Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with: (i)

Exercises · 4.15

Q. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide

(ii) iron(II) solution and

(iii)

H_2S

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Concept understanding — Inorganic Synthesis

Inorganic Synthesis – What It Really Means

Imagine you want to build a house. You need bricks, cement, steel, and a plan to put them together. Inorganic synthesis is exactly that — but for making chemical compounds that do not contain carbon-hydrogen bonds (the domain of organic chemistry). You take simple starting materials (elements or simple compounds) and, through a controlled chemical reaction, build a more complex inorganic product.

The intuition is simple: you are a chemist-craftsman. You decide what to make, choose the right ingredients, set the right conditions (temperature, pressure, solvent, time), and then isolate the pure product. The "synthesis" part is the entire journey from idea to pure substance.

The Precise Statement

Inorganic synthesis is the branch of chemistry concerned with the design, planning, and execution of chemical reactions to prepare inorganic compounds — including metals, alloys, coordination complexes, main-group compounds, solid-state materials, and nanomaterials — with controlled purity, structure, and properties.

It is not just "mixing chemicals." It involves:

Choosing the correct starting materials (precursors) — often simple salts, oxides, or elements.
Selecting a reaction method — solid-state heating, solution precipitation, electrochemical deposition, sol-gel, hydrothermal, etc.
Controlling reaction conditions — temperature, pressure, pH, concentration, atmosphere (inert gas, air, vacuum).
Purifying the product — recrystallization, distillation, sublimation, chromatography.
Characterising the product — proving you actually made what you intended (X-ray diffraction, spectroscopy, elemental analysis).

A Concrete Example: Making Copper(II) Sulfate Pentahydrate

You want to make the familiar blue crystal, $\ce{CuSO4·5H2O}$ .

Intuition: You have copper metal (a wire) and dilute sulfuric acid. Copper does not react with dilute acid directly — you need an oxidising agent. So you add nitric acid or simply heat copper with concentrated sulfuric acid.

Reaction:

\ce{Cu + 2H2SO4(conc.) -> CuSO4 + SO2 + 2H2O}

Then you evaporate the solution carefully. Blue crystals of $\ce{CuSO4·5H2O}$ appear.

What you did: You synthesised an inorganic compound from elemental copper and an acid. You controlled the concentration, temperature, and evaporation rate. You then filtered and dried the crystals.

Why It Matters

Inorganic synthesis is the foundation of:

Catalysts (e.g., $\ce{Pt}$ on alumina for car exhausts)
Electronic materials (silicon wafers, gallium arsenide for LEDs)
Medicinal compounds (cisplatin for cancer therapy)
Pigments (titanium dioxide white, Prussian blue)
Batteries (lithium cobalt oxide electrodes)

Without inorganic synthesis, modern technology would not exist.

A Common Misconception …

Why this formula?

Inorganic Synthesis: Why the Key Formulae Hold

Inorganic synthesis is the branch of chemistry concerned with the preparation of inorganic compounds — from simple salts to complex coordination compounds, organometallics, and solid-state materials. The key formulae in this field are not arbitrary; they arise from fundamental principles of stoichiometry, thermodynamics, kinetics, and coordination chemistry.

Let’s break down the reasoning behind the most important formulae.

1. The Yield Formula: Why It’s Not Just “Product/Reactant”

The most basic formula in any synthesis is:

\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%

Why this holds:

Theoretical yield is calculated from the limiting reagent — the reactant that runs out first. This is based on the law of conservation of mass and the stoichiometric coefficients from the balanced chemical equation.
Actual yield is always less than theoretical because of:
- Side reactions (competing pathways)
- Incomplete reactions (equilibrium limitations)
- Loss during purification (filtration, crystallization, etc.)
The formula is a ratio because yield is a fractional measure of efficiency — it tells you how much of the maximum possible product you actually obtained.

Key insight: The formula works only if you correctly identify the limiting reagent. For example, in the synthesis of $\ce{FeCl3}$ from $\ce{Fe}$ and $\ce{Cl2}$ , if you have 1 mol Fe and 2 mol $\ce{Cl2}$ , Fe is limiting (1:1.5 stoichiometry), so theoretical yield is based on Fe.

