Respuesta :
Answer: The ratio of [tex][Cu^{2+}]/[Cu^+][/tex] in an aqueous solution is [tex]5.75\times 10^{-35}M[/tex]
Explanation:
The substance having highest positive [tex]E^o[/tex] potential will always get reduced and will undergo reduction react.
Oxidation half reaction: [tex]Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-;E^o_{Cu^{2+}/Cu}=+0.34V[/tex]
Reduction half reaction: [tex]Cu^+(aq.)+e^-\rightarrow Cu(s);E^o_{Cu^+/Cu}=+0.13V[/tex] ( × 2)
Net cell reaction: [tex]2Cu^+(aq.)\rightarrow Cu^{2+}(aq.)+Cu(s)[/tex]
Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
To calculate the [tex]E^o_{cell}[/tex] of the reaction, we use the equation:
[tex]E^o_{cell}=E^o_{cathode}-E^o_{anode}[/tex]
[tex]E^o_{cell}=0.13-(0.34)=-0.21V[/tex]
To calculate the EMF of the cell, we use the Nernst equation, which is:
[tex]E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Cu^{2+}]}{[Cu^{+}]^2}[/tex]
Concentration of pure solids and liquids are taken as 1 in the ratio
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = +0.80 V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = -0.21 V
n = number of electrons exchanged = 2
[tex]\frac{[Cu^{2+}]}{[Cu^+]^2}[/tex] = Ratio of the concentration
Putting values in above equation, we get:
[tex]0.80=-0.21-\frac{0.059}{2}\times \log(\frac{[Cu^{2+}]}{[Cu^{+}]^2})\\\\\frac{[Cu^{2+}]}{[Cu^{+}]^2}=5.75\times 10^{-35}M[/tex]
Hence, the ratio of [tex][Cu^{2+}]/[Cu^+][/tex] in an aqueous solution is [tex]5.75\times 10^{-35}M[/tex]
