Chromate-dichromate interconversion

One oxidation state, two structures. Chromium(VI) in aqueous solution exists as either a lone tetrahedral $\text{CrO}_4^{2-}$ ion or as two such tetrahedra sharing one corner oxygen — a $\text{Cr–O–Cr}$ bridge — to give $\text{Cr}_2\text{O}_7^{2-}$. The dichromate ion forms by a condensation reaction: two chromate ions lose a molecule of water as they join. Because the oxidation number of chromium is $+6$ in both species, converting one into the other is not a redox change; only the degree of protonation and condensation of the chromium-oxygen anion changes. The equilibrium that links them is:

where: $\text{CrO}_4^{2-}$ is the chromate ion (yellow), $\text{H}^+$ is supplied by whatever acid or removed by whatever alkali is added, and $\text{Cr}_2\text{O}_7^{2-}$ is the dichromate ion (orange). Reading the equation as an equilibrium expression,

where: $K$ is the equilibrium constant for the interconversion at a fixed temperature, and the square brackets denote molar concentrations. The exponent $2$ on $[\text{H}^+]$ — a direct consequence of the balanced equation needing two protons per dichromate formed — is the algebraic reason the colour change looks so sharp, worked out below.

Le Chatelier's principle applied to pH. Adding acid raises $[\text{H}^+]$, a reactant, so the equilibrium shifts right: chromate is consumed and dichromate builds up, and in strongly acidic solution chromium(VI) exists almost entirely as $\text{Cr}_2\text{O}_7^{2-}$. Adding alkali neutralises $\text{H}^+$ ($\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$), lowering $[\text{H}^+]$, so the equilibrium shifts left: dichromate is converted back to chromate, and in strongly alkaline solution chromium(VI) exists almost entirely as $\text{CrO}_4^{2-}$. At intermediate pH both ions coexist, giving an intermediate yellow-orange shade.

Why it matters practically. The two forms differ in reactivity, so $\text{K}_2\text{Cr}_2\text{O}_7$'s oxidising titrations are always carried out in acidic medium, keeping the equilibrium pinned toward the reactive dichromate/acidic-chromate species; in alkaline conditions the equilibrium sits toward the much weaker oxidant, chromate, and the oxidising action largely disappears.

Problem. A student adds dilute sulphuric acid, drop by drop, to a yellow solution of $\text{K}_2\text{CrO}_4$ that starts near pH $9$, and continues until the pH reaches $6$. Explain the colour change observed and estimate, using the result above, how much the dichromate-favouring ratio $Q = [\text{Cr}_2\text{O}_7^{2-}]/[\text{CrO}_4^{2-}]^2$ has increased.

Step 1. At the start, near pH $9$, $[\text{H}^+]$ is low, so by $Q = K\,[\text{H}^+]^2$ the ratio $Q$ is small: chromium(VI) is present predominantly as $\text{CrO}_4^{2-}$, and the solution is yellow.

Step 2. As dilute $\text{H}_2\text{SO}_4$ is added, $[\text{H}^+]$ rises steadily and the pH falls from $9$ toward $6$, a drop of $\Delta\text{pH} = 3$ units. By Le Chatelier's principle applied to $2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}$, increasing $[\text{H}^+]$ (a reactant) pushes the equilibrium toward the products, converting $\text{CrO}_4^{2-}$ into $\text{Cr}_2\text{O}_7^{2-}$.

Step 3. Use $\log Q = \log K - 2\,\text{pH}$ to size the effect. Over a $3$-unit pH drop, $\log Q$ increases by $2 \times 3 = 6$, so $Q$ itself increases by a factor of $10^6$ — a million-fold swing toward dichromate for a modest change in acidity.

Step 4. State the observation. As more dichromate forms relative to chromate, the solution's colour moves from yellow, through intermediate yellow-orange shades, to the deep orange characteristic of a solution dominated by $\text{Cr}_2\text{O}_7^{2-}$. With enough acid, the colour change is essentially complete — pure yellow to pure orange — and chromium stays at $+6$ throughout: this is an equilibrium shift, not a redox reaction, and the million-fold change in $Q$ over just three pH units is why the transition looks abrupt rather than gradual.

Competitive stretch

Running the same acid addition in reverse — adding excess NaOH to an orange $\text{K}_2\text{Cr}_2\text{O}_7$ solution — neutralises $\text{H}^+$ and, by the same $Q = K[\text{H}^+]^2$ relation, collapses $Q$ back toward zero as pH rises, driving the equilibrium fully to the left and turning the solution yellow again; because chromium remains $+6$ throughout, this reverse shift is not a redox process either, only a lower-$[\text{H}^+]$ equilibrium position.

• Treating the interconversion as a redox reaction: it is not — chromium stays at $+6$ in both $\text{CrO}_4^{2-}$ and $\text{Cr}_2\text{O}_7^{2-}$, and no electrons are transferred; only the pH-dependent equilibrium position changes.
• Assuming the ratio $Q$ scales linearly with $[\text{H}^+]$: because $[\text{H}^+]$ is squared in the equilibrium expression, a one-unit pH change alters $Q$ a hundredfold, not tenfold — this is what makes the colour shift look sharp.
• Mixing up the direction: acid drives the system toward orange (dichromate), alkali drives it toward yellow (chromate) — matching $\text{H}^+$ sitting on the chromate/reactant side of the written equilibrium.
• Confusing this pH equilibrium with dichromate's separate oxidising half-reaction (reduction of $\text{Cr}_2\text{O}_7^{2-}$ to green $\text{Cr}^{3+}$ in the presence of a reducing agent) — that is a genuine, distinct redox process, not the interconversion itself.

Ask the tutor: Ask the AI tutor to walk through a numerical pH problem using $\log Q = \log K - 2\,\text{pH}$, or to compare the chromate and dichromate ion structures side by side.