What is the mass of copper metal produced at cathode during the passage of 2.03A current through the`CuSO_4` solution for 1 hour. Molar mass of Cu = 63.5 g `mol^-1`

Answer

To find the mass of copper metal produced at the cathode during the passage of a current through a copper sulfate solution, we can follow these steps:

### Step 1: Identify the Given Values
- Current (I) = 2.03 A
- Time (t) = 1 hour = 3600 seconds
- Molar mass of copper (Cu) = 63.5 g/mol

### Step 2: Calculate the Total Charge (Q) Passed
The total charge (Q) can be calculated using the formula:
\[ Q = I \times t \]
Substituting the values:
\[ Q = 2.03 \, \text{A} \times 3600 \, \text{s} \]
\[ Q = 7310.8 \, \text{C} \]

### Step 3: Determine the Charge Required for the Deposition of Copper
From electrochemistry, we know that the reduction of copper ions (Cu²⁺) to copper metal (Cu) at the cathode requires 2 moles of electrons (2F) for 1 mole of copper:
- Faraday's constant (F) = 96500 C/mol
- Charge required for 1 mole of copper:
\[ 2F = 2 \times 96500 \, \text{C} = 193000 \, \text{C} \]

### Step 4: Calculate the Amount of Copper Deposited
Using the total charge calculated in Step 2, we can find out how many moles of copper are deposited:
\[ \text{Moles of Cu} = \frac{Q}{2F} = \frac{7310.8 \, \text{C}}{193000 \, \text{C/mol}} \]
\[ \text{Moles of Cu} = 0.0379 \, \text{mol} \]

### Step 5: Calculate the Mass of Copper Deposited
Now, we can find the mass of copper deposited using the molar mass:
\[ \text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} \]
\[ \text{Mass of Cu} = 0.0379 \, \text{mol} \times 63.5 \, \text{g/mol} \]
\[ \text{Mass of Cu} = 2.4 \, \text{g} \]

### Final Answer
The mass of copper metal produced at the cathode is **2.4 grams**.

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