Q.Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
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🔒 Start your 14-day free trial to unlock the full solution →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, CuSOX4⋅5HX2O.
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:
Cu+2HX2SOX4(conc⋅)CuSOX4+SOX2+2HX2O
Then you evaporate the solution carefully. Blue crystals of CuSOX4⋅5HX2O 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., 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:
Percentage Yield=Theoretical YieldActual Yield×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 FeClX3 from Fe and ClX2, if you have 1 mol Fe and 2 mol ClX2, Fe is limiting (1:1.5 stoichiometry), so theoretical yield is based on Fe.
2. The Atom Economy Formula: Why It Measures “Greenness”
Atom Economy=Sum of Molecular Masses of All ReactantsMolecular Mass of Desired Product×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 A+BC) means less waste. A low atom economy (e.g., substitution reactions with leaving groups) means more byproducts.
Example: In the synthesis of NaCl from Na and ClX2:
2Na+ClX2→2NaCl
Atom economy = 2×22.99+70.902×58.44×100%=100% — because all atoms end up in the product.
3. The Solubility Product and Precipitation: Why Ksp Controls Synthesis
For a sparingly soluble salt like AgCl:
AgCl(s)AgX+(aq)+ClX−(aq)
Ksp=[AgX+][ClX−]
Why this holds:
- Ksp is an equilibrium constant derived from the law of mass action. It applies only to saturated solutions.
- In synthesis, you use Ksp to predict whether a precipitate will form when mixing solutions. If the ion product Q=[AgX+][ClX−] exceeds Ksp, precipitation occurs.
- The formula is temperature-dependent (because ΔG∘=−RTlnKsp). 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 Ksp 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:
LFSE=(−0.4×nt2g+0.6×neg)Δo
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 t2g (three orbitals) and the higher-energy eg (two orbitals).
- The splitting energy Δo is the energy difference between these sets.
- Electrons fill the t2g orbitals first (Hund’s rule), and each electron in t2g stabilizes the complex by −0.4Δo relative to the barycenter (average energy). Each electron in eg destabilizes by +0.6Δo.
- The formula explains why certain coordination numbers are preferred: for example, [Co(HX2O)X6]X2+ (high-spin d7) has LFSE = −0.8Δo, while [CoClX4]X2− (tetrahedral) has a smaller LFSE — so the octahedral form is more stable. …
The key idea is that potassium dichromate (K2Cr2O7) acts as a strong oxidising agent in acidic medium, where the dichromate ion (Cr2O72−) is reduced to Cr3+ (green). The half-reaction is:
Cr2O72−+14H++6e−→2Cr3++7H2O
Step 1 – Reaction with iodide (I−): Iodide is oxidised to iodine (I2). Balancing electrons (6e⁻ from dichromate, 2e⁻ per I2 molecule) gives:
Cr2O72−+14H++6I−→2Cr3++7H2O+3I2
Step 2 – Reaction with iron(II) solution (Fe2+): Fe2+ is oxidised to Fe3+ (1e⁻ each). Balancing:
Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+ …
Potassium dichromate in acidic medium acts as a strong oxidising agent because the dichromate ion (Cr2O72−) gets reduced to Cr3+, 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 (K2Cr2O7) comes from chromium in its +6 oxidation state. In acidic solution, the dichromate ion accepts electrons and gets reduced to the green Cr3+ ion. The half-reaction is:
Cr2O72−+14H++6e−→2Cr3++7H2O
This is a six-electron reduction. The standard reduction potential (E∘=+1.33 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 (CrO42−), which is a much weaker oxidant.
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 H+ and H2O.
Step-by-Step Ionic Equations
1. Reaction with Iodide (I−)
Iodide is oxidised to iodine. Each 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: 2I−→I2+2e− (but this gives only 2 electrons; we need 6)
- Multiply by 3: 6I−→3I2+6e−
- Reduction: Cr2O72−+14H++6e−→2Cr3++7H2O
Combined:
Cr2O72−+14H++6I−→2Cr3++3I2+7H2O
The iodine produced gives a brown colour in solution, or a violet colour if extracted into an organic solvent like chloroform.
To balance redox equations quickly: balance atoms other than H and O first, then balance O with H2O, then H with H+, and finally charge with electrons. Then make electrons equal in both halves.
2. Reaction with Iron(II) Solution (Fe2+)
Iron(II) is oxidised to iron(III). Each Fe2+ loses one electron. Since dichromate accepts six electrons, we need six Fe2+ ions.
Half-reactions:
- Oxidation: Fe2+→Fe3++e− (multiply by 6)
- 6Fe2+→6Fe3++6e−
- Reduction: Cr2O72−+14H++6e−→2Cr3++7H2O
Combined:
Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2O
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.
