We were taught that $\ce{Cr_2O_7^{2-}}$ (dichromate) in $\ce{H_2SO_4}$ gives $\ce{CrO_3}$

$$\ce{Cr_2O_7^{2-} +H_2SO_4 -> CrO_3 + ...}$$

Link: wikipedia

But we know that dichromate in acidic medium converts to $\rm Cr(III)$.

$$\ce{ Cr_2O^{2−}_7 + 14 H3O+ + 6 e− → 2 Cr^{3+} + 7 H2O}$$

Link: wikipedia

How is this possible?

  • $\begingroup$ Your question answers itself! With $H_2 SO4$, we get $CrO_3$. With $H_3 O^+$, we get $Cr^{3+}$. $\endgroup$ Commented Nov 2, 2016 at 4:40
  • $\begingroup$ It's a matter of concentration $\endgroup$ Commented Nov 2, 2016 at 4:40
  • $\begingroup$ The first reaction is in anhydrous $\ce{H2SO4}$. The second reaction is in an aqueous acidic solution. $\endgroup$
    – MaxW
    Commented Nov 2, 2016 at 5:35

1 Answer 1


Both statements on the behaviour of $\ce{Cr2O7^{2-}}$ are true, but you are confusing two different reactions.

The formation of chromium trioxide from (sodium) dichromate in sulfuric acd is not a redox reaction. There is no partner that could be oxidized by $\ce{Cr2O7^{2-}}$. Concentrated sulfuric acid is an oxidant itself and metals like copper are dissolved (oxidized to $\ce{Cu^{2+}}$) while not hydrogen is generated.

Remember also that sulfuric acid is a strong, oxidizing and dehydrating acid. This effect is used in the production of $\ce{CrO3}$, which can be considered the anhydride of chromic acid ($\ce{H2CrO4}$). Note that chromium in $\ce{Cr2O7^{2-}}$, $\ce{H2CrO4}$, and $\ce{CrO3}$ always has the same oxidation state.

The second reaction that you mentioned does happen too, and this is indeed part of a redox reaction in which $\ce{Cr(VI)}$ is reduced to $\ce{Cr(III)}$. However, this can only happen when a suitable reaction partner is present, which in turn is oxidized by $\ce{Cr2O7^{2-}}$.

  • $\begingroup$ Sorry I was not taught about any details of the reaction, only the reactions were taught. Very nice explanation! Thank you! $\endgroup$
    – Max Payne
    Commented Nov 3, 2016 at 10:54
  • $\begingroup$ @MaxPayne No worries, that's what we are here for :) $\endgroup$ Commented Nov 3, 2016 at 10:55

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