Quote and apply suitable electrode potentials, $E^\ominus$, to build two equations that demonstrate how $\text{MnO}_2$ catalyses this reaction. Give equation 1 and equation 2.
The uncatalysed decomposition equation for hydrogen peroxide is shown: $2\text{H}_2\text{O}_2(aq) \rightarrow 2\text{H}_2\text{O}(l) + \text{O}_2(g)$. Under certain conditions, this reaction is first order with respect to hydrogen peroxide, and the rate constant, $k$, is $2.0 \times 10^{-6}\,\text{s}^{-1}$ at $298\,\text{K}$. Calculate the initial rate at which a $0.75\,\text{mol dm}^{-3}$ hydrogen peroxide solution decomposes at $298\,\text{K}$. Initial rate $= \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\,\text{mol dm}^{-3}\,\text{s}^{-1}$
A four-step mechanism is proposed for the reaction between hydrogen peroxide and iodide ions in acidic solution. Step 1: $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{IO}^- + \text{H}_2\text{O}$. Step 2: $\text{H}^+ + \text{IO}^- \rightarrow \text{HIO}$. Step 3: $\text{HIO} + \text{I}^- \rightarrow \text{I}_2 + \text{OH}^-$. Step 4: $\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}$. Step 1 is the rate-determining step. State what is meant by the term rate-determining step.
Using this mechanism, construct a balanced equation for this reaction.
Deduce the order of reaction for each of the following: $\text{H}_2\text{O}_2 = \ldots\ldots\ldots\ldots$, $\text{I}^- = \ldots\ldots\ldots\ldots$, $\text{H}^+ = \ldots\ldots\ldots\ldots$.