Question

part ii

1. when a 1.0 m kcl solution is electrolyzed using silver electrodes, a precipitate forms at the anode. explain this result.

Explanation:

Identify species present in the system

In the aqueous \(1.0\text{ M }\mathrm{KCl}\) solution with silver (\(\mathrm{Ag}\)) electrodes, the species present at the anode are:

• Chloride ions: \(\mathrm{Cl^-}\)
• Water molecules: \(\mathrm{H_2O}\)
• Silver metal (active anode): \(\mathrm{Ag(s)}\)

Determine the reaction at the anode

The anode is the site of oxidation. We compare the standard reduction potentials (\(E^\circ\)) of the possible oxidation reactions to find which species is most easily oxidized (lowest reduction potential):

• Oxidation of \(\mathrm{Ag}\):

$$\mathrm{Ag(s) ightarrow Ag^+(aq) + e^-} \quad (E^\circ_{\text{red}} = +0.80\text{ V})$$

• Oxidation of \(\mathrm{Cl^-}\):

$$\mathrm{2Cl^-(aq) ightarrow Cl_2(g) + 2e^-} \quad (E^\circ_{\text{red}} = +1.36\text{ V})$$

• Oxidation of \(\mathrm{H_2O}\):

$$\mathrm{2H_2O(l) ightarrow O_2(g) + 4H^+(aq) + 4e^-} \quad (E^\circ_{\text{red}} = +1.23\text{ V})$$

Since silver metal has the lowest reduction potential, it is the easiest to oxidize. Thus, the silver anode undergoes oxidation:

$$\mathrm{Ag(s) ightarrow Ag^+(aq) + e^-}$$

Explain the formation of the precipitate

As \(\mathrm{Ag^+(aq)}\) ions are produced at the anode, they immediately encounter the high concentration of \(\mathrm{Cl^-(aq)}\) ions from the dissolved \(\mathrm{KCl}\) in the surrounding solution.

Silver chloride (\(\mathrm{AgCl}\)) is highly insoluble in water. The ions combine to form a white precipitate:

$$\mathrm{Ag^+(aq) + Cl^-(aq) ightarrow AgCl(s)}$$

Answer:

During the electrolysis of a \(1.0\text{ M }\mathrm{KCl}\) solution using active silver (\(\mathrm{Ag}\)) electrodes, the silver anode is oxidized because it has a lower oxidation potential than water or chloride ions:

$$\mathrm{Ag(s) ightarrow Ag^+(aq) + e^-}$$

The newly generated silver ions (\(\mathrm{Ag^+}\)) immediately react with the chloride ions (\(\mathrm{Cl^-}\)) present in the solution to form an insoluble white precipitate of silver chloride (\(\mathrm{AgCl}\)):

$$\mathrm{Ag^+(aq) + Cl^-(aq) ightarrow AgCl(s)}$$