In the lab, this reaction is often done in the presence of dilute H2SO4. Hydrochloric acid is avoided because chloride ions can also be oxidised by dichromate, interfering with the result.
3. Reaction with Hydrogen Sulphide (H2S) …
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 H2O.
- 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 (K2Cr2O7)
In acidic medium, dichromate ion (Cr2O72−) is a strong oxidising agent. It gets reduced to green Cr3+:
Cr2O72−+14H++6e−→2Cr3++7H2O
This is the reduction half-reaction used in all three cases below.
(i) Reaction with Iodide (I−)
Oxidation half-reaction:
2I−→I2+2e−
Reduction half-reaction (from above):
Cr2O72−+14H++6e−→2Cr3++7H2O
Balance electrons: Multiply oxidation half by 3:
6I−→3I2+6e−
Add both half-reactions:
Cr2O72−+14H++6I−→2Cr3++7H2O+3I2
Ionic equation:
Cr2O72−+14H++6I−→2Cr3++7H2O+3I2
(ii) Reaction with Iron(II) Solution (Fe2+)
Oxidation half-reaction:
Fe2+→Fe3++e−
Reduction half-reaction:
Cr2O72−+14H++6e−→2Cr3++7H2O
Balance electrons: Multiply oxidation half by 6:
6Fe2+→6Fe3++6e−
Add both half-reactions:
Cr2O72−+14H++6Fe2+→2Cr3++7H2O+6Fe3+
Ionic equation: …
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 (K2Cr2O7) 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 Cr3+ without balancing the oxygen with water and hydrogen ions.
Why it happens: Students memorise the half-reaction as:
Cr2O72−→2Cr3+
but forget that this is not balanced for charge or atoms.
How to Avoid:
Always write the balanced half-reaction in acidic medium:
Cr2O72−+14H++6e−→2Cr3++7H2O
- 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 I2 but getting the mole ratio wrong (e.g., 1:1 instead of 1:6).
Why it happens: Students forget that each Cr2O72− gains 6 electrons, while each I− loses 1 electron to form 21I2.
How to Avoid:
- Step 1: Write the oxidation half-reaction:
2I−→I2+2e−
- Step 2: Multiply by 3 to match the 6 electrons gained by dichromate:
6I−→3I2+6e−
- Step 3: Combine with the reduction half-reaction:
Cr2O72−+14H++6e−→2Cr3++7H2O
- Final balanced equation:
Cr2O72−+14H++6I−→2Cr3++3I2+7H2O
Key result: 1 mole K2Cr2O7 oxidises 6 moles of I−.
Common Mistake #3: Confusing Iron(II) with Iron(III) Products
The Mistake: Writing Fe2+→Fe3+ 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:
Fe2+→Fe3++e−
- Multiply by 6:
6Fe2+→6Fe3++6e−
- Combine:
Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2O
Key result: 1 mole K2Cr2O7 oxidises 6 moles of Fe2+.
Common Mistake #4: Wrong Product for H2S Oxidation
The Mistake: Writing H2S→S but forgetting that H2S is a gas and S is a solid (precipitate). Or writing SO2 or SO42− as the product.
Why it happens: Students over-oxidise H2S because they think "strong oxidiser" means all the way to sulfate. But in acidic medium, H2S is typically oxidised only to elemental sulfur (S), not to SO2 or SO42−.
How to Avoid:
- Oxidation half-reaction:
H2S→S+2H++2e−
- Multiply by 3 to match 6 electrons:
3H2S→3S+6H++6e−
- Combine:
Cr2O72−+14H++3H2S→2Cr3++3S+7H2O+6H+
- Simplify H+: 14H+−6H+=8H+ on left side.
- Final: …
- TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQQ.(CH3)3CHKMnO4XCu 573KY The number of sp3 and sp2 carbons in Y are respectively (A) 3, 1 (B) 1, 3 (C) 2, 2 (D) 4, 0
›Reveal solutionSolution
The reaction sequence oxidises isobutane to tert-butyl alcohol, then dehydrates it to isobutylene; Y has two sp³ and two sp² carbons, so the answer is (C).
The key is to track the carbon skeleton through two transformations. The starting molecule is isobutane, (CH3)3CH. The first step uses hot KMnO4, a strong oxidising agent. The second step uses copper at 573 K, a classic dehydration condition for alcohols.
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Identify the starting compound and the first reaction.
Isobutane has a tertiary carbon (the CH group) and three methyl groups. Hot KMnO4 oxidises alkanes at the most substituted carbon, converting a C–H bond to a C–OH bond. So the product X is tert-butyl alcohol: (CH3)3C−OH.
Why this works: The tertiary C–H is the weakest, so it’s attacked first; permanganate inserts oxygen.
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Determine the second reaction.
