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EEJ MAIN Chemistry QUESTION #7187
Question 1
In an acidic environment, hydrogen peroxide acts as an oxidant to convert $[Fe(CN)_{6}]^{4-}$ into $[Fe(CN)_{6}]^{3-}$. Conversely, in an alkaline medium, it reduces $[Fe(CN)_{6}]^{3-}$ back to $[Fe(CN)_{6}]^{4-}$. Identify the additional products generated in these two respective processes.
  • $H_{2}O$ and $(H_{2}O+O_{2})$✔️
  • $H_{2}O$ and $(H_{2}O+OH^{-})$
  • $(H_{2}O+O_{2})$ and $H_{2}O$
  • $(H_{2}O+O_{2})$ and $(H_{2}O+OH^{-})$
Correct Answer Explanation

The reactions are as follows:

  • Acidic Medium: $2[Fe(CN)_{6}]^{4-} + H_{2}O_{2} + 2H^{+} \rightarrow 2[Fe(CN)_{6}]^{3-} + 2H_{2}O$. Here, $H_{2}O_{2}$ is reduced to $H_{2}O$.
  • Alkaline Medium: $2[Fe(CN)_{6}]^{3-} + H_{2}O_{2} + 2OH^{-} \rightarrow 2[Fe(CN)_{6}]^{4-} + 2H_{2}O + O_{2}$. Here, $H_{2}O_{2}$ is oxidized to $O_{2}$.

Thus, the products are $H_{2}O$ and $(H_{2}O + O_{2})$.

[cite: 13]