Some Important Compounds of Transition Elements

Two orange and purple crystals dominate every Class 12 inorganic-chemistry answer sheet: potassium dichromate, $K_2Cr_2O_7$ , and potassium permanganate, $KMnO_4$ . Both are transition-metal oxo-salts in an unusually high oxidation state — chromium at +6, manganese at +7 — and that high oxidation state is precisely why each is a powerful oxidising agent in acidic solution. The syllabus asks for three linked things: how each salt is manufactured from its natural ore, how the chromate–dichromate equilibrium shifts with pH, and the balanced ionic half-equations that make these compounds the workhorse titrants of volumetric analysis. Master the electron count in each half-equation — six for dichromate, five for permanganate — and every redox titration question involving these two salts becomes a mechanical exercise in balancing rather than a guessing game.

The core insight

$K_2Cr_2O_7$ (Cr, +6 to +3, 6-electron change) and $KMnO_4$ (Mn, +7 to +2, 5-electron change) are strong oxidising agents in acidic medium, manufactured from chromite and pyrolusite ore respectively, whose reduction half-equations anchor every dichromate and permanganate titration.

The core idea

Potassium dichromate, $K_2Cr_2O_7$ , is manufactured from chromite ore ( $FeCr_2O_4$ ) by fusing it with sodium or potassium carbonate in air. This oxidises the chromium and converts iron into $Fe_2O_3$ , leaving sodium chromate in solution. Acidifying gives sodium dichromate, and adding KCl precipitates the less-soluble $K_2Cr_2O_7$ out of the mother liquor. The chromate ion (yellow, tetrahedral, $CrO_4^{2-}$ ) and the dichromate ion (orange, two corner-sharing tetrahedra, $Cr_2O_7^{2-}$ ) are simply the same +6 chromium species in different protonation states, related by a pH-dependent equilibrium:

2\,CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O

where: $CrO_4^{2-}$ is the chromate ion, stable in alkaline medium; $Cr_2O_7^{2-}$ is the dichromate ion, stable in acidic medium; $H^+$ drives the equilibrium to the right as acid is added. Adding alkali reverses it, turning an orange dichromate solution yellow. In acidic solution, dichromate oxidises other species while chromium itself is reduced from +6 to +3:

Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \qquad E^\circ = +1.33\ \text{V}

Potassium permanganate, $KMnO_4$ , is manufactured from pyrolusite ore ( $MnO_2$ ) by fusing it with KOH and an oxidant (fused with air or $KNO_3$ ) to give green potassium manganate, $K_2MnO_4$ (Mn, +6). This is then oxidised — electrolytically, or by passing chlorine or ozone gas through the solution — to purple-black permanganate, $KMnO_4$ (Mn, +7):

2MnO_4^{2-} + Cl_2 \rightarrow 2MnO_4^- + 2Cl^-

In acidic solution permanganate is an even stronger oxidiser, with manganese reduced from +7 all the way to +2:

MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \qquad E^\circ = +1.51\ \text{V}

In neutral or faintly alkaline medium the reduction stops at $MnO_2$ (+7 to +4), and in strongly alkaline medium at manganate, $MnO_4^{2-}$ (+7 to +6) — but only the acidic-medium, +2 pathway is examined in balancing detail. Both salts' industrial value rests on the same property that makes them examinable: their strong, well-defined oxidising action in acidic medium. $K_2Cr_2O_7$ is used in leather tanning and as an oxidiser in organic chemistry (converting alcohols to aldehydes or acids), while $KMnO_4$ is used as a disinfectant, a bleaching agent, and — most relevant here — as the titrant of choice in volumetric estimations of reducing agents such as $Fe^{2+}$ , oxalate, and hydrogen peroxide. In every one of these applications, the number that actually matters for exam purposes is the electron count in the acidic-medium half-equation, because that is what fixes the stoichiometric ratio in any titration calculation.

Figure

Bar chart comparing the oxidation state drop and electron count for dichromate versus permanganate in acidic medium

Why it works

There is no single formula to derive here — the chapter is descriptive chemistry — so the reasoning that matters is why these two ions are strong oxidisers and why the electron counts come out to 6 and 5. Both metals sit in an unusually high, electron-poor oxidation state (+6 for Cr in dichromate, +7 for Mn in permanganate). An ion that far above its stable lower oxidation states is "hungry" for electrons, and accepting them (being reduced) releases a large amount of energy — that is what a large positive $E^\circ$ means. The electron count in each half-equation is fixed purely by how far the metal must fall: chromium drops three units (+6 to +3) but there are two chromium atoms in $Cr_2O_7^{2-}$ , so $3 \times 2 = 6$ electrons; manganese drops five units in one step (+7 to +2) in $MnO_4^-$ , which has only one manganese atom, so $5 \times 1 = 5$ electrons. The oxygen and hydrogen atoms are balanced afterward using $H^+$ and $H_2O$ because the reaction happens in acidic solution, which supplies both. This is why $K_2Cr_2O_7$ can be weighed out directly as a primary standard (it is stable and non-hygroscopic), while $KMnO_4$ , although self-indicating because its own purple colour vanishes at the end point, slowly decomposes and must always be standardised before use.

