`CuO+H_(2) rarr Cu+H_(2)O` (reference to copper)

Answer

To analyze the reaction \( \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \) with reference to copper, we can follow these steps:

### Step 1: Identify the Reactants and Products
The reactants in the reaction are copper(II) oxide (CuO) and hydrogen gas (H₂). The products are elemental copper (Cu) and water (H₂O).

### Step 2: Determine the Oxidation States
Next, we need to assign oxidation states to each element in the reaction:

- In CuO:
- Oxygen (O) has an oxidation state of -2.
- To balance this, copper (Cu) must have an oxidation state of +2.

- In H₂ (hydrogen gas):
- Hydrogen (H) is in its elemental form, so its oxidation state is 0.

- In the products:
- In Cu (elemental copper), the oxidation state is 0.
- In H₂O:
- Oxygen (O) has an oxidation state of -2.
- Hydrogen (H) has an oxidation state of +1 (there are two H atoms, contributing +2 total).

### Step 3: Analyze the Changes in Oxidation States
Now, we analyze the changes in oxidation states during the reaction:

- Copper in CuO changes from +2 to 0 (reduction).
- Hydrogen in H₂ changes from 0 to +1 (oxidation).

### Step 4: Identify Oxidation and Reduction
From the changes in oxidation states:
- The reduction occurs when copper (Cu) is reduced from +2 in CuO to 0 in elemental Cu.
- The oxidation occurs when hydrogen (H) is oxidized from 0 in H₂ to +1 in H₂O.

### Step 5: Conclusion
In conclusion, with reference to copper, the reaction involves the reduction of copper from copper(II) oxide to elemental copper, while hydrogen is oxidized to form water.

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