A `110-`watt `100-`volt incandescent lamp is connectfedin series with an electrolytic cell containing cadmium sulphate solution. The mass of cadmium deposited by the current flowing for `1` hour (at mass of `Cd = 112`) is

Options

A. 16.06g
B. 38.12g
C. 1.906g
D. 3.812g

Answer

To solve the problem of finding the mass of cadmium deposited by the current flowing for 1 hour through an electrolytic cell, we can follow these steps:

### Step 1: Calculate the Current (I)
We know that power (P) is given by the formula:
\[ P = V \times I \]
Where:
- \( P = 110 \) watts
- \( V = 100 \) volts

Rearranging the formula to find the current (I):
\[ I = \frac{P}{V} = \frac{110 \text{ W}}{100 \text{ V}} = 1.1 \text{ A} \]

### Step 2: Calculate the Total Charge (Q)
The total charge (Q) passed through the circuit can be calculated using the formula:
\[ Q = I \times t \]
Where:
- \( I = 1.1 \) A
- \( t = 1 \text{ hour} = 3600 \text{ seconds} \)

Calculating Q:
\[ Q = 1.1 \text{ A} \times 3600 \text{ s} = 3960 \text{ C} \]

### Step 3: Determine the Number of Moles of Cadmium Deposited
The electrochemical reaction for cadmium deposition is:
\[ \text{Cd}^{2+} + 2e^- \rightarrow \text{Cd} \]
From this reaction, we see that 2 moles of electrons are required to deposit 1 mole of cadmium.

Using Faraday's law, the number of moles of cadmium deposited (n) can be calculated using:
\[ n = \frac{Q}{nF} \]
Where:
- \( n = 2 \) (because 2 moles of electrons are needed for 1 mole of cadmium)
- \( F = 96500 \text{ C/mol} \) (Faraday's constant)

Calculating n:
\[ n = \frac{3960 \text{ C}}{2 \times 96500 \text{ C/mol}} = \frac{3960}{193000} \approx 0.0205 \text{ mol} \]

### Step 4: Calculate the Mass of Cadmium Deposited
Now, we can calculate the mass (m) of cadmium deposited using the formula:
\[ m = n \times M \]
Where:
- \( M = 112 \text{ g/mol} \) (molar mass of cadmium)

Calculating m:
\[ m = 0.0205 \text{ mol} \times 112 \text{ g/mol} \approx 2.296 \text{ g} \]

### Final Answer
The mass of cadmium deposited by the current flowing for 1 hour is approximately:
\[ m \approx 2.296 \text{ g} \]