Passing an alcohol over copper at 573 K (about 300 °C) causes dehydration. For a tertiary alcohol like tert-butyl alcohol, elimination is very easy — it loses water to form an alkene. The only possible alkene here is isobutylene (2-methylpropene): (CH3)2C=CH2.
Why this works: Tertiary alcohols dehydrate via an E1 mechanism; the carbocation intermediate is stable, and the most substituted alkene (Saytzeff product) forms.
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Count the hybridisation in Y (isobutylene).
Draw the structure:
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- TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQQ.Composition of siderite ore is (A) CuCO3.Cu(OH)2 (B) FeCO3 (C) CuFeS2 (D) ZnCO3
›Reveal solutionSolution
Siderite is an iron carbonate ore. Its chemical formula is FeCO3, which corresponds to option (B).
The question asks for the composition of siderite ore. This is a straightforward recall from metallurgy — the study of how metals are extracted from their ores. Each ore has a distinct chemical identity, and siderite is one of the common iron ores.
The key idea: ores are naturally occurring minerals from which metals are extracted. Iron ores include hematite (Fe2O3), magnetite (Fe3O4), limonite (Fe2O3⋅3H2O), and siderite. The name "siderite" comes from the Greek word for iron, sideros, which hints at its composition.
Let’s go through the options one by one.
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Option (A): CuCO3⋅Cu(OH)2
This is malachite, a copper ore. The dot here indicates a hydrated or basic carbonate — copper carbonate combined with copper hydroxide. Not siderite.
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Option (B): FeCO3
This is iron(II) carbonate. Siderite is indeed ferrous carbonate. It forms in sedimentary environments and is an important but less common iron ore. The iron is in the +2 oxidation state.
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Option (C): CuFeS2 …
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- TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQQ.Consider the following reactions Cs + O2 (excess) → X Cs + O2 (limited) → X Na + O2 → Y Identify the correct statement about X and Y (A) Y is monoxide and X is superoxide (B) Y is peroxide and X is peroxide (C) Y is peroxide and X is superoxide (D) Y is superoxide and X is peroxide
›Reveal solutionSolution
The key idea is that alkali metals form different oxides depending on their size and the oxygen supply: larger metals (like Cs) form superoxides, while smaller metals (like Na) form peroxides. The correct option is (C).
The relevant concept here is the trend in oxide formation among alkali metals. As you go down Group 1, the metal cation becomes larger and less polarizing. This stabilizes larger, more complex oxygen anions:
- Small Li forms only the normal oxide (O²⁻).
- Na forms the peroxide (O₂²⁻) under normal conditions.
- K, Rb, and Cs form superoxides (O₂⁻) because the large cation stabilizes the big superoxide ion. Also, excess vs. limited oxygen doesn’t change the product for Cs — it always forms the superoxide.
Let’s work through it step by step:
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Identify the product for Cs with oxygen
Cesium is the largest alkali metal. Its cation (Cs⁺) is very large and has low charge density, which stabilizes the large superoxide ion (O₂⁻).
- With excess O₂: Cs + O₂ → CsO₂ (superoxide).
- With limited O₂: Cs still forms CsO₂ because the superoxide is the most stable oxide for Cs. So X = superoxide.
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Identify the product for Na with oxygen
Sodium is smaller than Cs. Its cation (Na⁺) has a higher charge density, which stabilizes the peroxide ion (O₂²⁻) better than the superoxide.
- Na + O₂ → Na₂O₂ (peroxide). So Y = peroxide.
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Match with the options …
- TG EAPCET 2023Set ap-2023-05-10-FN1 markMCQQ.Match the following List-I (Ore name) A. Calamine B. Copper pyrites C. Zincite D. Malachite List-II (Type of ore) I. Oxide II. Sulphide III. Carbonate-hydroxide IV. Carbonate The correct answer is (A) A – II, B – I, C – IV, D – III (B) A – IV, B – III, C – I, D – II (C) A – IV, B – II, C – I, D – III (D) A – III, B – II, C – IV, D – I
›Reveal solutionSolution
This is a matching problem linking ore names to their chemical type. The key is knowing the chemical composition of each ore: Calamine is zinc carbonate, Copper pyrites is a sulphide, Zincite is zinc oxide, and Malachite is a basic copper carbonate (carbonate-hydroxide). The correct match is A–IV, B–II, C–I, D–III, which corresponds to option (C).
The question tests your memory of common ores and their classification by chemical nature — oxide, sulphide, carbonate, or carbonate-hydroxide. In metallurgy, ores are grouped by the anion present: oxides (O²⁻), sulphides (S²⁻), carbonates (CO₃²⁻), and sometimes mixed salts like basic carbonates (carbonate-hydroxide). Getting this right is just a matter of recalling the formula of each ore.
Let’s go through each one.
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Calamine – This is an ore of zinc. Its chemical formula is ZnCO₃. That’s a pure carbonate — no hydroxide group. So it belongs to List-II: IV (Carbonate).