2. The Atom Economy Formula: Why It Measures “Greenness”

\text{Atom Economy} = \frac{\text{Molecular Mass of Desired Product}}{\text{Sum of Molecular Masses of All Reactants}} \times 100\%

Why this holds:

This formula was introduced by Barry Trost (1991) to quantify how much of the starting materials ends up in the product.
It is not a yield — it’s a theoretical maximum based on the balanced equation. It assumes 100% yield.
The denominator includes all reactants (including solvents if they are consumed, but usually only stoichiometric reagents).
A high atom economy (e.g., 100% for addition reactions like $\ce{A + B -> C}$ ) means less waste. A low atom economy (e.g., substitution reactions with leaving groups) means more byproducts.

Example: In the synthesis of $\ce{NaCl}$ from $\ce{Na}$ and $\ce{Cl2}$ :

$2\ce{Na} + \ce{Cl2} \rightarrow 2\ce{NaCl}$

Atom economy = $\frac{2 \times 58.44}{2 \times 22.99 + 70.90} \times 100\% = 100\%$ — because all atoms end up in the product.

3. The Solubility Product and Precipitation: Why $K_{sp}$ Controls Synthesis

For a sparingly soluble salt like $\ce{AgCl}$ :

\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}

K_{sp} = [\ce{Ag+}][\ce{Cl-}]

Why this holds:

$K_{sp}$ is an equilibrium constant derived from the law of mass action. It applies only to saturated solutions.
In synthesis, you use $K_{sp}$ to predict whether a precipitate will form when mixing solutions. If the ion product $Q = [\ce{Ag+}][\ce{Cl-}]$ exceeds $K_{sp}$ , precipitation occurs.
The formula is temperature-dependent (because $\Delta G^\circ = -RT \ln K_{sp}$ ). So you must control temperature to control precipitation.

Reasoning: The equilibrium constant arises from the balance between the lattice energy (holding the solid together) and the hydration energy (stabilizing ions in solution). A very small $K_{sp}$ means the solid is very stable — useful for gravimetric synthesis.

4. The Coordination Number and Ligand Field Stabilization Energy (LFSE)

For an octahedral complex, the LFSE is:

\text{LFSE} = \left( -0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g} \right) \Delta_o

Why this holds:

This formula comes from crystal field theory (CFT). In an octahedral field, the five $d$ orbitals split into two sets: the lower-energy $t_{2g}$ (three orbitals) and the higher-energy $e_g$ (two orbitals).
The splitting energy $\Delta_o$ is the energy difference between these sets.
Electrons fill the $t_{2g}$ orbitals first (Hund’s rule), and each electron in $t_{2g}$ stabilizes the complex by $-0.4\Delta_o$ relative to the barycenter (average energy). Each electron in $e_g$ destabilizes by $+0.6\Delta_o$ .
The formula explains why certain coordination numbers are preferred: for example, $\ce{[Co(H2O)6]^{2+}}$ (high-spin $d^7$ ) has LFSE = $-0.8\Delta_o$ , while $\ce{[CoCl4]^{2-}}$ (tetrahedral) has a smaller LFSE — so the octahedral form is more stable. …

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Potassium dichromate in acidic medium acts as a strong oxidising agent because the dichromate ion ( $\text{Cr}_2\text{O}_7^{2-}$ ) gets reduced to $\text{Cr}^{3+}$ , gaining six electrons. It oxidises iodide to iodine, iron(II) to iron(III), and hydrogen sulphide to sulphur.

Why Potassium Dichromate is an Oxidising Agent

The oxidising power of potassium dichromate ( $\text{K}_2\text{Cr}_2\text{O}_7$ ) comes from chromium in its +6 oxidation state. In acidic solution, the dichromate ion accepts electrons and gets reduced to the green $\text{Cr}^{3+}$ ion. The half-reaction is:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

This is a six-electron reduction. The standard reduction potential ( $E^\circ = +1.33\ \text{V}$ ) is high enough to oxidise many common reducing agents. The reaction is strongly favoured in acidic medium — in neutral or alkaline conditions, dichromate converts to chromate ( $\text{CrO}_4^{2-}$ ), which is a much weaker oxidant.

