In the following redox reaction `Cu(OH)_(2) (s) + N_(2) H_(4)(aq) rarr Cu(s) + N_(2)(g)` number of mole of `Cu(OH)_(2)` reduced by one mol of `N_(2) H_(4)` is -

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the question regarding the number of moles of Cu(OH)₂ reduced by one mole of N₂H₄ in the redox reaction:

**Given Reaction:**
\[ \text{Cu(OH)}_2 (s) + \text{N}_2\text{H}_4 (aq) \rightarrow \text{Cu} (s) + \text{N}_2 (g) \]

**Step 1: Identify Oxidation States**
- For Cu(OH)₂:
- Cu has an oxidation state of +2.
- OH has an oxidation state of -1.

- For N₂H₄:
- Let the oxidation state of nitrogen be \( x \).
- The equation for the oxidation state is:
\[ 2x + 4(1) = 0 \]
\[ 2x + 4 = 0 \]
\[ 2x = -4 \]
\[ x = -2 \]
- Therefore, nitrogen in N₂H₄ has an oxidation state of -2.

**Step 2: Determine Changes in Oxidation States**
- Cu changes from +2 (in Cu(OH)₂) to 0 (in Cu).
- N changes from -2 (in N₂H₄) to 0 (in N₂).

**Step 3: Calculate Electrons Transferred**
- For Cu:
- Change: +2 to 0 means it gains 2 electrons (reduction).

- For N:
- Change: -2 to 0 means it loses 4 electrons (oxidation).

**Step 4: Establish the Stoichiometry**
- From the above, we see that:
- 1 mole of N₂H₄ loses 4 electrons.
- 1 mole of Cu(OH)₂ gains 2 electrons.

- Therefore, to balance the electrons:
- 2 moles of Cu(OH)₂ will be reduced by 1 mole of N₂H₄.

**Final Answer:**
The number of moles of Cu(OH)₂ reduced by one mole of N₂H₄ is **2 moles**.

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