Potassium dichromate

Orange crystals that turn a test tube green the instant they meet a reducing agent — that colour change is potassium dichromate, K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7, doing chemistry in front of your eyes. It is one of only two named compounds (alongside KMnO4\text{KMnO}_4) that the ISC d- and f-block unit asks you to know in full preparation-and-property detail, and it earns that spot because it packs four exam-favourite ideas into one molecule: a distinctive two-tetrahedra structure, a multi-step industrial extraction from chromite ore, a clean six-electron oxidising half-reaction in acid, and a pH-driven equilibrium with its yellow cousin, the chromate ion. This article walks through all four, reusing the exact numbers and equations from your syllabus notes, and adds one fully-solved titration calculation so you can see the redox chemistry turned into a real laboratory answer.

The core insight

In Cr2O72−\text{Cr}_2\text{O}_7^{2-}, chromium sits at +6+6 inside two corner-shared CrO4\text{CrO}_4 tetrahedra, and in acidic solution it is reduced to green Cr3+\text{Cr}^{3+} by gaining 6e−6e^- per dichromate ion — the reaction exploited both in the industrial extraction from chromite ore and in redox titrations.

The core idea

Structure and oxidation state. Chromium in Cr2O72−\text{Cr}_2\text{O}_7^{2-} is in the +6+6 oxidation state. Each Cr atom is sp3sp^3 hybridised, at the centre of its own tetrahedron of oxygen atoms, and the two CrO4\text{CrO}_4 tetrahedra are fused by sharing a single corner oxygen — a Cr–O–Cr bridge. That bridged, two-tetrahedra shape is what distinguishes dichromate from the single-tetrahedron chromate ion, CrO42−\text{CrO}_4^{2-}.

Extraction from chromite ore (FeCr2O4\text{FeCr}_2\text{O}_4). Three stages carry Cr from +3+3 in the ore to +6+6 in the final salt. First, alkaline oxidising fusion converts the ore to yellow sodium chromate:

4FeCr2O4+8Na2CO3+7O2→Δ8Na2CrO4+2Fe2O3+8CO24\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \xrightarrow{\Delta} 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2

where: atmospheric O2\text{O}_2 is the oxidant that lifts Cr from +3+3 to +6+6, and insoluble Fe2O3\text{Fe}_2\text{O}_3 is removed by filtration. Second, acidification shifts the chromate–dichromate equilibrium toward orange dichromate; third, treatment with KCl crystallises out K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 because it is less soluble than the sodium salt.

Oxidising nature. Dichromate's oxidising power is sharply pH-dependent — the same feature that makes permanganate tricky, but here the story is simpler because dichromate itself only exists in acidic-to-neutral solution:

Cr2O72−+14H++6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

where: each of the two Cr atoms is reduced from +6+6 to +3+3 (a 3e−3e^- change each, 6e−6e^- total), 14H+14\text{H}^+ balance the seven oxygens as water, and green Cr3+\text{Cr}^{3+} is the product colour you should always name. In basic medium dichromate cannot exist at all — it converts entirely to the much weaker oxidant chromate — so oxidations with dichromate are always run after acidification.

Chromate–dichromate equilibrium. In water the yellow and orange forms interconvert with pH:

2CrO42−(yellow)+2H+⇌Cr2O72−(orange)+H2O2\text{CrO}_4^{2-}(\text{yellow}) + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}(\text{orange}) + \text{H}_2\text{O}

Acid pushes the equilibrium right (yellow to orange); alkali pushes it left (orange to yellow) — this equilibrium is the reason the extraction sequence above needs an acidification step at all.

Figure
Two CrO4 tetrahedra sharing one corner oxygen to form the dichromate ion, next to a single separate CrO4 tetrahedron labelled chromate ion
Dichromate: two tetrahedra fused by an oxygen bridge dichromate ion, Cr2O7 2- Cr bridge O Cr orange, Cr at +6, corner-sharing bridge chromate ion, CrO4 2- Cr yellow, single tetrahedron, no bridge
Derivation ABalancing the dichromate half-reaction

The goal is to build the half-reaction Cr2O72−+14H++6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} from scratch using only the oxidation-state change and atom balancing, so that the electron count and the acid coefficient are not memorised numbers but numbers you can rebuild in an exam under pressure, and so that the same method extends to any acidic-medium oxidant such as permanganate.

Step 1. Write the skeleton and find the oxidation-state change. Dichromate, Cr2O72−\text{Cr}_2\text{O}_7^{2-}, contains two Cr atoms; oxygen is −2-2 throughout, so each Cr is at +6+6 (check: 2(+6)+7(−2)=−22(+6) + 7(-2) = -2, matching the ion charge). The product is Cr3+\text{Cr}^{3+}, so each Cr atom is reduced from +6+6 to +3+3, a gain of 3e−3e^- per Cr atom.

Step 2. Balance the chromium and the electrons. There are two Cr atoms on the left, so the right side needs 2Cr3+2\text{Cr}^{3+}, and the total electron gain is 2×3e−=6e−2 \times 3e^- = 6e^-. Write Cr2O72−+6e−→2Cr3+\text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} as the running skeleton.

