How long will it taje 5 ampere of current to deposit 2 g of copper from a solution of copper sulphate ?
(Given : CE of copper =32, F=96500 coulomb)

Answer

To solve the problem of how long it will take for a 5 ampere current to deposit 2 grams of copper from a copper sulfate solution, we can use Faraday's laws of electrolysis. Here’s the step-by-step solution:

### Step 1: Understand the Given Data
- Current (I) = 5 A (amperes)
- Mass of copper deposited (m) = 2 g
- Electrochemical equivalent (E) of copper = 32 g/mol
- Faraday's constant (F) = 96500 C/mol (coulombs per mole)

### Step 2: Calculate the Number of Moles of Copper Deposited
To find the number of moles of copper (Cu) deposited, we can use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of copper is 63.5 g/mol. Therefore:
\[ \text{Number of moles of Cu} = \frac{2 \text{ g}}{63.5 \text{ g/mol}} \approx 0.0315 \text{ mol} \]

### Step 3: Determine the Number of Electrons Transferred
Copper (Cu) typically undergoes a reduction reaction where one mole of Cu^2+ ions gains 2 electrons to form one mole of copper metal. Thus, the number of electrons (n) transferred per mole of copper is 2.

### Step 4: Calculate the Total Charge Required
Using Faraday's law, the total charge (Q) required to deposit the copper can be calculated using the formula:
\[ Q = n \times F \times \text{Number of moles} \]
Substituting the values:
\[ Q = 2 \times 96500 \text{ C/mol} \times 0.0315 \text{ mol} \approx 6074.5 \text{ C} \]

### Step 5: Calculate the Time Required
Using the formula relating charge, current, and time:
\[ Q = I \times T \]
Rearranging for time (T):
\[ T = \frac{Q}{I} \]
Substituting the values:
\[ T = \frac{6074.5 \text{ C}}{5 \text{ A}} \approx 1214.9 \text{ seconds} \]

### Step 6: Convert Time to Minutes
To convert seconds into minutes, divide by 60:
\[ T \approx \frac{1214.9 \text{ seconds}}{60} \approx 20.25 \text{ minutes} \]

### Final Answer
It will take approximately **20.25 minutes** to deposit 2 grams of copper from a copper sulfate solution using a 5 ampere current.

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