Redox and electrochemistry — Question 136

Answer
Oxygen states are −1, −2 and 0 respectively. Oxidation: ${\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$. Reduction: ${\mathrm{H}}_2{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to 2{\mathrm{H}}_2\mathrm{O}$. Peroxide both loses and gains electrons in different reacting molecules.

Working
In H₂O₂, two H atoms contribute +2, so the two O atoms together contribute −2. In water oxygen is −2, whereas elemental O₂ is 0. The peroxide molecules converted to O₂ are oxidised and act as reducing agent. Those converted to H₂O are reduced and act as oxidising agent. Adding the half-equations cancels H⁺ and electrons and reproduces the supplied net equation.

Marking criteria

• Determines oxygen in H₂O₂ as −1. [1 mark]
• Determines oxygen in H₂O as −2 and in O₂ as 0. [1 mark]
• Writes the balanced peroxide oxidation half-equation. [1 mark]
• Writes the balanced peroxide reduction half-equation. [1 mark]
• Links peroxide oxidation/electron loss to its reducing-agent role. [1 mark]
• Links peroxide reduction/electron gain to its oxidising-agent role. [1 mark]