In which of the following reactions, the underlined substance has been reduced ?

Options

A. `ul(Cu)+CuO rarr CO_(2)+Cu`
B. `ul(CuO)+2HCl rarr CuCl_(2)+H_(2)O`
C. `ul(4H)_(ul(2))ul(O)+3Frarr 4H_(2)+Fe_(3)O_(4)`
D. `ul(C)+4HNO_(3)rarr CO_(2)+2H_(2)O+4NO_(2)`

Explanation

To determine which substance has been reduced in the given reactions, we need to analyze the oxidation states of the substances involved in each reaction. A substance is considered reduced if it gains electrons (or hydrogen) or loses oxygen.

### Step-by-Step Solution:

1. **Analyze the First Reaction:**
- Reaction: Cu + CuO → CO2 + Cu
- The underlined substance is Cu.
- Oxidation state of Cu (elemental form) = 0.
- In the products, Cu remains as Cu (oxidation state = 0).
- **Conclusion:** Cu is neither reduced nor oxidized.

2. **Analyze the Second Reaction:**
- Reaction: CuO + 2HCl → CuCl2 + H2O
- The underlined substance is CuO.
- In CuO, Cu has an oxidation state of +2 and O has -2.
- In the products, Cu in CuCl2 also has an oxidation state of +2.
- **Conclusion:** Cu is neither reduced nor oxidized.

3. **Analyze the Third Reaction:**
- Reaction: 4H2O + 3Fe → 4H2 + Fe3O4
- The underlined substance is H2O.
- In H2O, O has an oxidation state of -2.
- In the products, H2 has an oxidation state of 0 (as it is in elemental form).
- **Analysis:** H2O is losing oxygen (it is being converted to H2), which indicates that it is being reduced.
- **Conclusion:** H2O is reduced.

4. **Analyze the Fourth Reaction:**
- Reaction: C + O2 → CO2
- The underlined substance is C.
- In elemental form, C has an oxidation state of 0.
- In CO2, C has an oxidation state of +4.
- **Conclusion:** C is being oxidized (gaining oxygen).

### Final Answer:
The substance that has been reduced in the reactions is **H2O** in the third reaction.