User talk:Dtlima

The Decomposition of Hydrogen Peroxide H2O2, with Manganese(IV) Oxide MnO2 as the catalyst.

Hydrogen Peroxide spontaneously decomposes according to the reaction

$$2 H_2O_2(l) \xrightarrow[catalyst]{MnO_2(s)} 2 H_2O(l) + O_2(g)$$

In the presence of solid MnO2, H2O2 will first be oxidized and then reduced according to the reactions

Oxidation of H2O2: $$\begin{align} 2 MnO_2(s) + H_2O_2(aq) + 6 H^+(aq) & \rightarrow 2 Mn^{3+}(aq) + O_2(g) + 4 H_2O(l), & {E_{cell}^o}_{ox of H_2O_2} = +0.25V \\ MnO_2(s) + H_2O_2(aq) + 2H^+(aq) & \rightarrow Mn^{2+}(aq) + O_2(g) + 2 H_2O(l), & {E_{cell}^o}_{ox} = +0.53V \end{align} $$ Reduction of H2O2: $$   \begin{align} 2 Mn^{3+}(aq) + H_2O_2(l) + 2 H_2O(l) & \rightarrow 2 MnO_2(s) + 6 H^+(aq), & {E_{cell}^o}_{red of H2O_2} = +0.83V \\ Mn^{2+}(aq) + H_2O_2(l) & \rightarrow MnO_2(s) + 2 H^+(aq), & {E_{cell}^o}_{red} = +0.55V \end{align} $$

The Gibbs energy of these reactions may be calculated like so: $$   \begin{align} \Delta{G_{cell}^o}_{ox} = -nFE_{cell}^o & = -(2 mol e^-)(96485.34 \frac{C}{mol e^-})(+0.53 or +0.25 \frac{J}{C}) = -102 kJ or -48.2 kJ \\ \Delta{G_{cell}^o}_{red} & = -(2 mol e^-)(96485.34 \frac{C}{mol e^-})(+0.83 or +0.55 \frac{J}{C}) = -160 kJ or -106 kJ   \end{align} $$

Note: MnO2 is reduced to both Mn(II) and Mn(III) and both Mn(II) and Mn(III) are oxidized to MnO2, leading to a more complex situation.

It is apparent that the first step of catalysis is the limiting reaction (I think?). I am wondering how much catalyst MnO2 I will need. I could figure this out if I had the reaction constant k; or we could do a test in a pressure chamber and see how long it takes for the reaction to go to completion depending on the amount of MnO2 added.