A curent of `3A` was passed for 1 hour through an electrolyte solution of `A_(x) B_(y)` in water. If `2.977g ` of `A(` atomic weight `106.4)` was deposited at cathode and `B` was a monovalent ion, the formula of electrolyte was

Options

A. `AB_(2)`
B. `AB`
C. `AB_(3)`
D. `AB_(4)`

Explanation

To solve the problem, we need to determine the formula of the electrolyte \( A_xB_y \) based on the information provided. Here are the steps to arrive at the solution:

### Step 1: Calculate the total charge (Q) passed through the electrolyte
The current (I) is given as 3 A and the time (t) is 1 hour. We need to convert the time into seconds:
\[ t = 1 \text{ hour} = 3600 \text{ seconds} \]
Now, we can calculate the total charge (Q) using the formula:
\[ Q = I \times t = 3 \, \text{A} \times 3600 \, \text{s} = 10800 \, \text{C} \]

### Step 2: Use Faraday's first law of electrolysis
According to Faraday's first law, the weight of the substance deposited (W) is directly proportional to the charge (Q) passed through the electrolyte:
\[ W = Z \times Q \]
Where \( Z \) is the electrochemical equivalent, which can be expressed as:
\[ Z = \frac{\text{Equivalent weight}}{96500} \]
We know that the weight of \( A \) deposited is 2.977 g and the atomic weight of \( A \) is 106.4 g/mol.

### Step 3: Calculate the equivalent weight of \( A \)
The equivalent weight (E) can be calculated using the formula:
\[ E = \frac{\text{Atomic weight}}{n} \]
Where \( n \) is the number of electrons transferred per ion. Since \( A \) is deposited at the cathode, we can assume that \( n = y \) (the valency of \( A \)).

### Step 4: Set up the equation
Substituting the values into the equation:
\[ 2.977 = \left(\frac{106.4}{y}\right) \times \frac{Q}{96500} \]
Substituting \( Q = 10800 \):
\[ 2.977 = \left(\frac{106.4}{y}\right) \times \frac{10800}{96500} \]

### Step 5: Solve for \( y \)
Rearranging the equation gives:
\[ y = \frac{106.4 \times 10800}{2.977 \times 96500} \]
Calculating the right side:
\[ y \approx \frac{1146492}{287.5} \approx 3999.99 \approx 4 \]

### Step 6: Determine the value of \( x \)
Since \( B \) is a monovalent ion, it has a valency of 1. Therefore, for the formula \( A_xB_y \):
- \( x = 1 \) (because \( B \) is monovalent)

### Step 7: Write the final formula
Combining the values of \( x \) and \( y \):
\[ \text{The formula of the electrolyte is } AB_4 \]

### Final Answer
The formula of the electrolyte is \( AB_4 \).

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