-1
$\begingroup$

This question already has an answer here:

I have the following equation, already written, with a few coefficients to write.

$\ce{2MnO4^- + a H2O2 + b H+ -> 2Mn^2+ +c O2 + dH2O}$

I have put a = 1, b = 6, c = 3 and d = 4.

The number of atoms and the charges appear balanced to me. It is wrong, so where am I mistaking?

EDIT:I have just noticed that the exercise says "under acidic conditions": may it help?

$\endgroup$

marked as duplicate by Mithoron, airhuff, Todd Minehardt, M.A.R., pentavalentcarbon Mar 19 '18 at 14:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If they were expecting a unique set of coefficients, there is some bad news. See the answer(s). $\endgroup$ – Oscar Lanzi Mar 18 '18 at 22:56
  • $\begingroup$ @OscarLanzi Yes, they expected just a set of coefficients; that's what I can guess from the answers who only gave me a=5, b=6, c=5 and d=8. $\endgroup$ – TheBarbarios Mar 23 '18 at 20:59
1
$\begingroup$

The solution is not unique.

$\ce{2 MnO_4^- + 1 H_2O_2 + 6 H^+}\rightarrow \ce{2 Mn^{2+} + 3 O_2 + 4 H_2O}$

And

$\ce{2 MnO_4^- + 3 H_2O_2 + 6 H^+}\rightarrow \ce{2 Mn^{2+} + 4 O_2 + 6 H_2O}$

And

$\ce{2 MnO_4^- + 5 H_2O_2 + 6 H^+}\rightarrow \ce{2 Mn^{2+} + 5 O_2 + 8 H_2O}$

You really have two reactions:

$\ce{2 MnO4- + 6 H^+ -> 2 Mn^2+ +} \frac 52 \ce{O2 + 3 H2O}$

And

$\ce{H2O2 ->} \frac 12 \ce{O2 + H2O}$

The coefficients you get by adding these together depend on how much permanganate and hydrogen peroxide are assumed to react.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.