Chemistry - Redox Reactions Question with Solution | TestHub

The problem involves two redox reactions. First, Fe³⁺ is reduced by hydrazine. Second, the resulting Fe²⁺ is re-oxidized by nitric acid. We need to find the volume of nitric acid required.

Step 1: Determine the n-factor for hydrazine (N₂H₄) in its oxidation to N₂.

The oxidation state of N in N₂H₄ is -2. In N₂, it is 0.

Change in oxidation state per N atom = $0-(-2)=2$.

Since there are two N atoms in N₂H₄, the total change is $2\times 2=4$.

So, the n-factor for N₂H₄ is 4.

Step 2: Calculate milliequivalents of Fe³⁺ (which equals milliequivalents of Fe²⁺ formed).

Milliequivalents of hydrazine = Molarity $\times$ Volume $\times$ n-factor

Milliequivalents of hydrazine = $0.1M\times 40mL\times 4=16meq$

Therefore, milliequivalents of Fe²⁺ = 16 meq.

Step 3: Determine the n-factor for nitric acid (HNO₃) in its reduction to NO.

The oxidation state of N in HNO₃ is +5. In NO, it is +2.

Change in oxidation state = $5-2=3$.

So, the n-factor for HNO₃ is 3.

Step 4: Calculate the volume of nitric acid required.

Milliequivalents of Fe²⁺ = Milliequivalents of HNO₃

For the oxidation of Fe²⁺ to Fe³⁺, the n-factor for Fe²⁺ is 1 (change from +2 to +3).

So,

$16meq=MolarityofHN{O}_3\times VolumeofHN{O}_3\times n-factorofHN{O}_3$

$16=\frac{2}{3}M\times VmL\times 3$

$16=2\times V$

$V=\frac{16}{2}=8mL$

The volume of nitric acid required is 8 mL.