Chromium and Manganese Chemistry

Introduction

Chromium and manganese have several oxidation states and vivid oxoanion chemistry. Chromate and dichromate interconvert with pH while keeping Cr at +6. Permanganate contains Mn at +7 and is an oxidant whose reduction product depends strongly on medium. These examples show why oxidation state and protonation or condensation state must be tracked separately.

Core explanation

Chromate, CrO₄²⁻, contains Cr(VI): four O atoms contribute −8, and ion charge −2 requires Cr +6. Dichromate, Cr₂O₇²⁻, also contains Cr(VI) at each of its two equivalent chromium centres: seven O atoms contribute −14 and total Cr must be +12. Their pH-dependent equilibrium can be written 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. In more acidic solution, dichromate becomes more favoured; in more basic solution, chromate becomes more favoured. Chromate solutions are commonly yellow and dichromate solutions orange. The colour shift is not a Cr(VI)-to-Cr(III) redox change because oxidation state remains +6 on both sides.

Cr(VI) oxoanions are oxidising under suitable conditions. In acid, a representative dichromate reduction half-reaction is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Check charge: left −2 +14 −6 = +6; right 2×(+3) = +6. The exact Cr(III) product in water is hydrated and may form complexes, but Cr³⁺ is convenient net-ionic shorthand. The six-electron change follows two Cr centres going from +6 to +3, three electrons each.

Permanganate MnO₄⁻ has Mn(VII), because four oxide oxygens total −8 and ion charge −1 requires Mn +7. In strongly acidic solution, a common reduction half-reaction is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Charge balances: −1 +8 −5 = +2. In neutral or mildly basic water, reduction often produces MnO₂(s) instead, with Mn(IV): MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. In strongly basic settings, manganate MnO₄²⁻ with Mn(VI) can be relevant. A question asking for permanganate's product without stating pH is therefore under-specified.

Manganese has a broad range of oxidation states, including common Mn²⁺. Cr³⁺ is also important and can form coloured coordination complexes. Their high oxoanion states are stabilised by metal–oxygen bonding, not by free Cr⁶⁺ or Mn⁷⁺ aqua ions. Oxidising strength depends on the full half-reaction and medium. The chromium chromate/dichromate equilibrium additionally illustrates how an acid–base or condensation change can move oxygen atoms between oxoanions while leaving formal Cr state unchanged.

Step-by-step reasoning

1. Assign Cr or Mn oxidation state from oxygen and overall ion charge. 2. For chromate/dichromate, write the acid-dependent equilibrium and verify Cr stays +6. 3. For a requested reduction, identify the pH and target product before adding electrons. 4. Balance O with H₂O, H with H⁺ in acid or OH⁻ in base, then balance charge with electrons. 5. Check total atoms and charge and distinguish colour change from proof of redox.

Visual explanation

Draw yellow CrO₄²⁻ units on the basic side and orange Cr₂O₇²⁻ on the acidic side with an H⁺ arrow, labelling Cr +6 throughout. Below draw purple MnO₄⁻ branching to Mn²⁺ in acid and MnO₂(s) in neutral/basic conditions, with electron counts five and three on the corresponding arrows.

Real-world analogy

Two people can join into a pair without either changing their rank. Two chromate units condense into dichromate while each Cr keeps oxidation state +6. By contrast, reducing permanganate changes the manganese electron bookkeeping, more like changing rank itself. The analogy separates association from redox.

Real-world example

Potassium permanganate has been used in redox titrations because its intense purple colour makes an endpoint visible under suitable acidic conditions. The analyst must use the correct MnO₄⁻ → Mn²⁺ five-electron stoichiometry for the acid medium. Applying a neutral-medium MnO₂ product would give the wrong titration factor.

Why?

Why does acid favour dichromate over chromate? H⁺ is a reactant in 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O. Increasing its activity shifts equilibrium toward the dichromate side, without requiring electron transfer at chromium.

Common misconception

“Orange dichromate must contain a higher oxidation state than yellow chromate” is false: both have Cr(VI). Another error is assuming permanganate always yields Mn²⁺; its reduced product depends on medium and reaction conditions.

Worked example

Balance permanganate reduction to Mn²⁺ in acid. Begin MnO₄⁻ → Mn²⁺. Add 4H₂O to products for four O, then 8H⁺ to reactants for H. Charge is now +7 on the left before electrons and +2 on the right, so add 5e⁻ to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The five-electron count also matches Mn +7 to +2.

Quick check

1. Does the chromate-to-dichromate conversion change Cr oxidation state? Answer: No. Cr is +6 in CrO₄²⁻ and each Cr is +6 in Cr₂O₇²⁻. The change is a proton-dependent condensation equilibrium, not chromium redox.

Exam focus

Always assign oxidation states before calling a colour change redox. For chromium, write the balanced chromate/dichromate equilibrium with 2H⁺. For permanganate, state the medium and product before balancing the half-reaction. Verify both atom and charge balance; a memorised equation is useful only if it matches the conditions.

Advanced insight

Chromium(VI) speciation includes hydrogen chromate as well as chromate and dichromate, so the two-form sketch is a teaching simplification. At a particular total concentration and pH, the distribution depends on multiple equilibria. Similarly, manganese redox can involve kinetic pathways and intermediate oxidation states not shown in the net half-reaction.

Summary

Chromate and dichromate both contain Cr(VI) and interconvert with pH; their yellow/orange difference is not a redox change. Permanganate contains Mn(VII) and reduces to different products depending on medium, commonly Mn²⁺ in strong acid or MnO₂ in neutral/basic conditions. Balanced half-reactions and explicit pH are essential.

Practice questions

1. Determine Cr oxidation state in Cr₂O₇²⁻. Answer: Seven O atoms total −14, so two Cr atoms total +12 to give charge −2. Each Cr is +6.

2. How many electrons are accepted when one MnO₄⁻ becomes Mn²⁺ in acid? Answer: Five. Manganese falls from +7 to +2, gaining five electrons; the balanced half-reaction also contains five e⁻.

3. Why does adding acid to yellow chromate solution tend to produce orange colour without an oxidant or reductant? Answer: H⁺ shifts the chromate/dichromate equilibrium toward orange dichromate. Both ions contain Cr(VI), so the shift needs no electron transfer.