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A problem on redox reactions invites the reader to balance the actual reaction from which the redox simplification arose. After some hunting I found the balanced equation and attempted to re-derive the coefficients using:

$$\ce{a H_2SO_4 + b KMnO_4 + cH_2O_2 -> dK_2SO_4 + eMnSO_4 + fH_2O + gO}$$

We can quickly derive some valid relations by looking at each element:

$$ \begin{array}{cll}\\ \ce{H}&\quad a+c &=& f\\ \ce{O}&\quad 4a+4b+2c&=& 4d+4e+f+g\\ \ce{S}&\quad a&=& d+e\\ \ce{K}&\quad b&=& 2d\\ \ce{Mn}&\quad b&=& e \end{array} $$

This can be massaged a little but because there are 5 equations and 7 unknowns it seems that the system is underdetermined.

If we let $a=3,~b=2,~c=5,~d=1,~e=2,~f=8,~g=10$ the equations all work.

My question is:

Could we arrive at these values without guessing or actually doing an experiment to measure (say) the oxygen evolved?

FWIW, I notice an old paper - J. Am. Chem. Soc. 1889, 11 (6), 94-98 - of which I can only see the first page seems to take a similar approach and gets two possible equations, maybe for the reason above.


marked as duplicate by Mithoron, aventurin, Community May 19 '18 at 20:43

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    $\begingroup$ Shouldn't it be $g\ce{O2}$? Nascent oxygen doesn't exist, so is that an error in the reaction product? $\endgroup$ – Gaurang Tandon May 19 '18 at 9:29
  • $\begingroup$ @GaurangTandon: I don't think it matters for purposes of balancing, if you look at the paper, the authors put it this way. Just divide by two to get the amount of oxygen gas. $\endgroup$ – daniel May 19 '18 at 9:31
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    $\begingroup$ We could arrive at these values just as well as any other possible values, of which there are plenty. This is a sum of two reactions, and it could be balanced to anything. Related: chemistry.stackexchange.com/questions/64202/… $\endgroup$ – Ivan Neretin May 19 '18 at 9:57
  • $\begingroup$ @IvanNeretin: This seems like it may be an answer to the question, thanks. $\endgroup$ – daniel May 19 '18 at 12:33
  • $\begingroup$ chemistry.stackexchange.com/questions/63369/… was marked as duplicate of that ^ $\endgroup$ – Mithoron May 19 '18 at 18:16

It seems to me that this is simply a redox reaction in the presence of acid.

If you write the 2 half reactions, and balance in the presence of acid (sulfuric acid), you end up with the same answer (with 5 O2 instead of 10 O on the product side).

This seems to be more of a mathematical manipulation rather than a chemistry problem, IMHO.


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