Oxidation of `Cu_(3)P " to " CuSO_(4) " and " H_(3)PO_(4)` involves the loss of

Options

A. 3 electrons
B. 5 electrons
C. 9 electrons
D. 11 electrons

Answer

To solve the problem of determining the total loss of electrons during the oxidation of \( Cu_3P \) to \( CuSO_4 \) and \( H_3PO_4 \), we will follow these steps:

### Step 1: Identify the oxidation states of the elements in \( Cu_3P \)
- In \( Cu_3P \), the oxidation state of copper (Cu) is +1 and phosphorus (P) is -3.

### Step 2: Identify the oxidation states of the elements in the products \( CuSO_4 \) and \( H_3PO_4 \)
- In \( CuSO_4 \), the oxidation state of copper (Cu) is +2.
- In \( H_3PO_4 \), the oxidation state of phosphorus (P) is +5.

### Step 3: Calculate the change in oxidation state for copper
- Copper changes from +1 in \( Cu_3P \) to +2 in \( CuSO_4 \).
- The change in oxidation state for copper is \( +2 - (+1) = +1 \).
- This indicates that 1 electron is lost by each copper atom.

### Step 4: Calculate the change in oxidation state for phosphorus
- Phosphorus changes from -3 in \( Cu_3P \) to +5 in \( H_3PO_4 \).
- The change in oxidation state for phosphorus is \( +5 - (-3) = +5 + 3 = +8 \).
- This indicates that 8 electrons are lost by phosphorus.

### Step 5: Calculate the total loss of electrons
- Total loss of electrons = Loss from copper + Loss from phosphorus
- Total loss = 1 (from Cu) + 8 (from P) = 9 electrons.

### Final Answer
The oxidation of \( Cu_3P \) to \( CuSO_4 \) and \( H_3PO_4 \) involves the loss of **9 electrons**.

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