For balancing redox reactions in an acidic medium I follow this procedure:
1. Writing the half reaction
2. Balance the atoms other than $\ce{O}$ and $\ce{H}$
3. Balance the oxygen by adding water
4. Balance the hydrogen by adding protons
5. Make the number of electrons equal and add the half reaction.

Is there any difference if I change the medium if so what are the differences ?

  • $\begingroup$ You're good. That procedure works in any and every case. $\endgroup$ May 2, 2017 at 3:09

1 Answer 1


The five points to pay attention are correct; and I would like to add a sixth one: depending on the reaction condition, namely if the reaction occurs in acidic, neutral, or basic solution, the outcome and the reaction product(s) of the half reaction may differ.

Take potassium permanganate, for example. Under aqueous acidic conditions, this is a very powerful oxidiser, and formally $\ce{Mn^{7+}}$ is reduced all the way down to $\ce{Mn^{2+}}$. This half reaction, starting from purple coloured solution of $\ce{KMnO4}$ to practical colourless $\ce{MnSO4}$ (assuming $\ce{H2SO4}$ as the acid present) may be written as

$$ \ce{MnO4^- + 5e^- + 8H^+ -> Mn^{2+} + 4H2O} $$

Under neutral conditions, the product however is manganese dioxide (formally $\ce{Mn^{4+}}$), which is poorly soluble in water and precipitates

$$ \ce{MnO4^- + 3e^- + 2H2O -> MnO2 + 4OH^-} $$

Eventually under (very) basic conditions, the reaction will stop already at a green colour in form of oxidation state $\ce{Mn^{6+}}$:

$$ \ce{MnO4^- + e^- -> MnO4^{2-}} $$

Paying attention to this has practical, preparative relevance, too. It offers one way to attune the oxidation power of potassium permanganate. A moderated reactivity often correlates with selectivity of the reactions.

This pH-dependence of oxidation-reduction reactions correlates with electrode potentials which often are already tabulated, like here or in the CRC Handbook of Chemical and Physics (keyword Nernst equation).

As a side note: These equations sometimes do not tell the complete story. $\ce{MnO2}$, for example, may act / may be used itself as oxidiser. The simple sum formula however hides well that the oxidising power of this material depends how this material was prepared, too.


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