H2O2→2H++O2+2e; E∘=−0.68V This equation represents which of the following behaviour of H2O2

$H_{2} O_{2} \to 2 H^{+} + O_{2} + 2 e; E^{\circ} = - 0.68 V$ This equation represents which of the following behaviour of $H_{2} O_{2}$

As $H_{2} O_{2}$ is loosing electrons so it is acting as reducing agent

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