Redox with Dichromate in Acid

Introduction

Dichromate contains two chromium atoms, so its electron count requires attention to both atoms. In the stated acidic reduction, each chromium changes from +6 to +3. One dichromate ion therefore accepts six electrons in total, not three. The balanced half-reaction connects this oxidation-state change to water, H⁺ and charge conservation.

Core explanation

Assign chromium's oxidation state in Cr₂O₇²⁻: 2Cr + 7(−2) = −2, so 2Cr = +12 and each Cr is +6. The product Cr³⁺ has chromium +3. Since there are two chromium atoms, write 2Cr³⁺ on the product side. Seven oxygen atoms become 7H₂O, requiring 14H⁺ on the reactant side. Before adding electrons the left charge is −2 + 14 = +12, and the right charge is 2(+3) = +6. Add six electrons on the left to give equal charge +6. Thus Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

Pair this half with Fe²⁺ → Fe³⁺ + e⁻. Multiply the iron half by six. The net equation is Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺. The charge is +24 on each side: left −2 + 14 + 12, right +6 + 18. Each dichromate ion corresponds to six Fe²⁺ ions when these are the only oxidant and reductant and the stated products form.

The same six-electron half can combine with a two-electron oxidation. For example, Sn²⁺ → Sn⁴⁺ + 2e⁻ releases two electrons per tin ion. Multiply the tin half by three to supply six electrons. The overall equation is Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 7H₂O + 3Sn⁴⁺. Check charge: left −2 + 14 + 6 = +18; right +6 + 12 = +18. The dichromate:tin(II) ratio is 1:3, not 1:6, because each Sn²⁺ supplies two electrons.

Do not confuse two chromium atoms with a diatomic element. Cr₂O₇²⁻ is a polyatomic ion with oxygen and overall −2 charge. Its chromium +6 assignment is formal. The six-electron count is a property of the particular reduction to two Cr³⁺ ions. If a question specifies a different chromium product or medium, rederive the half-reaction rather than carrying this ratio over mechanically.

As with other aqueous redox equations, counter-ions may appear in a molecular equation but are absent from the net ionic core if unchanged. Potassium dichromate is a common reagent source; K⁺ remains +1 and does not accept the six electrons. Likewise, sulfate or chloride may be present depending on the acid and salts used, but only species actually changing or needed to balance O and H belong in the simplest net equation.

Dichromate compounds also have important safety controls in real laboratories because hexavalent chromium species can pose serious health and environmental hazards. Classroom calculations should not be interpreted as instructions to handle the reagent without appropriate procedures. The chemistry lesson here is the stoichiometric electron count, which can be understood from equations without an experiment.

Step-by-step reasoning

1. Use 2Cr + 7(−2) = −2 to find Cr at +6 in dichromate. 2. Write two Cr³⁺ products, add 7H₂O and 14H⁺ for O and H. 3. Add six electrons to the left to balance charge. 4. Match six electrons with the reductant's oxidation half-reaction. 5. Combine and check every atom and the total ionic charge.

Visual explanation

Draw two parallel chromium arrows from +6 down to +3. Label each descent “3e⁻ gained”. Bracket the two arrows and write “6e⁻ per Cr₂O₇²⁻”. Then show six Fe²⁺ arrows each rising by one, or three Sn²⁺ arrows each rising by two. The picture prevents forgetting the ion's two chromium atoms.

Real-world analogy

A container has two receivers, each needing three units. The container's total requirement is six units. Six one-unit donors or three two-unit donors can meet that requirement. Dichromate's two chromium centres create the same accounting pattern for formal electrons, although the actual molecular pathway may be more complex.

Real-world example

In analytical redox calculations, a measured amount of acidic dichromate can determine the amount of an iron(II) or tin(II) reductant from the balanced ratio. The ratio changes with the reductant's electrons released per ion. A careful analyst writes the net ionic equation before using a titre or concentration.

Why?

Why does dichromate require six rather than three electrons? Each of its two chromium atoms falls by three oxidation-state units. The ion's total change is 2 × 3 = 6. Subscripts multiply the per-atom change before any coefficient from the overall equation is applied.

Common misconception

“Chromium +6 to +3 means dichromate accepts three electrons.” That counts only one chromium atom. Cr₂O₇²⁻ contains two chromium atoms and yields two Cr³⁺ ions, so its half-reaction consumes six electrons.

Worked example

How much Fe²⁺ corresponds to 0.0030 mol dichromate in the stated acidic reaction? The net ionic equation has six Fe²⁺ per Cr₂O₇²⁻. Therefore n(Fe²⁺) = 6 × 0.0030 = 0.018 mol. This assumes Fe²⁺ is the only reductant measured and that chromium becomes Cr³⁺. A 1:3 ratio would apply to a two-electron donor such as Sn²⁺, not to Fe²⁺.

Quick check

1. How many electrons does one Cr₂O₇²⁻ ion accept when it becomes two Cr³⁺ ions in acid? Answer: Six electrons, because each of two chromium atoms falls from +6 to +3.

Exam focus

Show the 2Cr calculation and write 2Cr³⁺ before balancing water and H⁺. State the medium and final chromium state. Use the reductant's actual electron count to derive its coefficient, then verify charge.

Advanced insight

The six-electron half-reaction is a net thermodynamic and stoichiometric description. Detailed reduction pathways can involve intermediate chromium oxidation states and ligand changes. The balanced equation is valid for amount calculations when the net reactants and products are correct, even though it does not specify every intermediate.

Summary

In acid, Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. Two Cr atoms each fall from +6 to +3, giving six accepted electrons per dichromate. This pairs with six Fe²⁺ ions or three Sn²⁺ ions under the stated product assumptions.

Practice questions

1. Find chromium's oxidation state in Cr₂O₇²⁻. Answer: +6, since 2Cr − 14 = −2. 2. What is the acidic dichromate:Sn²⁺ ratio if Sn²⁺ becomes Sn⁴⁺? Answer: 1:3, because one dichromate accepts six electrons and each tin(II) supplies two. 3. Why should K⁺ from potassium dichromate not be included in the simplest net ionic redox equation? Answer: K⁺ remains +1 and is a spectator; it does not participate in the oxidation-state change.