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In alkaline medium $H_2O_2$ reacts with $Fe^{3+}$ and $Mn^{2+}$ separately to give

$\begin{array}{1 1}(a)\;Fe^{4+}\;and\;Mn^{4+}\\(b)\;Fe^{2+}\;and\;Mn^{4+}\\(c)\;Fe^{2+}\;and\;Mn^{2+}\\(d)\;Fe^{4+}\;and\;Mn^{2+}\end{array}$

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$K_3Fe(CN)_6$ is reduced by $H_2O_2$ in alkaline medium.
$2K_3Fe(CN)_6+2KOH+H_2O_2\rightarrow 2K_4Fe(CN)_6+2H_2O+O_2$
$Mn(OH)_2$ is oxidised by $H_2O_2$ in alkaline medium.
$Mn(OH)_2+H_2O_2\rightarrow MnO_2+2H_2O$
Hence (c) is the correct answer.
answered Jan 24, 2014 by sreemathi.v
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