redox - What role does H2O2 have in copper (II) acetate formation?

Mixing solid copper, 5% vinegar, and hydrogen peroxide ($\ce{H2O2}$) causes copper acetate to form. The process will occur very slowly without hydrogen peroxide. Adding $\ce{H2O2}$ speeds the formation of copper acetate.

How does Hydrogen Peroxide promote the formation of Copper (II) Acetate?

Answer

The redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ cannot oxidize $\ce{Cu}$ to $\ce{Cu^2+}$:

$$\begin{alignat}{2} \ce{Cu^2+ + 2e- \;&<=> Cu}\quad &&E^\circ = +0.340\ \mathrm{V}\\ \ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V} \end{alignat}$$

Thus, non-oxidizing acids such as acetic acid cannot directly oxidize copper.

However, $\ce{Cu}$ can be oxidized by $\ce{O2}$:

$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad E^\circ = +1.229\ \mathrm{V}$$

Therefore, copper is slowly oxidized in acetic acid in contact with air.

The oxidation can be increased by addition of oxidizing agents such as hydrogen peroxide:

$$\ce{H2O2 + 2H+ + 2e- <=> 2H2O}\quad E^\circ = +1.763\ \mathrm{V}$$