Watch out

A common mistake is to forget that the reduction of dichromate consumes 14 H⁺ ions. If the medium is not sufficiently acidic, the reaction slows down or stops. Always write the full ionic equation with $\text{H}^+$ and $\text{H}_2\text{O}$ .

Step-by-Step Ionic Equations

1. Reaction with Iodide ( $\text{I}^-$ )

Iodide is oxidised to iodine. Each $\text{I}^-$ loses one electron, while the dichromate ion accepts six electrons, so six iodide ions are needed to supply those six electrons.

Half-reactions:

Oxidation: $2\text{I}^- \rightarrow \text{I}_2 + 2e^-$ (but this gives only 2 electrons; we need 6)
Multiply by 3: $6\text{I}^- \rightarrow 3\text{I}_2 + 6e^-$
Reduction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Combined:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{I}^- \rightarrow 2\text{Cr}^{3+} + 3\text{I}_2 + 7\text{H}_2\text{O}

The iodine produced gives a brown colour in solution, or a violet colour if extracted into an organic solvent like chloroform.

Tip

To balance redox equations quickly: balance atoms other than H and O first, then balance O with $\text{H}_2\text{O}$ , then H with $\text{H}^+$ , and finally charge with electrons. Then make electrons equal in both halves.

2. Reaction with Iron(II) Solution ( $\text{Fe}^{2+}$ )

Iron(II) is oxidised to iron(III). Each $\text{Fe}^{2+}$ loses one electron. Since dichromate accepts six electrons, we need six $\text{Fe}^{2+}$ ions.

Half-reactions:

Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (multiply by 6)
$6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-$
Reduction: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}$

Combined:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}

This is a classic titration reaction used to estimate iron(II) in solution. The colour change from orange (dichromate) to green (Cr³⁺) marks the endpoint.

Note

In the lab, this reaction is often done in the presence of dilute $\text{H}_2\text{SO}_4$ . Hydrochloric acid is avoided because chloride ions can also be oxidised by dichromate, interfering with the result.

3. Reaction with Hydrogen Sulphide ( $\text{H}_2\text{S}$ ) …

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Method: Oxidation–Reduction (Redox) Half-Reaction Method

This method breaks the overall reaction into two half-reactions — oxidation and reduction — then balances atoms and charge step-by-step.

Steps:

Identify the oxidising agent (gets reduced) and the reducing agent (gets oxidised).
Write the half-reaction for reduction of dichromate.
Write the half-reaction for oxidation of the given species.
Balance atoms other than H and O.
Balance oxygen by adding $H_2O$ .
Balance hydrogen by adding $H^+$ (acidic medium).
Balance charge by adding electrons ( $e^-$ ).
Multiply half-reactions so electrons cancel.
Add the half-reactions and simplify.

Oxidising Action of Potassium Dichromate ( $K_2Cr_2O_7$ )

In acidic medium, dichromate ion ( $Cr_2O_7^{2-}$ ) is a strong oxidising agent. It gets reduced to green $Cr^{3+}$ :

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

This is the reduction half-reaction used in all three cases below.

(i) Reaction with Iodide ( $I^-$ )

Oxidation half-reaction:

2I^- \rightarrow I_2 + 2e^-

Reduction half-reaction (from above):

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

Balance electrons: Multiply oxidation half by 3:

6I^- \rightarrow 3I_2 + 6e^-

Add both half-reactions:

Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2

Ionic equation:

\boxed{Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 7H_2O + 3I_2}

(ii) Reaction with Iron(II) Solution ( $Fe^{2+}$ )

Oxidation half-reaction:

Fe^{2+} \rightarrow Fe^{3+} + e^-

Reduction half-reaction:

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

Balance electrons: Multiply oxidation half by 6:

6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-

Add both half-reactions:

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

Ionic equation: …

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Here is a breakdown of the common mistakes students make when answering this question, along with the correct reasoning and exam-ready solutions.

The Core Concept (Why Students Slip Up)

The key to this question is stoichiometry and medium. Potassium dichromate ( $K_2Cr_2O_7$ ) is a strong oxidising agent only in acidic medium. In neutral or basic medium, its oxidising power is much weaker. Students often forget to specify the medium or balance the half-reactions incorrectly.

