Find the number of electrons involved in the electro-deposition of 63.5 g of copper from a solution of copper sulphate is:

Answer

To find the number of electrons involved in the electro-deposition of 63.5 g of copper from a solution of copper sulfate, we can follow these steps:

### Step 1: Determine the molar mass of copper
The molar mass of copper (Cu) is given as 63.5 g/mol. Since we are dealing with 63.5 g of copper, we can conclude that this corresponds to 1 mole of copper.

### Step 2: Identify the oxidation state of copper in copper sulfate
In copper sulfate (CuSO₄), copper exists in the +2 oxidation state (Cu²⁺). This means that each copper ion will gain 2 electrons during the electro-deposition process.

### Step 3: Calculate the number of equivalents
The number of equivalents (n) can be calculated using the formula:
\[ \text{Number of equivalents} = \text{Number of moles} \times n \text{-factor} \]

Here, the number of moles of copper is 1 (from Step 1), and the n-factor for copper (Cu²⁺) is 2 (since it gains 2 electrons). Therefore:
\[ \text{Number of equivalents} = 1 \text{ mole} \times 2 = 2 \text{ equivalents} \]

### Step 4: Calculate the number of electrons
The total number of electrons involved in the electro-deposition can be calculated using the formula:
\[ \text{Number of electrons} = \text{Number of equivalents} \times N_A \]
where \( N_A \) is Avogadro's number, approximately \( 6.022 \times 10^{23} \) particles/mol.

Substituting the values:
\[ \text{Number of electrons} = 2 \text{ equivalents} \times 6.022 \times 10^{23} \text{ electrons/equivalent} \]
\[ \text{Number of electrons} = 12.044 \times 10^{23} \text{ electrons} \]

### Final Answer
The number of electrons involved in the electro-deposition of 63.5 g of copper from a solution of copper sulfate is approximately \( 12.044 \times 10^{23} \) electrons.

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