A 100 watt, 110 volt incandescent lamp is connected in series with an electrolytic cell containing CdSO4 solution. What weight of Cd will be deposited by current flowing for 10 hr.? At.wt. of Cd = 112.4

Explanation

To solve the problem of determining the weight of cadmium (Cd) deposited by the current flowing through an electrolytic cell for 10 hours, we will follow these steps:

### Step 1: Calculate the Current (I)
The power (P) of the lamp is given as 100 watts, and the voltage (V) is 110 volts. We can use the formula:

\[ I = \frac{P}{V} \]

Substituting the values:

\[ I = \frac{100 \text{ W}}{110 \text{ V}} = 0.909 \text{ A} \]

### Step 2: Convert Time from Hours to Seconds
The time (t) is given as 10 hours. We need to convert this into seconds:

\[ t = 10 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 36000 \text{ seconds} \]

### Step 3: Determine the Number of Electrons Involved
In the electrolysis of cadmium sulfate (CdSO₄), cadmium ions (Cd²⁺) gain 2 electrons (2e⁻) to deposit as metallic cadmium (Cd). Therefore, the number of electrons (n) involved in the reaction is 2.

### Step 4: Calculate the Total Charge (Q)
Using the formula for charge:

\[ Q = I \times t \]

Substituting the values:

\[ Q = 0.909 \text{ A} \times 36000 \text{ s} = 32724 \text{ C} \]

### Step 5: Calculate the Number of Moles of Cadmium Deposited
Using Faraday's law, the weight of cadmium deposited (W) can be calculated using the formula:

\[ W = \frac{(Q \cdot \text{atomic weight})}{n \cdot F} \]

Where:
- \(F\) (Faraday's constant) = 96500 C/mol
- Atomic weight of Cd = 112.4 g/mol
- \(n\) = 2 (number of electrons)

Substituting the values:

\[ W = \frac{(32724 \text{ C} \cdot 112.4 \text{ g/mol})}{2 \cdot 96500 \text{ C/mol}} \]

Calculating this gives:

\[ W = \frac{3670800.8}{193000} \approx 19.06 \text{ g} \]

### Final Answer
The weight of cadmium deposited after 10 hours is approximately **19.06 grams**.

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