H₂O₂ 2H⁺ + O₂ + 2e^- ; E⁰ = - 0.68V

Which of the following nature of $H_{2}$ $O_{2}$ isrepresented by the above equation?

Acidic

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Bleaching

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Oxidising

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Reducing

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Solution

The correct option is D
Reducing

Loss of electrons & lower the Electrode potential value

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Similar questions

H_{2} O_{2} \to 2 H^{+} + O_{2} + 2 ¯ e; E^{\circ} = - 0.68 V

H_{2} O_{2}

E^{o}

H_{2} O_{2} \to 2 H^{+} + O_{2} + 2 e^{-}

- 0.68 V

H_{2} O_{2}

H_{2} O_{2} \to 2 H^{+} + O_{2} + 2 e^{-}

H_{2} O_{2}

H_{2} O_{2}

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

H_{2} O_{2} - 2 e^{-} \to O_{2} + 2 H^{+}

H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

H_{2} O_{2} + 2 O H^{-} 2 e^{-} \to O_{2} + 2 H_{2} O

H_{2} O_{2}

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