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In aqueous alkaline solution, two electrons reduction of $\mathrm{HO}_{2}^{-}$ gives
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$\mathrm{HO}^{-}$
In aqueous alkaline solution, two electron reduction of $\mathrm{HO}_{2}^{-}$ gives OH The reaction is as follows:
$\begin{array}{l}
\mathrm{H}_{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HO}_{2}^{-} \\
\mathrm{HO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O}+2 e^{-} \rightleftharpoons 3 \mathrm{OH}
\end{array}$
$\begin{array}{l}
\mathrm{H}_{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HO}_{2}^{-} \\
\mathrm{HO}_{2}^{-}+\mathrm{H}_{2} \mathrm{O}+2 e^{-} \rightleftharpoons 3 \mathrm{OH}
\end{array}$
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