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🧪 Chemistry  ·  Redox Reactions  ·  NEET & JEE

Assign the oxidation state of each element in [Fe(CN)<sub>6</sub>]4- and determine the oxidation state of iron:

Answer: Fe = +2 (CN- is -1 each; 6(-1) + Fe = -4; Fe = +2).

  • A Fe = +2 (CN- is -1 each; 6(-1) + Fe = -4; Fe = +2)
  • B Fe = +3, the oxidation state found instead in the ferricyanide ion
  • C Fe = 0, the oxidation state found instead in neutral iron carbonyls
  • D Fe = +4, an oxidation state not consistent with this complex's overall charge

Correct answer: A. Fe = +2 (CN- is -1 each; 6(-1) + Fe = -4; Fe = +2)

Explanation: In [Fe(CN)₆]⁴⁻: each CN⁻ has a charge of -1; 6 CN⁻ contribute -6. Overall charge = -4. So Fe + (-6) = -4; Fe = +2. Ferrocyanide contains Fe(II).

Electron Transfer: Zn + Cu²⁺ → Zn²⁺ + CuZnloses 2e⁻OXIDISED (Zn → Zn²⁺)2e⁻Cu²⁺gains 2e⁻REDUCED (Cu²⁺ → Cu)Zn = reducing agentCu²⁺ = oxidising agent

Electron transfer in a redox reaction: zinc loses electrons (oxidised, acts as reducing agent) and copper(II) ions gain those same electrons (reduced, acts as oxidising agent).

Concept context

Master electron transfer in chemistry: assign oxidation states, balance half-reactions, identify oxidising and reducing agents, and connect redox to everyday reactions.

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