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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?

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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