Write the balanced overall chemical equation for the reduction of nitrobenzene to aniline using iron filings and hydrochloric acid. Also state the oxidation states of nitrogen in nitrobenzene and aniline.

medium 2 marks Class 12 · Chemistry · Amines

Answer

C_{6} H_{5} N O_{2} + 3Fe + 6HCl \to C_{6} H_{5} N H_{2} + 3FeC l_{2} + 2 H_{2} O

; N oxidation state: +3 in nitrobenzene, –3 in aniline

The correct answer is $C_{6} H_{5} N O_{2} + 3Fe + 6HCl \to C_{6} H_{5} N H_{2} + 3FeC l_{2} + 2 H_{2} O$ ; N oxidation state: +3 in nitrobenzene, –3 in aniline — State oxidation numbers: In –NO₂ group, N is +3; in –NH₂ group, N is –3.

Step-by-step solution

Write the unbalanced reduction: nitrobenzene + Fe + HCl → aniline + FeCl₂ + H₂O.
Balance the benzene ring atoms (C and H) – they already match on both sides.
Balance chlorine: each FeCl₂ requires 2 Cl⁻, so 3 FeCl₂ need 6 Cl⁻, provided by 6 HCl.
Balance iron: 3 Fe on reactant side give 3 FeCl₂ on product side.
Balance oxygen: nitrobenzene has 2 O, which become 2 H₂O on product side; thus add 2 H₂O.
Check hydrogen atoms: left side has 6 H from HCl; right side has 2 H in aniline (NH₂) + 4 H in 2 H₂O = 6 H – balanced.
State oxidation numbers: In –NO₂ group, N is +3; in –NH₂ group, N is –3.

Common mistake: Forgetting to balance the iron and chloride ions or using the wrong number of HCl molecules

Concept tested: Reduction of nitro compounds to amines

Published 5 September 2026 · How this solution was prepared and checked

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