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 at the anode

Using the Electrolysis with Active Anodes knowledge point

$$ \text{Species present: } \text{Ag}(s), \text{Cl}^-(aq), \text{H}_2\text{O}(l) $$

Determine the oxidation reaction

Using the Electrolysis with Active Anodes knowledge point

$$
LATEXBLOCK0
$$

$$ \text{Ag}(s) \text{ oxidizes preferentially due to its lower oxidation potential.} $$

Formulate the precipitation reaction

Using the Precipitation Reactions knowledge point

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

Answer:

During the electrolysis of a \(1.0\text{ M }\text{KCl}\) solution using active silver (\(\text{Ag}\)) electrodes, the silver anode undergoes oxidation preferentially over water and chloride ions because it has a lower oxidation potential:

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

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

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