In an electrolysis experiment, current was passed for `5h` through two cells connected in series. The first cell contains a solution of gold and second contains copper sulphate solution. In the first cell, `9.85g ` of gold was deposited. If the oxidation number of gold is `+3`, find the amount of copper deposited at the cathode of the second cell. Also calculate the magnitude of the current in ampere, `(` Atomic weight of `Au` is 1197 and atomic weight of `Cu` is `63.5)`.

Explanation

To solve the problem step by step, we will follow the principles of electrolysis and the relationships between mass, equivalent mass, and current.

### Step 1: Calculate the equivalent mass of gold (Au)
The equivalent mass of an element can be calculated using the formula:
\[ \text{Equivalent mass} = \frac{\text{Atomic mass}}{\text{n}} \]
where \( n \) is the number of electrons transferred per atom during the reaction. For gold (Au) with an oxidation state of +3:
- Atomic weight of Au = 197 g/mol
- n = 3

Thus, the equivalent mass of gold is:
\[ \text{Equivalent mass of Au} = \frac{197}{3} = 65.67 \, \text{g/equiv} \]

### Step 2: Calculate the equivalent mass of copper (Cu)
For copper (Cu) with an oxidation state of +2:
- Atomic weight of Cu = 63.5 g/mol
- n = 2

Thus, the equivalent mass of copper is:
\[ \text{Equivalent mass of Cu} = \frac{63.5}{2} = 31.75 \, \text{g/equiv} \]

### Step 3: Use the second law of electrolysis
According to the second law of electrolysis, the ratio of the mass of gold deposited to the mass of copper deposited is equal to the ratio of their equivalent masses:
\[ \frac{\text{mass of Au}}{\text{mass of Cu}} = \frac{\text{Equivalent mass of Au}}{\text{Equivalent mass of Cu}} \]
Given that the mass of gold deposited is 9.85 g, we can set up the equation:
\[ \frac{9.85}{\text{mass of Cu}} = \frac{65.67}{31.75} \]

### Step 4: Solve for the mass of copper deposited
Rearranging the equation to find the mass of copper deposited:
\[ \text{mass of Cu} = \frac{9.85 \times 31.75}{65.67} \]
Calculating this gives:
\[ \text{mass of Cu} = \frac{312.3875}{65.67} \approx 4.76 \, \text{g} \]

### Step 5: Calculate the current (I)
Using the first law of electrolysis:
\[ W = Z \cdot I \cdot t \]
Where:
- \( W \) = mass of copper deposited = 4.76 g
- \( Z \) = equivalent weight of Cu = 31.75 g/equiv
- \( t \) = time in seconds = 5 hours = 5 × 3600 = 18000 seconds

Rearranging for current \( I \):
\[ I = \frac{W}{Z \cdot t} \]
Substituting the values:
\[ I = \frac{4.76}{31.75 \times 18000} \]
Calculating this gives:
\[ I = \frac{4.76}{570300} \approx 0.00834 \, \text{A} \text{ or } 8.34 \, \text{mA} \]

### Final Answers
- Mass of copper deposited: **4.76 g**
- Current: **0.00834 A or 8.34 mA**