I'm probably missing a simple component of this reaction, but for some reason I not seeing it.

These are two steps that compose a radical process of the Briggs-Rauscher reaction (that are autocatalytic):

$IO_3^- + HIO_2+H^+ \rightarrow 2IO_2 \cdot +H_2O$
$IO_2 \cdot + Mn^{2+} + H_2O \rightarrow HIO_2 + Mn(OH)^{2+}$

This process is autocatalytic in the sense that 2 $HIO_2$ is produced for each one consumed. However, I'm just not seeing that for some reason. Like I said, it's probably a simple thing I overlooked. So, I'm looking for an explanation for what's going on here.


1 Answer 1


In your first reaction, $\ce{HIO2}$ and $\ce{HIO3}$ react to form $\ce{2IO2.}$ and $\ce{H2O}$.

In your second reaction, $\ce{IO2\cdot}$ is being reduced back to $\ce{HIO2}$ by $\ce{Mn^{2+}}$. Since the first reaction produced $\ce{2IO2.}$, this second reaction needs to happen twice for each time the first reaction occurs. The pair of reactions is really:

$$\ce{ IO3- + HIO2 + H+ -> 2IO2. +H2O }$$ $$2\left(\ce{IO2. +Mn^{2+} +H2O-> HIO2 + Mn(OH)^{2+}} \right)$$

When you combine these reactions into an overall reaction, the instances of $\ce{IO2.}$ cancel and you are left with a net gain of $\ce{HIO2}$.

$$\ce{IO3- + H+ +2Mn^{2+} + H2O-> HIO2 + 2Mn(OH)^{2+}}$$

Of course, the net reaction does not hint at the autocatalytic behavior. To emphasize that the reaction is autocatalytic, you could write the overall reaction in a not quite simplest form:

$$\ce{IO3- + HIO2 + H+ +2Mn^{2+} + H2O-> 2HIO2 + 2Mn(OH)^{2+}}$$


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