Which component is being reduced in the reaction ? `CuO+H_(2)overset(Delta)toCu+H_(2)O`

Options

A. Cu
B. `H_(2)`
C. `CuO`
D. Both CuO and `H_(2)`

Explanation

To determine which component is being reduced in the reaction \( \text{CuO} + \text{H}_2 \overset{\Delta}{\rightarrow} \text{Cu} + \text{H}_2\text{O} \), we can follow these steps:

### Step 1: Identify the oxidation states of the elements involved
- In copper(II) oxide (CuO), copper (Cu) has an oxidation state of +2 and oxygen (O) has an oxidation state of -2.
- In the product copper (Cu), the oxidation state is 0 (as it is in its elemental form).
- In water (H2O), hydrogen (H) has an oxidation state of +1 and oxygen (O) has an oxidation state of -2.

### Step 2: Determine the changes in oxidation states
- Copper (Cu) changes from +2 in CuO to 0 in elemental Cu. This is a reduction because the oxidation state decreases.
- Hydrogen (H) remains at +1 in both reactants (H2) and products (H2O), so it is not undergoing any change.

### Step 3: Identify the component being reduced
- Since copper (Cu) is the only component that undergoes a decrease in oxidation state (from +2 to 0), it is the component that is being reduced in this reaction.

### Conclusion
The component being reduced in the reaction \( \text{CuO} + \text{H}_2 \overset{\Delta}{\rightarrow} \text{Cu} + \text{H}_2\text{O} \) is **copper (Cu)**.

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