Question

Which of the following equation depicts reducing nature of $H_{2}O_{2}$?

• A. $2\lbrack Fe(CN){6}\rbrack^{4 -} + 2H^{+} + H{2}O_{2} \rightarrow 2\lbrack Fe(CN){6}\rbrack^{3 -} $$$+ 2H{2}O$$
• B. $I_{2} + H_{2}O_{2} + 2OH^{-} \rightarrow 2I^{-} + 2H_{2}O + O_{2}$
• C. $Mn^{2 +} + H_{2}O_{2} \rightarrow Mn^{4 +} + 2OH^{-}$
• D. $PbS + 4H_{2}O_{2} \rightarrow PbSO_{4} + 4H_{2}O$

Answer 1

Answer: (B)

$I_{2} + H_{2}O_{2} + 2OH^{-} \rightarrow 2I^{-} + 2H_{2}O + O_{2}$

Answer 2

: Reducing nature of $H_{2}O_{2}$ means it reduces other substance and itself gets oxidised. In such reactions $O_{2}$is evolved.

In (1) $Fe^{2 +}$gets oxidised to $Fe^{3 +}$

In (2), $I_{2}$gets reduced to I

In (3), $Mn^{2 +}$gets oxidised to $Mn^{4 +}$

In (4), $Pb\overset{(2 - 1)}{s}$gets oxidised to $\overset{( + 6)}{PbSo_{4}}$