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n-factor of Stannous chloride acting like reducing agent in acidic medium is

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answer is B.

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Detailed Solution

n-factor = number of electrons lost or gained per formula unit in the redox change.

In SnCl₂, tin is in the +2 oxidation state (Sn²⁺). When it acts as a reducing agent in acidic medium it is oxidized to Sn⁴⁺. The half-reaction is:

Sn²⁺ → Sn⁴⁺ + 2 e⁻

This shows tin loses 2 electrons per Sn atom.

Therefore, n-factor = 2 for SnCl₂ acting as a reducing agent.

Note: Sometimes confusion arises by mixing up n-factor with simple valency or other per-molecule counting conventions. For redox n-factor always counts electrons transferred in the half-reaction.

Product	Mn oxidation state	Electrons gained per Mn (7 → state)
MnO₄²⁻ (manganate)	+6	1
MnO₂	+4	3
Mn₂O₃	+3	4
Mn²⁺	+2	5

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