In calculating n-factor for Na 2 S 2 O 3 in its reaction with I 2 (where products include Na 2 S 4 O 6 ), the n-factor of Na 2 S 2 O 3 is:

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In calculating n-factor for Na2S2O3 in its reaction with I2 (where products include Na2S4O6), the n-factor of Na2S2O3 is:

Answer: 1.

Correct answer: B. 1

Explanation: In S₂O₃²⁻: S is +2. In S₄O₆²⁻: S is +2.5. Each Na₂S₂O₃ loses 0.5 electrons (per S₂O₃²⁻ unit, 2 S atoms each go from +2 to +2.5 = 1 electron lost total per mole). n-factor = 1.

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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In calculating n-factor for Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> in its reaction with I<sub>2</sub> (where products include Na<sub>2</sub>S<sub>4</sub>O<sub>6</sub>), the n-factor of Na<sub>2</sub>S<sub>2</sub>O<sub>3</sub> is:

Concept context

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