Question

A silver concentration cell is set up at 25°C as shown below:

The AgCl(s) is in excess in the left compartment.
a. Label the anode and cathode, and describe the direction of the electron flow.
b. Determine the value of K_sp for AgCl at 25°C.

Short answer

(a) Anode, cathode, and the direction of electron flow can be labelled as follows:

(b) The value of K_sp for AgCl at 25°C is $6.45\times {10}^9$

Step-by-step solution

Labelling of anode and cathode

Anode and cathode can be labelled as:

Direction of the electron flow

The ability of species to undergo reduction is called reduction potential. The reduction electrode potential of Ag⁺ should be higher than AgCl. Thus, it undergoes oxidation and reduction. In the left compartment, oxidation occurs and acts as an anode, whereas the right compartment acts as a cathode with reduction. The flow of electrons is from oxidized to reduced part.

Determining the value of Ksp for AgCl at 25°C

The overall reaction can be written as:

$(A{g}^{+}{)}_{c1}\left(aq\right)+C{l}^{-}\left(aq\right)+(A{g}^{+}{)}_{c2}\left(aq\right)\to AgCl\left(s\right)+Ag\left(s\right)+e^{-}$

The overall cell potential can be calculated using,

${E^o}_{cell}={E^o}_{cathode}-{E^o}_{anode}$

On substituting the values,

$$\begin{gathered} {E^o}_{cell}=0.8 V-0.22 V \\ =0.58 V \end{gathered}$$

$$\begin{gathered} E_{cell}={E^o}_{cell}-\frac{0.0591}{n}\log K_{sp} \\ \log K_{sp}=\frac{n({E^o}_{cell}-E_{cell})}{0.0591} \\ \log K_{sp}=\frac{(0.58-0)}{0.0591} \\ \log K_{sp}=9.81 \\ {K}_{sp}=6.45\times {10}^9 \end{gathered}$$