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Copper pyrites – Also called chalcopyrite, this is the most common copper ore. Its formula is CuFeS₂. The presence of S²⁻ makes it a sulphide. So it matches List-II: II (Sulphide).
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Zincite – Another zinc ore, but this one is ZnO. That’s straightforward — an oxide. So it goes with List-II: I (Oxide). …
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- TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQQ.The oxidation state of Cr in CrO5 is (A) 3 (B) 5 (C) 10 (D) 6
›Reveal solutionSolution
The compound CrO5 contains peroxide linkages, which means not all oxygen atoms have an oxidation state of -2. Accounting for these, the oxidation state of Cr is +6.
The oxidation state of an element in a compound represents the hypothetical charge it would have if all bonds were ionic. While there are general rules for assigning oxidation states, it's crucial to understand that these rules have exceptions, especially when dealing with peroxides or superoxides.
For oxygen, the most common oxidation state is -2. However, in peroxides (containing an O-O bond), each oxygen atom has an oxidation state of -1. In superoxides, it's -1/2. When calculating oxidation states, if the standard assumption (e.g., O = -2) leads to an impossibly high or low oxidation state for the central atom, it's a strong indicator that an exception to the general rules is present, and the compound's structure needs to be considered.
Let's determine the oxidation state of Cr in CrO5:
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Initial Calculation (and why it's incorrect):
If we assume all five oxygen atoms have an oxidation state of -2 (the most common state for oxygen), then for the neutral compound CrO5:
Let the oxidation state of Cr be x.
x+5×(−2)=0
x−10=0
x=+10
This result, +10, is impossible for chromium. Chromium is a d-block element in Group 6, and its maximum possible oxidation state is +6 (corresponding to the total number of valence electrons it can lose). An oxidation state of +10 would imply the loss of more electrons than it possesses, which is chemically unfeasible. This tells us that our initial assumption about all oxygen atoms being -2 is incorrect.
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Understanding the Structure of CrO5:
The compound CrO5 has a characteristic "butterfly" structure. This structure reveals the presence of peroxide linkages.
It consists of:
- One oxygen atom double-bonded to chromium (a normal oxide oxygen).
- Two peroxide linkages, each containing two oxygen atoms. This means there are four oxygen atoms involved in O-O bonds. So, out of the five oxygen atoms, one is a normal oxide, and four are peroxide oxygens.
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Assigning Correct Oxidation States to Oxygen:
- The oxygen atom that is double-bonded to chromium (Cr=O) has an oxidation state of -2. …
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- TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQQ.The reagent used in the determination of the chemical oxygen demand (COD) of water sample is (A) Potassium permanganate (B) Sulphuric acid (C) Acidified Potassium dichromate (D) Potassium dichromate
›Reveal solutionSolution
The chemical oxygen demand (COD) of a water sample is determined by oxidizing organic and inorganic matter using a strong oxidizing agent in an acidic medium. The primary reagent used for this purpose is acidified potassium dichromate.
The Chemical Oxygen Demand (COD) test is a crucial measure of water quality, indicating the amount of oxygen required to chemically oxidize all oxidizable organic and inorganic substances present in a water sample. This test provides a rapid estimate of the pollution load in wastewater.
The core idea behind COD determination is to subject the water sample to a powerful chemical oxidation process. A strong oxidizing agent is used to break down the complex organic molecules into simpler, stable inorganic end products like carbon dioxide and water. The amount of oxidant consumed is then measured, which is directly proportional to the oxygen equivalent of the oxidizable matter.
Here's why a specific reagent system is chosen:
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Need for a Strong Oxidizing Agent: To ensure nearly complete oxidation of a wide range of organic compounds, a very strong oxidizing agent is required.
- Potassium dichromate (K2Cr2O7) is an excellent choice because it is a powerful oxidant, stable, and can oxidize almost all types of organic compounds (except for a few aromatic hydrocarbons and pyridine) to carbon dioxide and water. Its oxidation efficiency is high and consistent.
- Potassium permanganate (KMnO4), while also an oxidant, is generally less effective than dichromate for COD determination. Its oxidizing power can vary with pH, and it does not oxidize all organic compounds as completely as dichromate, leading to lower and less reliable COD values. It is more commonly used in other specific titrations or for measuring oxygen consumed in milder conditions (e.g., Permanganate Value).
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Role of Acidification: The oxidizing power of potassium dichromate is significantly enhanced in an acidic environment.
- Sulphuric acid (H2SO4) is added to provide the necessary acidic medium. In this acidic environment, the dichromate ion (Cr2O72−) is converted to chromic acid, which is a much stronger oxidizing agent. The reaction proceeds as follows: Cr2O72−+14H++6e−→2Cr3++7H2O …
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