Consider the electrochemical reaction between Ag (s) and `Cl_(2) (g)` electrodes in 1 litre of 0. 1 M KCl aqueous solution . Solubility product of AgCl is `1.8 xx 10^(-10)` and F = 96500 C/mol . At 1 `mu A` current , calculate the time required to start observing the AgCl precipitation in the galvanic cell .

Options

A. 173 s
B. 346 s
C. `1.25 xx 10^(6)` s
D. `1.25 xx 10^(5) s`

Explanation

To solve the problem, we need to calculate the time required to start observing the precipitation of AgCl in the galvanic cell when a current of 1 μA is passed. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction and Conditions
The electrochemical reaction involves the dissociation of AgCl into Ag⁺ and Cl⁻ ions. We are given the solubility product (Ksp) of AgCl and the concentration of Cl⁻ ions from the KCl solution.

### Step 2: Calculate the Concentration of Ag⁺ Ions
Using the solubility product expression:
\[ K_{sp} = [Ag^+][Cl^-] \]
Given:
- \( K_{sp} = 1.8 \times 10^{-10} \)
- \( [Cl^-] = 0.1 \, M \)

We can rearrange the equation to find the concentration of Ag⁺ ions (S):
\[ S = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.1} = 1.8 \times 10^{-9} \, M \]

### Step 3: Calculate the Mass of AgCl that will Precipitate
To find the mass (W) of AgCl that corresponds to the concentration of Ag⁺ ions, we use the formula:
\[ W = S \times V \times M_{AgCl} \]
Where:
- \( V = 1 \, L \) (volume of the solution)
- \( M_{AgCl} = 143.5 \, g/mol \)

Calculating W:
\[ W = (1.8 \times 10^{-9} \, mol/L) \times (1 \, L) \times (143.5 \, g/mol) \]
\[ W = 2.58 \times 10^{-7} \, g \]

### Step 4: Use Faraday's Law to Calculate Time
According to Faraday's first law of electrolysis:
\[ W = Z \times I \times t \]
Where:
- \( Z = \frac{M_{AgCl}}{n} \)
- \( I = 1 \, \mu A = 1 \times 10^{-6} \, A \)
- \( n = 1 \) (since 1 electron is involved in the reduction of Ag⁺ to Ag)

Calculating Z:
\[ Z = \frac{143.5 \, g/mol}{96500 \, C/mol} = 0.001484 \, g/C \]

Substituting into Faraday's law:
\[ 2.58 \times 10^{-7} = 0.001484 \times (1 \times 10^{-6}) \times t \]

### Step 5: Solve for Time (t)
Rearranging the equation to solve for t:
\[ t = \frac{2.58 \times 10^{-7}}{0.001484 \times 1 \times 10^{-6}} \]
\[ t = \frac{2.58 \times 10^{-7}}{1.484 \times 10^{-9}} \]
\[ t \approx 173 \, seconds \]

### Final Answer
The time required to start observing the AgCl precipitation in the galvanic cell is approximately **173 seconds**.