Hydrogen peroxide oxidises Fe(CN)64− to Fe(CN)63− in acidic medium but reduces Fe(CN)63 to Fe(CN)64− in alkaline medium. The other products formed are, respectively

A
$(H_{2} O - O_{2}) and (H_{2} O + {OH}^{-})$
B
$H_{2} O and (H_{2} O + O_{2})$
C
$H_{2} O and (H_{2} O + {OH}^{-})$
D
$(H_{2} O + O_{2}) and H_{2} O$

Solution:

$2 {[Fe (CN)_{6}]}^{4 -} + H_{2} O_{2} + 2 H^{+} \to 2 {[Fe (CN)_{6}]}^{3 -} + 2 H_{2} O \begin{array}{l} 2 {[Fe (CN)_{6}]}^{3 -} + H_{2} O_{2} + 2 {OH}^{-} \to \\ 2 {[Fe (CN)_{6}]}^{4 -} + O_{2} + 2 H_{2} O \end{array}$

Hydrogen peroxide oxidises ${[Fe (CN)_{6}]}^{4 -} to {[Fe (CN)_{6}]}^{3 -}$ in acidic medium but reduces ${[Fe (CN)_{6}]}^{3} to {[Fe (CN)_{6}]}^{4 -}$ in alkaline medium. The other products formed are, respectively

Solution:

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