sblock Elements 1 Question 19
19. Hydrogen peroxide oxidises $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ to $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ in acidic medium but reduces $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-}$ to $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$ in alkaline medium. The other products formed are, respectively.
(2018 Main)
(a) $\left(\mathrm{H}_2 \mathrm{O}+\mathrm{O}_2\right)$ and $\mathrm{H}_2 \mathrm{O}$
(b) $\left(\mathrm{H}_2 \mathrm{O}+\mathrm{O}_2\right)$ and $\left(\mathrm{H}_2 \mathrm{O}+\mathrm{OH}^{-}\right)$
(c) $\mathrm{H}_2 \mathrm{O}$ and $\left(\mathrm{H}_2 \mathrm{O}+\mathrm{O}_2\right)$
(d) $\mathrm{H}_2 \mathrm{O}$ and $\left(\mathrm{H}_2 \mathrm{O}+\mathrm{OH}^{-}\right)$
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Answer:
Correct Answer: 19. (c)
Solution: Both reactions in their complete format are written below
(i) In acidic medium,
$$ \left[\mathrm{Fe}^{2+}(\mathrm{CN})_6\right]^4+\mathrm{H}_2 \stackrel{-}{O}_2+2 \mathrm{H}^{+} \longrightarrow\left[\mathrm{Fe}^{3+}(\mathrm{CN})_6\right]^{3-}+2 \mathrm{H}_2 \mathrm{O}^{-2} $$
(ii) In alkaline medium,
$$ \begin{aligned} & {\left[\mathrm{Fe}^{3+}(\mathrm{CN})_6\right]^{3-}+\mathrm{H}_2 \mathrm{O}_2^{-1}+2 \mathrm{OH}^{-} \longrightarrow} \ & {\left[\mathrm{Fe}^{2+}(\mathrm{CN})_6\right]^{+}+\mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}} \ & \end{aligned} $$
Hence, $\mathrm{H}_2 \mathrm{O}$ (for reaction (i)) and $\mathrm{O}_2+\mathrm{H}_2 \mathrm{O}$ (for reaction (ii)) are produced as by product.
BETA
