A `5A` current in passed through a solution of zinc sulphate for `40 min`. The amount of zinc deposited at the cathode is

Options

A. (a) 40.65 g
B. (b) 0.4065g
C. (c) 4.065g
D. (d) 65.04 g

Explanation

To determine the amount of zinc deposited at the cathode when a current of 5 A is passed through a zinc sulfate solution for 40 minutes, we can use Faraday's first law of electrolysis. Here’s a step-by-step solution:

### Step 1: Understand the formula
According to Faraday's first law of electrolysis, the amount of substance deposited (W) is given by the formula:
\[ W = Z \cdot I \cdot T \]
where:
- \( W \) = mass of the substance deposited (in grams)
- \( Z \) = electrochemical equivalent (in grams per coulomb)
- \( I \) = current (in amperes)
- \( T \) = time (in seconds)

### Step 2: Convert time to seconds
The time given is 40 minutes. We need to convert this into seconds:
\[ T = 40 \text{ minutes} \times 60 \text{ seconds/minute} = 2400 \text{ seconds} \]

### Step 3: Determine the electrochemical equivalent (Z)
The electrochemical equivalent (Z) can be calculated using the formula:
\[ Z = \frac{E}{96500} \]
where \( E \) is the equivalent mass of zinc. The equivalent mass of zinc is calculated as:
\[ E = \frac{\text{Atomic mass of zinc}}{\text{Number of electrons transferred}} \]
The atomic mass of zinc is approximately 65.38 g/mol, and since zinc ions (Zn²⁺) gain 2 electrons to become zinc metal, we have:
\[ E = \frac{65.38}{2} = 32.69 \text{ g/equiv} \]

Now substituting this into the equation for Z:
\[ Z = \frac{32.69}{96500} \]

### Step 4: Substitute values into the formula
Now we can substitute the values of \( Z \), \( I \), and \( T \) into the formula for \( W \):
\[ W = Z \cdot I \cdot T \]
\[ W = \left(\frac{32.69}{96500}\right) \cdot 5 \cdot 2400 \]

### Step 5: Calculate W
Now we perform the calculation:
1. Calculate \( Z \):
\[ Z = \frac{32.69}{96500} \approx 0.000338 \text{ g/C} \]

2. Calculate \( W \):
\[ W = 0.000338 \cdot 5 \cdot 2400 \]
\[ W = 0.000338 \cdot 12000 \]
\[ W \approx 4.056 \text{ grams} \]

### Step 6: Final result
The amount of zinc deposited at the cathode is approximately:
\[ W \approx 4.065 \text{ grams} \]

Thus, the correct answer is **4.065 grams**.

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