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Question

(D) In $SO_2$,the oxidation state of $S$ is $+4$. Since $S$ can be oxidized to $+6$,$SO_2$ acts as a strong reducing agent.$SO_2$ reduces $Cl_2$ water

Vedclass · Accepted Answer

all of these

Vedclass · Answer

(D) In $SO_2$,the oxidation state of $S$ is $+4$. Since $S$ can be oxidized to $+6$,$SO_2$ acts as a strong reducing agent.$SO_2$ reduces $Cl_2$ water to $HCl$ $(SO_2 + Cl_2 + 2H_2O \rightarrow H_2SO_4 + 2HCl)$.$SO_2$ reduces acidified $Cr_2O_7^{2-}$ to $Cr^{3+}$ $(Cr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O)$.$SO_2$ reduces acidified $MnO_4^-$ to $Mn^{2+}$ $(2MnO_4^- + 5SO_2 + 2H_2O \rightarrow 2Mn^{2+} + 5SO_4^{2-} + 4H^+)$.Therefore,$SO_2$ can reduce all of these.

$SO_2$ can reduce:

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