Worked example

Write the balanced ionic equation for the oxidation of $Fe^{2+}$ to $Fe^{3+}$ by acidified potassium dichromate, and identify the oxidation state change of chromium.

Step 1. Write the given reduction half-equation for dichromate in acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ . Chromium goes from +6 (in $Cr_2O_7^{2-}$ ) to +3 (in $Cr^{3+}$ ), a gain of 3 electrons per Cr atom; with two Cr atoms that is 6 electrons total, matching the equation as written.

Step 2. Write the oxidation half-equation for iron: $Fe^{2+} \rightarrow Fe^{3+} + e^-$ , a loss of one electron per iron atom.

Step 3. Balance electrons between the two halves. The reduction half needs 6 electrons; each oxidation event supplies only 1, so multiply the iron half by 6: $6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-$ .

Step 4. Add the two half-equations and cancel the 6 electrons appearing on each side:

Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 7H_2O + 6Fe^{3+}

Step 5. Verify by checking charge balance on both sides. Left: $(-2) + 14(+1) + 6(+2) = +24$ . Right: $2(+3) + 6(+3) = +24$ . Charges match, and atom counts (2 Cr, 14 H, 7 O, 6 Fe) match too, so the equation is balanced.

Chromium is reduced from +6 to +3 while iron is oxidised from +2 to +3 — this is the exact reaction on which dichromate titrations for estimating $Fe^{2+}$ concentration are built.

Competitive stretch

The parallel permanganate reaction — oxidation of oxalate, $C_2O_4^{2-}$ , to $CO_2$ — proceeds by the same method. Each carbon in oxalate is at +3 and rises to +4 in $CO_2$ , so each oxalate ion supplies 2 electrons: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$ . The reduction half needs 5 electrons per manganese, so the LCM of 5 and 2 is 10: multiply the reduction half by 2 and the oxidation half by 5, then add and cancel the 10 electrons, giving $2MnO_4^- + 16H^+ + 5C_2O_4^{2-} \rightarrow 2Mn^{2+} + 8H_2O + 10CO_2$ . Verify by charge: left $2(-1)+16(+1)+5(-2) = +4$ , right $2(+2) = +4$ — balanced. Manganese falls from +7 to +2, the reaction underlying the standard $KMnO_4$ –oxalate titration, where the disappearance of permanganate's own purple colour marks the end point without any external indicator.

Common traps

Reversing chromate and dichromate stability: chromate ( $CrO_4^{2-}$ , yellow) is the alkaline form, dichromate ( $Cr_2O_7^{2-}$ , orange) is the acidic form — students often swap these.
Using the wrong electron count for permanganate depending on medium: acidic medium gives Mn $^{2+}$ (5-electron change), but neutral/faintly alkaline medium gives $MnO_2$ (3-electron change); mixing these up is a common numerical slip.
Misassigning the oxidation state of Cr in $Cr_2O_7^{2-}$ as +7 (confusing it with Mn in permanganate); it is always +6 for chromium in both chromate and dichromate.
Balancing atoms but forgetting to check charge — always verify both mass balance and charge balance after combining half-equations.

Ask the tutor: Stuck balancing a redox equation involving dichromate or permanganate, or want to check your electron count step by step — ask the AI tutor to walk through it with you.

Why does KMnO4 need to be standardised before a titration but K2Cr2O7 does not?

$K_2Cr_2O_7$ is chemically stable, does not absorb moisture from the air, and can be obtained in high purity, so a freshly weighed sample can be used directly as a primary standard. $KMnO_4$ , by contrast, slowly decomposes in solution (partly catalysed by light and by trace $MnO_2$ ), so its exact concentration drifts over time. That means every fresh batch of $KMnO_4$ solution must first be titrated against a known primary standard, such as oxalic acid, before it can be trusted to measure anything else.

Why does the colour of a dichromate solution change when I add alkali to it?

Adding alkali removes $H^+$ from solution, which pulls the chromate-dichromate equilibrium $2\,CrO_4^{2-} + 2H^+ \rightleftharpoons Cr_2O_7^{2-} + H_2O$ to the left by Le Chatelier's principle. The orange dichromate ions are converted into yellow chromate ions, so the solution visibly turns from orange to yellow. Adding acid does the reverse, pulling the equilibrium right and turning the solution orange again — this colour change is a simple, visual way to test understanding of the equilibrium.

Why is permanganate a stronger oxidising agent than dichromate?

Compare the standard electrode potentials: $+1.51\ \text{V}$ for $MnO_4^-/Mn^{2+}$ versus $+1.33\ \text{V}$ for $Cr_2O_7^{2-}/Cr^{3+}$ . A higher (more positive) $E^\circ$ means a greater tendency to gain electrons, i.e. a stronger oxidising agent, so permanganate oxidises species that dichromate cannot touch. This is also why permanganate must be handled more carefully — its stronger pull for electrons makes it less stable in storage, decomposing slowly even in the dark.