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.

Redox Reaction

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?

  • 1
    $\begingroup$ Maybe this answer is of any help. $\endgroup$ Jun 16, 2014 at 2:20
  • $\begingroup$ Actually, $Mn^{7+}$ is reduced to $Mn^{6+}$, $Mn^{4+}$ and $Mn^{2+}$ and $SO_3^{2+}$ is oxidized to $SO_4^{2+}$. $\endgroup$
    – LDC3
    Aug 4, 2014 at 5:00

1 Answer 1


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).

With this and the Nernst equation you can relate the concentrations of any set of species that form a balanced reaction. In a complicated mix like this you don't know a priori the concentration of any species, only the certain sums (for instance, for all sulfur-containing ions) but you still can, perhaps iteratively, solve the corresponding set of algebraic equations and find the individual concentrations. All the equilibrium equations have to be satisfied simultaneously; if they aren't the system is not in equilibrium.

(In this answer I've used the word "concentration" but the equations written with concentrations are only approximately satisfied; to account for second-order effects the notion of "activity" was introduced, and while it is an essentially empirical notion, it's still useful if the dependence of activity on concentration for a particular problem has been tabulated.)


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