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 loss of electrons during the oxidation of \( \text{Cu}_3\text{P} \) to \( \text{CuSO}_4 \) and \( \text{H}_3\text{PO}_4 \), we will follow these steps:

### Step 1: Write the balanced chemical equation.
The oxidation reaction can be represented as:
\[ \text{Cu}_3\text{P} + 6 \text{H}_2\text{SO}_4 \rightarrow 3 \text{CuSO}_4 + \text{H}_3\text{PO}_4 + 6 \text{H}_2\text{O} \]

### Step 2: Determine the oxidation states of elements in reactants and products.
- In \( \text{Cu}_3\text{P} \):
- Copper (Cu) has an oxidation state of +1 (since there are 3 Cu atoms contributing to a total of +3 charge, and phosphorus balances this with -3).
- Phosphorus (P) has an oxidation state of -3.

- In \( \text{CuSO}_4 \):
- Copper (Cu) has an oxidation state of +2.
- Sulfate ion \( \text{SO}_4^{2-} \) has sulfur (S) in +6 oxidation state and oxygen (O) in -2.

- In \( \text{H}_3\text{PO}_4 \):
- Phosphorus (P) has an oxidation state of +5.
- Each hydrogen (H) is +1 and oxygen (O) is -2.

### Step 3: Calculate the change in oxidation states.
- For Copper:
- Change from +1 (in \( \text{Cu}_3\text{P} \)) to +2 (in \( \text{CuSO}_4 \)).
- Each Cu loses 1 electron, and since there are 3 Cu atoms:
\[ \text{Total loss of electrons from Cu} = 3 \times 1 = 3 \text{ electrons} \]

- For Phosphorus:
- Change from -3 (in \( \text{Cu}_3\text{P} \)) to +5 (in \( \text{H}_3\text{PO}_4 \)).
- The change in oxidation state is:
\[ +5 - (-3) = 8 \text{ electrons} \]

### Step 4: Calculate the total loss of electrons.
- Total loss of electrons during the oxidation process:
\[ \text{Total loss} = \text{Loss from Cu} + \text{Loss from P} = 3 + 8 = 11 \text{ electrons} \]

### Conclusion
The oxidation of \( \text{Cu}_3\text{P} \) to \( \text{CuSO}_4 \) and \( \text{H}_3\text{PO}_4 \) involves the loss of **11 electrons**.

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