Common Mistake #1: Forgetting the Acidic Medium

The Mistake: Writing the reaction without mentioning $H^+$ ions or writing the product as $Cr^{3+}$ without balancing the oxygen with water and hydrogen ions.

Why it happens: Students memorise the half-reaction as:

Cr_2O_7^{2-} \rightarrow 2Cr^{3+}

but forget that this is not balanced for charge or atoms.

How to Avoid:

Always write the balanced half-reaction in acidic medium:

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

Check: Left side: 2 Cr, 7 O, 14 H, charge = -2 + 14 = +12. Right side: 2 Cr, 7 O, 14 H, charge = +6. The 6 electrons balance the charge.
Exam tip: If the question says "acidified potassium dichromate", you must include $H^+$ in the ionic equation.

Common Mistake #2: Incorrect Stoichiometry with Iodide ( $I^-$ )

The Mistake: Writing the product as $I_2$ but getting the mole ratio wrong (e.g., 1:1 instead of 1:6).

Why it happens: Students forget that each $Cr_2O_7^{2-}$ gains 6 electrons, while each $I^-$ loses 1 electron to form $\frac{1}{2}I_2$ .

How to Avoid:

Step 1: Write the oxidation half-reaction:

2I^- \rightarrow I_2 + 2e^-

Step 2: Multiply by 3 to match the 6 electrons gained by dichromate:

6I^- \rightarrow 3I_2 + 6e^-

Step 3: Combine with the reduction half-reaction:

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

Final balanced equation:

Cr_2O_7^{2-} + 14H^+ + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O

Key result: 1 mole $K_2Cr_2O_7$ oxidises 6 moles of $I^-$ .

Common Mistake #3: Confusing Iron(II) with Iron(III) Products

The Mistake: Writing $Fe^{2+} \rightarrow Fe^{3+}$ but forgetting to balance the charge or writing the wrong number of electrons.

Why it happens: Students think it's a simple 1-electron transfer but forget to account for the 6 electrons from dichromate.

How to Avoid:

Oxidation half-reaction:

Fe^{2+} \rightarrow Fe^{3+} + e^-

Multiply by 6:

6Fe^{2+} \rightarrow 6Fe^{3+} + 6e^-

Combine:

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

Key result: 1 mole $K_2Cr_2O_7$ oxidises 6 moles of $Fe^{2+}$ .

Common Mistake #4: Wrong Product for $H_2S$ Oxidation

The Mistake: Writing $H_2S \rightarrow S$ but forgetting that $H_2S$ is a gas and $S$ is a solid (precipitate). Or writing $SO_2$ or $SO_4^{2-}$ as the product.

Why it happens: Students over-oxidise $H_2S$ because they think "strong oxidiser" means all the way to sulfate. But in acidic medium, $H_2S$ is typically oxidised only to elemental sulfur ( $S$ ), not to $SO_2$ or $SO_4^{2-}$ .

How to Avoid:

Oxidation half-reaction:

H_2S \rightarrow S + 2H^+ + 2e^-

Multiply by 3 to match 6 electrons:

3H_2S \rightarrow 3S + 6H^+ + 6e^-

Combine:

Cr_2O_7^{2-} + 14H^+ + 3H_2S \rightarrow 2Cr^{3+} + 3S + 7H_2O + 6H^+

Simplify $H^+$ : $14H^+ - 6H^+ = 8H^+$ on left side.
Final: …

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Q.Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

Why Potassium Dichromate is an Oxidising Agent

Step-by-Step Ionic Equations

1. Reaction with Iodide (I−\text{I}^-I−)

2. Reaction with Iron(II) Solution (Fe2+\text{Fe}^{2+}Fe2+)

3. Reaction with Hydrogen Sulphide (H2S\text{H}_2\text{S}H2​S) …

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1. Reaction with Iodide ( $\text{I}^-$ )

2. Reaction with Iron(II) Solution ( $\text{Fe}^{2+}$ )

3. Reaction with Hydrogen Sulphide ( $\text{H}_2\text{S}$ ) …

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