = 1.3 \times 10^{-16}[/tex]
s = [tex]3.19 \times 10^{-6}[/tex]
Therefore, concentration of [tex]Cu^{2+}[/tex] will be calculated as follows.
[tex]Cu^{2+}[/tex] = 2s
= [tex]2 \times 3.19 \times 10^{-6}[/tex]
= [tex]6.38 \times 10^{-6}[/tex] M
Now, we will calculate the value of [tex]E_{cell}[/tex] as follows.
[tex]E_{cell} = E^{o}_{cell} - \frac{0.0591}{2} \times log \frac{6.38 \times 10^{-6}}{1}[/tex]
= [tex]0 - \frac{0.0591}{2} \times log \frac{6.38 \times 10^{-6}}{1}[/tex]
= 0.1535 V
Thus, we can conclude that the potential of given cell is 0.1535 V.
