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# $H_2O_2\rightarrow 2H^++O_2+2e^-$.$E^{\large\circ}=-0.68v$.The given equation represents the following nature of $H_2O_2$

$\begin{array}{1 1}(a)\;\text{Bleaching powder}\\(b)\;\text{Acidic}\\(c)\;\text{Reducing}\\(d)\;\text{Oxidising}\end{array}$

Released electrons causes for reduction of other compounds.
Hence (c) is the correct answer.