Redox Reactions: common doubts, answered
The questions students ask most often about Redox Reactions, each with a short answer. For the full chapter, read the Redox Reactions notes.
About the chapter
How is Redox Reactions usually tested in NEET?
Expect questions on oxidation numbers in tricky species such as peroxides, superoxides, hydrides and S₄O₆²⁻, on picking out the oxidising and reducing agents, and on spotting disproportionation. Others ask how many electrons are transferred, as in permanganate or dichromate reactions, or for one coefficient in a balanced equation. Electrode potential ideas introduced here return in Class 12 electrochemistry.
Oxidation and reduction: the classical view
Read this section in the notes →Why was the oxygen-and-hydrogen definition of oxidation replaced?
Because it misses many reactions that clearly belong together. Originally oxidation meant adding oxygen or removing hydrogen, and reduction meant the opposite, yet sodium burning in chlorine involves neither element. Describing oxidation as loss of electrons and reduction as gain covers all such cases, and for covalent compounds the idea is extended further using changes in oxidation number.
Redox as electron transfer
Read this section in the notes →What is the difference between an oxidising agent and a reducing agent?
The oxidising agent is the substance that is itself reduced, and the reducing agent is the one that is itself oxidised. In Zn + Cu²⁺ → Zn²⁺ + Cu, zinc loses electrons, so it is oxidised and acts as the reducing agent; Cu²⁺ gains them and acts as the oxidising agent. Each is named for what it does to the other substance.
Why must oxidation and reduction always happen together?
Electrons cannot vanish or appear from nowhere, so whenever one species loses electrons another must take them up. Every oxidation needs a matching reduction, and the electrons lost must equal the electrons gained. Half-reactions are written separately only to make balancing easier; neither half can actually take place on its own in a chemical reaction.
Competitive electron transfer
Read this section in the notes →Why does zinc displace copper from solution but copper cannot displace zinc?
Zinc gives up electrons more readily than copper, so a zinc rod in copper sulphate solution dissolves as Zn²⁺ while copper metal deposits on it. The reverse transfer is unfavourable, so a copper rod in zinc sulphate solution shows no change. Comparing many such metal pairs ranks them by how easily they lose electrons, which grows into the electrochemical series.
Oxidation number and its rules
Read this section in the notes →How do you find the oxidation number of an element in a compound or ion?
Assign the known values first, with hydrogen +1 and oxygen −2, and make the total equal the charge on the species. For H₂SO₄, 2(+1) + x + 4(−2) = 0, so sulphur is +6. For Cr₂O₇²⁻, 2x + 7(−2) = −2, so chromium is +6. Watch for peroxides, superoxides and metal hydrides, where oxygen and hydrogen take unusual values.
Why is the oxidation number of oxygen not always −2?
Oxygen is −1 in peroxides such as H₂O₂ and Na₂O₂, where each oxygen is bonded to another oxygen, and −½ in superoxides such as KO₂. With fluorine, the only element more electronegative than oxygen, it becomes positive: +2 in OF₂ and +1 in O₂F₂. In nearly all other compounds oxygen takes −2.
When is the oxidation number of hydrogen −1?
Hydrogen is −1 in hydrides of active metals, such as NaH, LiH and CaH₂, because the metal is far less electronegative and hydrogen is present as the hydride ion H⁻. When hydrogen is bonded to non-metals, as in H₂O, HCl and NH₃, it is +1. Checking what hydrogen is attached to prevents the habit of writing +1 for it everywhere.
How do you find the oxidation number of carbon in organic compounds?
Take hydrogen as +1 and oxygen as −2 and solve for carbon. Carbon is −4 in CH₄, −2 in CH₃OH, 0 in HCHO, +2 in HCOOH and +4 in CO₂, which traces the stepwise oxidation of methane to carbon dioxide. In molecules with several carbons in different surroundings, this method gives only an average, and the individual atoms may differ from it.
Stock notation and redox by oxidation number
Read this section in the notes →What is Stock notation?
Stock notation shows a metal's oxidation number as a Roman numeral in brackets after its name or symbol, such as iron(II) oxide for FeO and iron(III) oxide for Fe₂O₃. It is used for elements that have more than one common oxidation state, where a plain name would be ambiguous. Likewise Hg₂Cl₂ is mercury(I) chloride and HgCl₂ is mercury(II) chloride.
Combination, decomposition and displacement
Read this section in the notes →Is every decomposition reaction a redox reaction?
