Question

Calculate the activity of KI, x, in the following electrochemical cell if the potential is +0.350 V. Ag|AgCl(s), NaCl(aq,a=0.1)||KI(aq,a=x), I2(s) | Pt

Answer 1

Answer: The activity KI for the given electrochemical cell is 0.0310

Explanation:

The given chemical cell follows:

$Ag|AgCl(s),NaCl(aq.,a=0.1)||KI(aq,a=x)|I_2(s)|Pt$

Oxidation half reaction: $Ag(s)+Cl^-(aq.,)\rightarrow AgCl(s)+e^-;E^o_{AgCl/Cl^-}=0.22V$

Reduction half reaction: $\frac{1}{2}I_2(s)+e^-\rightarrow I^-(aq.);E^o_{I_2/I^-}=0.54V$

Net cell reaction: $Ag(s)+Cl^-(aq.)+\frac{1}{2}I_2(s)\rightarrow AgCl(s)+I^-(aq.)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.54-(0.22)=0.32V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{a_{I^-}}{a_{Cl^-}}$

where,

$E_{cell}$ = electrode potential of the cell = 0.350 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.32 V

n = number of electrons exchanged = 1

$a_{Cl^-}=0.1$

$a_{I^-}=x$

Putting values in above equation, we get:

$0.350=0.32-\frac{0.059}{1}\times \log(\frac{x}{0.1})$

$x=0.0310$

Hence, the activity KI for the given electrochemical cell is 0.0310