Question

Match the reactions given in column I with their respective oxidant/reductant given in column II.

Answer 1

$2e^{-}+I_2⟶2I^{\ominus }$ (Reduction and acts as an oxidant)
$I_2⟶2I{O}_3^{\ominus }+10e^{-}$ (Oxidation and acts as a reductant)
$2x=0$ $2x-12=-2$
$2x=10$
So, $I_2$ acts both as an oxidant and a reductant.
No change in the oxidation number of either of the conjugate pairs.
${Ba}^{2+}{Cl}_2^{2-}⟶{Ba}^{2+}S{O}_4^{2-}$
${Na}_2^{2+}S{O}_4⟶2\stackrel{+2}{Na}Cl$
(None is an oxidant or a reductant)
In a conjugate pair, the oxidant has a higher oxidation number.
For $Al{Cl}_3:{Al}^{3+}+3e^{-}⟶{Al}^0$
(Reduction and acts as an oxidant)
For $Na:Na⟶{Na}^{\oplus }+e^{-}$
(Oxidation and acts as a reductant)
For $S{O}_2:4e^{-}+S{O}_2⟶S^0$
(Reduction and acts as an oxidant)
For $H_{2}S:S^{2-}⟶S^0+2e^{-}$
(Oxidation and acts as a reductant)
Hence, the answer is $A\to 3,B\to 1,C\to 4,D\to 2$.