I can't find a method to balance a reaction: the heating of potassium dichromate, which forms potassium dichromate (III), dichromate oxide (III) and oxygen. I can't add water products as $\ce{H+}$, $\ce{{}^{-}OH}$ because we are not in solution. So, how can I balance it?

Who oxidises in this redox? The oxygen inside the salt? The oxygen of $\ce{H2O}$ which is in the air?


I think that part of your difficulty lies in incorrect naming of the products which are simply potassium chromate(VI) and chromium(III) oxide together with oxygen. The equation then becomes:

$$\ce {4K2Cr2O7 -> 4K2CrO4 + 2Cr2O3 + 3O2}$$

As for what is oxidized and reduced; the chromium is reduced from chromium(VI) to chromium(III) in going from $\ce{K2Cr2O7}$ to $\ce{Cr2O3}$ and some of the oxygen is oxidized from −2 in the dichromate to 0 in the oxygen gas.

| improve this answer | |
  • $\begingroup$ Thank you, you're right there was an error in the products. But is there a method to balance it or you did it by trial and error? $\endgroup$ – user9824 Nov 16 '14 at 21:57
  • 2
    $\begingroup$ Small nitpick, the oxidation state in potassium chromate is not three, but six: $$\ce{4\overset{+I}{K}_2\overset{+VI}{Cr}_2\overset{-II}{O}_7 -> 4\overset{+I}{K}_2\overset{+VI}{Cr}\overset{-II}{O}_4 + 2\overset{+III}{Cr}_2\overset{-II}{O}_3 + 3\overset{\pm0}{O}_2}$$ $\endgroup$ – Martin - マーチン Nov 17 '14 at 2:51
  • $\begingroup$ The "nitpick" is absolutely correct and I have ammended my answer accordingly. Apologies for not proof-reading my answer carefully enough. The comment helpfully shows how I ballanced the equation. Provided you know the reactants and products (which you did) look for the oxidation numbers which change; in this case some of the chromium: down 3 and some of the oxygen: up 2. To balance the equation these must match, so for every 3 oxygen atoms in O2 there must be two chromium atoms in Cr2O3. Hence 2Cr2O3 + 3O2. You then adjust the other two to make this so. I have run out of characters. Sorry. $\endgroup$ – ed kinsella Nov 17 '14 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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