# Multiple solutions in the balancing of redox reactions

While looking at the reaction between potassium permanganate and hydrogen peroxide in the presence of sulfuric acid a friend and I were attempting to find a general equation to yield the coefficients of the balanced reaction depending on varying levels of acidity. I solved using half reactions and came up with equations for each coefficient in terms of the mole amounts for the manganese(II) ions and manganese(IV) oxide produced. My friend however, went about the problem by setting up a system of equations from the initial skeleton equation and yielded answers which depended on 3 variables.

From this, his method gave the answer

$$\ce{H2O2 + 4MnO4- + 6H+ -> 4O2 + Mn^2+ + 3MnO2 + 4H2O}$$

as a possibility, while my method yielded

$$\ce{7H2O2 + 4MnO4- + 6H+ -> 7O2 + Mn^2+ + 3MnO2 + 10H2O}$$

from the same ratio of manganese(IV) oxide to manganese(II). His reaction appears to balance correctly. Is there something he is missing that prevents that reaction from occurring with his coefficients, or does the half reaction method hide some possible answers?

• What are the half reactions?
– Zhe
Nov 28, 2016 at 2:08

The first thing that you should realise is that under specific conditions (here: acidic ones) only one product is formed in this reaction. Permanganate is not reduced to both manganese(IV) oxide and manganese(II) cations, it will either cleanly give manganese(IV) oxide (if the acid concentration is very low) or cleanly give manganese(II) cations (if the acid concentration is sufficient). The only way to generate both is to have sufficient acid at first which is subsequently used up by the reaction and drops below a threshold. Thus, if you were titrating and notice the dark precipitate of $$\ce{MnO2}$$, immediately stop, throw away the analysis and start again as it has become worthless.
\begin{align}\ce{MnO4- + 8 H+ + 5 e- &-> Mn^2+ + 4 H2O}\tag{Red1}\\[0.6em] \ce{H2O2 &-> O2 + 2 H+ + 2e-}\tag{Ox}\\[0.6em] \ce{2 MnO4- + 6 H+ + 5 H2O2 &-> 2 Mn^2+ + 8 H2O + 5 O2}\tag{Redox1}\\[1.6em] \ce{MnO4- + 3 e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red2}\\[0.6em] \ce{H2O2 &-> O2 + 2 H+ + 2 e-}\tag{Ox}\\[0.6em] \ce{2 MnO4- + 2 H+ + 3 H2O2 &-> 2 MnO2 + 4 H2O}\tag{Redox2}\end{align}