# In a redox reaction, why does more than one oxidation state of an element form?

Consider the reaction 100ml 10M $\ce{NaOH_{(aq)}} +$ 100ml 10M $\ce{NaHSO3_{(aq)} +}$ 100mL 0.01M $\ce{KMnO4_{(aq)}}$

In the picture posted here, the left columns give information on the Molarity of species present before 100mL 0.01 M $\ce{KMnO4}$ is added. The right column is after $\ce{KMnO4}$ is added. In the resulting solution, $\ce{Mn}$ is oxidized to various states: $\ce{Mn^2+ Mn^4+ Mn^6+ Mn^7+}$ (where $\ce{Mn^7+}$ is the original form).

I would like to know, why do different oxidation states of $\ce{Mn}$ appear? How can we predict the concentration of oxidation states?

• Maybe this answer is of any help. Jun 16, 2014 at 2:20
• Actually, $Mn^{7+}$ is reduced to $Mn^{6+}$, $Mn^{4+}$ and $Mn^{2+}$ and $SO_3^{2+}$ is oxidized to $SO_4^{2+}$.
– LDC3
Aug 4, 2014 at 5:00

They all appear because they are all finite amount of energy apart, but note that some concentrations are many orders of magnitude greater than others, and the concentration of $\ce{MnO4-}$ is so low that the table shows a zero.
Predicting the concentration requires being able to read tables of potentials. In standard tables like http://bilbo.chm.uri.edu/CHM112/tables/redpottable.htm you will find equations such as $$\ce{MnO2{}+4H+ +4e- \rightleftharpoons Mn + 2H2O}\quad (1.23\mathrm{V})$$ and $$\ce{Mn^{2+}{}+2e^- \rightleftharpoons Mn}\quad (-1.185\mathrm{V})$$ (the numbers in parentheses are the standard reduction potentials, $E^\circ$, always measured in volts).