No. A decomposition counts as redox only if some oxidation numbers change. When CaCO₃ is heated to CaO and CO₂, calcium stays +2, carbon +4 and oxygen −2, so nothing is oxidised or reduced. By contrast, 2H₂O → 2H₂ + O₂ is redox, since hydrogen falls from +1 to 0 and oxygen rises from −2 to 0.
Why can chlorine displace bromine and iodine from their salts but not the reverse?
Oxidising power among the halogens runs F₂ > Cl₂ > Br₂ > I₂. A halogen can take electrons from the halide ions of any halogen below it, so chlorine oxidises Br⁻ and I⁻ to Br₂ and I₂, but iodine cannot oxidise chloride or bromide. Fluorine is so strong that it even oxidises water, so its displacement reactions are not carried out in aqueous solution.
Disproportionation and fractional oxidation numbers
Read this section in the notes →What is a disproportionation reaction?
It is a redox reaction in which one element, starting from a single oxidation state, is partly oxidised and partly reduced. In 2H₂O₂ → 2H₂O + O₂, oxygen goes from −1 down to −2 in water and up to 0 in O₂. Chlorine does the same in cold alkali, where Cl₂ at 0 forms both Cl⁻ at −1 and ClO⁻ at +1.
Why can't ClO₄⁻ undergo disproportionation?
An element can disproportionate only if it can move both up and down from its present oxidation state. Chlorine in ClO₄⁻ is already at +7, its highest possible state, so it cannot be oxidised further and has nothing to pair with its reduction. ClO⁻, ClO₂⁻ and ClO₃⁻, with chlorine at +1, +3 and +5, have room in both directions and can disproportionate.
Why doesn't fluorine disproportionate in alkali like chlorine does?
Fluorine is the most electronegative element and never takes a positive oxidation number, so it has no higher state to move into. In alkali it can only be reduced, to F⁻, while oxygen from the hydroxide or water is oxidised instead, giving OF₂ or oxygen. Chlorine, bromine and iodine can take positive states, so they do disproportionate in alkali.
How can an oxidation number be a fraction, as in S₄O₆²⁻ or Fe₃O₄?
A fractional value is an average over atoms that are really in different states. In S₄O₆²⁻ the average for sulphur is +2.5, but the two end sulphurs are +5 and the two middle ones are 0. In Fe₃O₄ the average is +8/3 because the solid holds one Fe²⁺ for every two Fe³⁺. No individual atom actually has a fractional oxidation number.
Balancing redox equations
Read this section in the notes →How do you balance a redox equation in acidic medium by the half-reaction method?
Split the reaction into oxidation and reduction halves. In each, balance atoms other than O and H, add H₂O to balance oxygen and H⁺ to balance hydrogen, then add electrons to balance charge. Multiply the halves so the electrons match, add them and cancel common terms. The dichromate half, Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, shows six electrons taken per ion.
How do you balance a redox equation in basic medium?
First balance it exactly as for an acidic solution, using H⁺ and H₂O. Then add the same number of OH⁻ ions to both sides as there are H⁺ ions; each H⁺ and OH⁻ on one side combine into H₂O. Finally cancel water molecules that appear on both sides. Any final equation that still contains H⁺ cannot be right for a basic solution.
Redox titrations and limits of the concept
Read this section in the notes →Why is KMnO₄ called a self-indicator?
Permanganate solution is deep purple, while the Mn²⁺ it forms in acid is almost colourless. During a titration each drop loses its colour as long as reducing agent remains; once that is used up, the next drop leaves a faint pink tint that does not fade, marking the end point. No separate indicator is needed, unlike dichromate titrations, which use diphenylamine.
Why is starch used as the indicator in iodometric titrations?
In iodometry, iodine is titrated with thiosulphate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. The yellow-brown of iodine fades gradually and is hard to judge near the end, so starch is added, which gives an intense blue with even traces of iodine. The end point is the moment the blue colour disappears, showing all the iodine has become iodide.
Redox couples and electrode processes
Read this section in the notes →What does a more negative standard electrode potential mean?
It means the reduced form of that couple gives up electrons easily, so it is a stronger reducing agent. Potentials are measured against the standard hydrogen electrode, fixed at 0.00 V. A more positive value means the oxidised form readily accepts electrons and is a stronger oxidising agent. Zinc, with a negative value, reduces Cu²⁺, whose couple has a positive value.
In which direction do electrons and current flow in a Daniell cell?
Electrons travel from the zinc electrode to the copper electrode through the external wire, because zinc is oxidised and Cu²⁺ is reduced. Conventional current is defined as flowing the opposite way, from copper to zinc. The salt bridge allows ions to move between the two solutions, keeping each electrically neutral so the cell can continue to work.
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