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Assignment

A current of 0.2 ampere is passed in an aqueous solution of copper(II) sulfate ([tex]\text{CuSO}_4[/tex]) for one hour. Calculate the mass of copper deposited at the cathode.

Sagot :

GinnyAnswer
Certainly! Let's tackle the problem step-by-step to find the mass of copper deposited at the cathode.

### Step 1: Gather Known Information
- Current ([tex]\( I \)[/tex]) = 0.2 amperes (A)
- Time ([tex]\( t \)[/tex]) = 1 hour
- Molar mass of copper (Cu) = 63.546 grams per mole (g/mol)
- Faraday's constant ([tex]\( F \)[/tex]) = 96485 coulombs per mole (C/mol)
- Copper usually deposits from a [tex]\(\text{Cu}^{2+}\)[/tex] ion, which means each mole of copper requires 2 moles of electrons (2 Faradays’ of charge per mole of copper).

### Step 2: Convert Time into Seconds
Since time is given in hours, we first convert it to seconds:
[tex]\[ t = 1 \text{ hour} \times 3600 \text{ seconds/hour} = 3600 \text{ seconds} \][/tex]

### Step 3: Calculate the Total Charge Passed
Using the formula for charge ([tex]\( Q \)[/tex]), which is given by:
[tex]\[ Q = I \times t \][/tex]
Where:
- [tex]\( Q \)[/tex] is the charge in coulombs (C)
- [tex]\( I \)[/tex] is the current in amperes (A)
- [tex]\( t \)[/tex] is the time in seconds (s)

Plug in the known values:
[tex]\[ Q = 0.2 \text{ A} \times 3600 \text{ s} = 720 \text{ C} \][/tex]

### Step 4: Determine the Mass of Copper Deposited

According to Faraday's laws of electrolysis, the mass ([tex]\( m \)[/tex]) of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity ([tex]\( Q \)[/tex]) that passes through the electrolyte and can be given by:
[tex]\[ m = \left( \frac{Q \times M}{F \times z} \right) \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the substance deposited (in grams)
- [tex]\( M \)[/tex] is the molar mass of the substance (in grams per mole)
- [tex]\( F \)[/tex] is Faraday's constant (96485 C/mol)
- [tex]\( z \)[/tex] is the number of electrons transferred per ion (for [tex]\(\text{Cu}^{2+}\)[/tex], [tex]\( z = 2 \)[/tex])

Substitute in the known values:
[tex]\[ m = \left( \frac{720 \text{ C} \times 63.546 \text{ g/mol}}{96485 \text{ C/mol} \times 2} \right) \][/tex]

### Step 5: Simplify the Expression
Perform the division to find the mass:
[tex]\[ m = \left( \frac{720 \times 63.546}{96485 \times 2} \right) \][/tex]
[tex]\[ m \approx 0.237 \text{ grams} \][/tex]

### Conclusion
The mass of copper deposited at the cathode after passing a current of 0.2 A for one hour is approximately 0.237 grams.
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