On passing C ampere of electricity through an electrolyte solution
for t seconds, m gram metal deposits on cathode. The eq. wt. of
metal is

Options

A. `E=(Cxxt)/(mxx96500)`
B. `E=(Cxxm)/(txx96500)`
C. `E=(96500xxm)/(Cxxt)`
D. `E=(Cxxtxx96500)/(m)`

Explanation

To find the equivalent weight of the metal deposited on the cathode, we can use Faraday's laws of electrolysis. Here’s the step-by-step solution:

### Step 1: Understand the Given Information
We are given:
- Current (I) = C amperes
- Time (t) = t seconds
- Weight of the deposited metal (W) = m grams

### Step 2: Use the Formula for Equivalent Weight
The equivalent weight (E) of the metal can be calculated using the formula:
\[ E = \frac{W}{Q} \]
where \( Q \) is the quantity of electricity passed through the electrolyte.

### Step 3: Calculate the Quantity of Electricity (Q)
The quantity of electricity (Q) can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values we have:
\[ Q = C \times t \]

### Step 4: Substitute Q into the Equivalent Weight Formula
Now, substituting \( Q \) into the equivalent weight formula:
\[ E = \frac{W}{Q} = \frac{m}{C \times t} \]

### Step 5: Include the Faraday Constant
To express the equivalent weight in terms of the Faraday constant (96500 C/mol), we modify the formula:
\[ E = \frac{W \times 96500}{Q} \]
Substituting \( Q \):
\[ E = \frac{m \times 96500}{C \times t} \]

### Final Formula for Equivalent Weight
Thus, the equivalent weight of the metal is given by:
\[ E = \frac{m \times 96500}{C \times t} \]

### Conclusion
The equivalent weight of the metal deposited on the cathode is:
\[ E = \frac{m \times 96500}{C \times t} \]

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