A constant current was passed through a solution of `AuCl_(4)^(c-)` ion between gold electrodes. After a period of `10.0 mi n`, the increase in the weight of cathode was `1.314g`. The total charge passed through solution is `(` atomic weight of `AuCl_(4)^(c-)=339)`

Options

A. `1.16 xx 10^(-2)F`
B. `3.5xx10^(-2)F`
C. `2xx10^(-2)F`
D. `4xx10^(-3)F`

Explanation

To solve the problem step by step, we will use Faraday's laws of electrolysis.

### Step 1: Identify the relevant information
We have the following data:
- Increase in weight of the cathode (W) = 1.314 g
- Atomic weight of AuCl₄⁻ = 339 g/mol
- Time = 10 minutes (not directly needed for the calculation of charge)
- The reduction reaction at the cathode involves Au³⁺ ions.

### Step 2: Determine the equivalent weight of gold
The equivalent weight (E) of gold can be calculated using the formula:
\[ E = \frac{\text{Atomic weight of Au}}{n} \]
where \( n \) is the number of electrons transferred in the reduction reaction. For Au³⁺ to Au, \( n = 3 \).
- Atomic weight of Au = 197 g/mol
- Therefore, equivalent weight of Au:
\[ E = \frac{197}{3} \approx 65.67 \, \text{g/equiv} \]

### Step 3: Relate weight to charge using Faraday's first law
According to Faraday's first law of electrolysis:
\[ W = \frac{E}{F} \times Q \]
where:
- \( W \) = weight deposited (1.314 g)
- \( E \) = equivalent weight of gold (65.67 g/equiv)
- \( F \) = Faraday's constant (approximately 96500 C/equiv)
- \( Q \) = total charge passed through the solution

Rearranging the formula to find \( Q \):
\[ Q = \frac{W \times F}{E} \]

### Step 4: Substitute the values into the equation
Now substituting the known values:
\[ Q = \frac{1.314 \, \text{g} \times 96500 \, \text{C/equiv}}{65.67 \, \text{g/equiv}} \]

### Step 5: Calculate \( Q \)
Calculating \( Q \):
1. Calculate the numerator:
\[ 1.314 \times 96500 \approx 126,000.1 \, \text{C} \]
2. Divide by the equivalent weight:
\[ Q \approx \frac{126000.1}{65.67} \approx 1925.5 \, \text{C} \]

### Step 6: Convert to Farads
Since we are asked for the charge in Farads, we can express it as:
\[ Q \approx 0.02 \, \text{F} \]

### Final Answer
The total charge passed through the solution is approximately:
\[ Q \approx 2 \times 10^{-2} \, \text{F} \]