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Q. Statement 1: In the redox reaction $8H^{+} (aq) +4 NO^{-}_{3} +6Cl^{-}+Sn(s) \to snCl^{2-}_{6} +4NO_{2} +4H_{2}O,$ the reducing agent is Sn(s).
Statement 2: In balancing half-reaction, $S_{2}O^{2-}_{3} \to$S(s), the number of electrons added on the left is 4.

Redox Reactions

Solution:

Sn $(0 \to + 4)$
It undergoes oxidation, so it is the reducing agent
$S_{2}O^{2-}_{3}+6H^{+}+4e^{-} \to 2S +3H_{2}O$