Question

Hydrogen peroxide oxidises $\left[Fe\right(CN{)}_6{]}^{4-}$ to $\left[Fe\right(CN{)}_6{]}^{3-}$ in acidic medium but reduces $\left[Fe\right(CN{)}_6{]}^{3-}$ to $\left[Fe\right(CN{)}_6){]}^{4-}$ in alkaline medium. The other products formed, repectively are:

• A. $H_{2}O$ and $(H_{2}O+O{H}^{-})$
• B. $(H_{2}O+O_2)$ and $H_{2}O$
• C. $(H_{2}O+O_2)$ and $(H_{2}O+O{H}^{-})$
• D. $H_{2}O$ and $(H_{2}O+O_2)$

Answer 1

The correct option is D $H_{2}O$ and $(H_{2}O+O_2)$

In acidic medium,
$\left[Fe\right(CN{)}_6{]}^{4-}+H_2\stackrel{-1}{O_2}+2H^{+}\to \left[Fe\right(CN{)}_6{]}^{3-}+2H_2\stackrel{-2}{O}$
Here $Fe$ is getting oxidised from (+2 to +3) and $O$ of $H_2{O}_2$ is getting reduced from (-1 to -2).

In alkaline medium,
$\left[Fe\right(CN{)}_6{]}^{3-}+H_2\stackrel{-1}{O_2}+2O{H}^{-}\to \left[Fe\right(CN{)}_6{]}^{4-}+\stackrel{0}{O_2}+2H_{2}O$
Here $Fe$ is getting reduced from (+3 to +2) and $O$ of $H_2{O}_2$ is getting oxidised from (-1 to 0).