How much metal will be deposited when a current of 12 ampere with 75% efficiency is passed through the cell for 3 h? (Given: Z= `4xx10^(-4)` )

Options

A. 32.4 g
B. 38.8 g
C. 36.0 g
D. 22.4 g

Explanation

To solve the problem of how much metal will be deposited when a current of 12 amperes with 75% efficiency is passed through the cell for 3 hours, we can follow these steps:

### Step 1: Understand the formula
The amount of metal deposited (W) can be calculated using the formula:
\[ W = Z \cdot I \cdot T \cdot \text{Efficiency} \]
where:
- \( W \) = mass of the deposited metal (in grams)
- \( Z \) = electrochemical equivalent (in grams per coulomb)
- \( I \) = current (in amperes)
- \( T \) = time (in seconds)
- Efficiency = efficiency of the process (as a decimal)

### Step 2: Convert time from hours to seconds
Given that the time is 3 hours, we need to convert this into seconds:
\[ T = 3 \, \text{hours} \times 3600 \, \text{seconds/hour} = 10800 \, \text{seconds} \]

### Step 3: Calculate the efficiency as a decimal
The efficiency is given as 75%. To convert this percentage into a decimal:
\[ \text{Efficiency} = \frac{75}{100} = 0.75 \]

### Step 4: Substitute the values into the formula
We know:
- \( Z = 4 \times 10^{-4} \, \text{g/C} \)
- \( I = 12 \, \text{A} \)
- \( T = 10800 \, \text{s} \)
- Efficiency = 0.75

Now, substituting these values into the formula:
\[ W = (4 \times 10^{-4}) \cdot 12 \cdot 10800 \cdot 0.75 \]

### Step 5: Perform the calculations
1. Calculate \( 12 \cdot 10800 \):
\[ 12 \cdot 10800 = 129600 \]

2. Now multiply by \( Z \) and efficiency:
\[ W = (4 \times 10^{-4}) \cdot 129600 \cdot 0.75 \]
\[ W = (4 \times 0.75) \cdot 129600 \times 10^{-4} \]
\[ W = 3 \cdot 129600 \times 10^{-4} \]
\[ W = 388800 \times 10^{-4} \]
\[ W = 38.88 \, \text{grams} \]

### Final Answer
The amount of metal deposited is approximately **38.88 grams**.

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