How would you obtain a sol of Agl, the particles of which migrate toward cathode under the electric field ?

Options

A. By adding little excess of `KI` to `AgNO_(3)` solution
B. By adding little excess of `AgNO_(3)` to `KI` solution
C. By mixing equal volumes of `0.010 M AgNO_(3)` and `0.010 M KI`
D. None of these

Explanation

To obtain a sol of AgI (silver iodide) in which the particles migrate toward the cathode under an electric field, follow these steps:

### Step-by-Step Solution:

1. **Understand the Requirement**:
- We need to prepare a sol of AgI where the particles are positively charged. In an electric field, positively charged particles will migrate towards the cathode.

2. **Choose the Reactants**:
- The two main reactants required are silver nitrate (AgNO3) and potassium iodide (KI).

3. **Prepare the Reaction Mixture**:
- Take a solution of silver nitrate (AgNO3) and add it to a solution of potassium iodide (KI). It is important to use AgNO3 in slight excess to ensure complete reaction.

4. **Observe the Reaction**:
- When AgNO3 reacts with KI, a precipitation reaction occurs, forming silver iodide (AgI) as a solid. The reaction can be represented as:
\[ \text{AgNO}_3 (aq) + \text{KI} (aq) \rightarrow \text{AgI} (s) + \text{KNO}_3 (aq) \]

5. **Formation of Colloidal Sol**:
- The AgI particles that form will adsorb silver ions (Ag⁺) onto their surface. This adsorption of positive ions gives the AgI particles a net positive charge.

6. **Stabilization of Colloidal Sol**:
- To maintain the stability of the colloidal sol, ensure that the concentration of AgNO3 is sufficient to keep the Ag⁺ ions in solution, which helps to prevent the AgI particles from aggregating.

7. **Finalizing the Sol**:
- The resulting sol of AgI will have positively charged particles that will migrate towards the cathode when an electric field is applied.