Respuesta :
Answer: The activity KI for the given electrochemical cell is 0.0310
Explanation:
The given chemical cell follows:
[tex]Ag|AgCl(s),NaCl(aq.,a=0.1)||KI(aq,a=x)|I_2(s)|Pt[/tex]
Oxidation half reaction: [tex]Ag(s)+Cl^-(aq.,)\rightarrow AgCl(s)+e^-;E^o_{AgCl/Cl^-}=0.22V[/tex]
Reduction half reaction: [tex]\frac{1}{2}I_2(s)+e^-\rightarrow I^-(aq.);E^o_{I_2/I^-}=0.54V[/tex]
Net cell reaction: [tex]Ag(s)+Cl^-(aq.)+\frac{1}{2}I_2(s)\rightarrow AgCl(s)+I^-(aq.)[/tex]
Oxidation reaction occurs at anode and reduction reaction occurs at 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]
Putting values in above equation, we get:
[tex]E^o_{cell}=0.54-(0.22)=0.32V[/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{a_{I^-}}{a_{Cl^-}}[/tex]
where,
[tex]E_{cell}[/tex] = electrode potential of the cell = 0.350 V
[tex]E^o_{cell}[/tex] = standard electrode potential of the cell = +0.32 V
n = number of electrons exchanged = 1
[tex]a_{Cl^-}=0.1[/tex]
[tex]a_{I^-}=x[/tex]
Putting values in above equation, we get:
[tex]0.350=0.32-\frac{0.059}{1}\times \log(\frac{x}{0.1})[/tex]
[tex]x=0.0310[/tex]
Hence, the activity KI for the given electrochemical cell is 0.0310