Step 3. Balance the oxygen using water. The left side has 7 oxygen atoms (all in dichromate), so the right side needs 7H2O7\text{H}_2\text{O} to account for them: Cr2O72−+6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}.

Step 4. Balance the hydrogen using H+\text{H}^+ (acidic medium). The 7H2O7\text{H}_2\text{O} on the right carries 1414 hydrogen atoms, so 14H+14\text{H}^+ must be added to the left: Cr2O72−+14H++6e−→2Cr3++7H2O\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}.

Step 5. Check both mass and charge balance as a final verification. Atoms: 2 Cr, 7 O and 14 H appear on each side. Charge: left is (−2)+14(+1)+6(−1)=+6(-2) + 14(+1) + 6(-1) = +6; right is 2(+3)+0=+62(+3) + 0 = +6. Both sides match, confirming the half-reaction is fully balanced and ready to combine with any oxidation half-reaction by multiplying to equalise electrons.

Worked example

Problem: 25.0 mL25.0\ \text{mL} of an acidified FeSO4\text{FeSO}_4 solution is exactly titrated by 20.0 mL20.0\ \text{mL} of 0.0200 M0.0200\ \text{M} K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 solution, using diphenylamine as indicator. Find the molarity of the FeSO4\text{FeSO}_4 solution.

Step 1. Write the balanced ionic equation (from the core-idea section): Cr2O72−+6Fe2++14H+→2Cr3++6Fe3++7H2O\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}. The mole ratio at the endpoint is 11 mol dichromate to 66 mol Fe2+\text{Fe}^{2+}.

Step 2. Find moles of K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 used: moles =M×V=0.0200 mol/L×0.0200 L=4.00×10−4 mol= M \times V = 0.0200\ \text{mol/L} \times 0.0200\ \text{L} = 4.00\times10^{-4}\ \text{mol}.

Step 3. Scale by the mole ratio to get moles of Fe2+\text{Fe}^{2+}: moles Fe2+=6×4.00×10−4=2.40×10−3 mol\text{Fe}^{2+} = 6 \times 4.00\times10^{-4} = 2.40\times10^{-3}\ \text{mol}.

Step 4. Divide by the volume of the FeSO4\text{FeSO}_4 solution to get its molarity: M=2.40×10−3 mol0.0250 L=0.0960 MM = \dfrac{2.40\times10^{-3}\ \text{mol}}{0.0250\ \text{L}} = 0.0960\ \text{M}. The FeSO4\text{FeSO}_4 solution is 0.0960 M0.0960\ \text{M}, and at the endpoint the orange dichromate colour gives way to the diphenylamine indicator's sharp violet-blue, since Cr3+\text{Cr}^{3+} alone is only faintly green in dilute solution.

Common traps

  • Drawing dichromate as a single tetrahedron — that shape belongs to chromate, CrO42−\text{CrO}_4^{2-}; dichromate is two corner-sharing tetrahedra joined by a Cr–O–Cr bridge.
  • Skipping the intermediate chromate stage when writing the extraction sequence: it is chromite ore to sodium chromate (fusion) to sodium dichromate (acidification) to K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 (KCl double decomposition), not a one-step conversion.
  • Mixing up which species is oxidised and which is reduced: in the extraction, Cr(+3) in the ore is oxidised to Cr(+6) by O2\text{O}_2; in a titration, dichromate Cr(+6) is the oxidant and is itself reduced to Cr(+3).
  • Assuming dichromate oxidises just as well in any medium — it exists (and oxidises strongly) only in acidic solution; in base it converts entirely to the weaker oxidant chromate.

Ask the tutor: if you would like more practice balancing dichromate redox half-reactions or a side-by-side comparison of KMnO4\text{KMnO}_4 versus K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 as oxidising agents.

Why does dichromate need acid to oxidise Fe2+ but permanganate can also work without it?

Dichromate's own existence as Cr2O72−\text{Cr}_2\text{O}_7^{2-} depends on acid — in neutral or basic solution it converts to the much weaker chromate ion, so without acid there is barely any dichromate left to oxidise anything. Permanganate, by contrast, is a distinct ion at every pH and simply changes which reduction product it gives (Mn2+ in acid, MnO2 in neutral, MnO4 2- in base). So for dichromate, acid is not just a rate booster, it is a precondition for the oxidant to exist in its strong form at all.

Why is potassium dichromate preferred over potassium permanganate as a primary standard?

K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 can be obtained very pure, is stable on storage, is not hygroscopic and does not decompose in light, so a weighed sample gives a solution of exactly known concentration without any prior standardisation. KMnO4\text{KMnO}_4, in contrast, is slightly unstable, reacts with trace organic matter and light, and its solutions typically need to be standardised themselves before use. That reliability is why dichromate is the benchmark solution in many Fe2+ estimations.

Does the colour change in a dichromate titration mean I do not need an indicator?

Not for a precise endpoint. The colour shift from orange dichromate through green Cr3+ is real but gradual and hard to judge sharply by eye, especially with dilute solutions. An indicator such as diphenylamine gives a sudden, unambiguous colour change (to violet-blue) exactly at the equivalence point, which is why it is used routinely alongside the natural colour change rather than instead of it.

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