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🧪 Chemistry  ·  Redox Reactions  ·  NEET & JEE

During iodometric back-titration, Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is used to titrate liberated I<sub>2</sub>. The reaction is: 2S<sub>2</sub>O<sub>3</sub><sup>2-</sup> + I<sub>2</sub> → S<sub>4</sub>O<sub>6</sub><sup>2-</sup> + 2I-. What is the role of Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> here?

Answer: Reducing agent.

  • A Oxidising agent
  • B Reducing agent
  • C Both oxidising and reducing agent
  • D Catalyst

Correct answer: B. Reducing agent

Explanation: S₂O₃²⁻ loses electrons (S goes from +2 to +2.5 on average in S₄O₆²⁻); I₂ gains electrons (reduced to I⁻). Therefore Na₂S₂O₃ is the reducing agent.

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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