EPFO APFC 2023 General Studies · Question 90 of 120

Correct — A, (a) The oxidation state of hydrogen increases while that of oxygen decreases. Work out the four oxidation states the equation involves and the answer follows in one line. On the left both reactants are elements in their free state — hydrogen as H₂ and oxygen as O₂ — and an element in its free state is assigned an oxidation state of zero by definition, because the two atoms sharing each bond are identical and neither can be said to have taken electron density from the other. On the right, in H₂O, oxygen is much the more electronegative of the two elements, so by the usual convention the shared electrons of each bond are counted to oxygen: oxygen is assigned −2 and each hydrogen +1, and the two hydrogens at +1 with one oxygen at −2 sum to zero, as they must for a neutral molecule. So hydrogen moves from 0 to +1, which is an increase, and oxygen moves from 0 to −2, which is a decrease. That is exactly what option (a) says. The vocabulary attached to those movements is worth stating in the same breath, because the rest of the option set is written in it. An increase in oxidation state is oxidation and a decrease is reduction, so in this reaction hydrogen is oxidised and oxygen is reduced. The species that is oxidised is the reducing agent, so hydrogen is the reducing agent here, and oxygen, being reduced, is the oxidising agent. Counting electrons confirms the bookkeeping: four hydrogen atoms each give up one unit of electron density, four units in all, and two oxygen atoms each take two, which is also four. One caution about what these numbers mean. Water is a covalent molecule and contains no H⁺ or O²⁻ ions; the oxidation states are a formal accounting device that assigns the shared electrons of each bond to the more electronegative atom, and they are useful precisely because they let a reaction like this one be recognised as a redox reaction at all.

• (b)The oxidation state of hydrogen decreases while that of oxygen increases. — This reverses the direction of both changes. It would require hydrogen to fall from 0 to a negative oxidation state and oxygen to rise from 0 to a positive one, which is what happens only when hydrogen is bonded to something less electronegative than itself or oxygen to something more electronegative than itself. Both cases exist and are worth knowing — hydrogen is −1 in a metal hydride such as sodium hydride, and oxygen is +2 in oxygen difluoride, since fluorine is the only element more electronegative than oxygen — but neither describes water, where oxygen is the more electronegative partner and takes the negative value. Notice also that this option makes the same claim as option (d) in different words: saying hydrogen's oxidation state falls is saying hydrogen is reduced. Two options asserting the same thing in a single-answer question must both be wrong, which narrows the field before any chemistry is done.
• (c)There is no change in the oxidation state of hydrogen as well as that of oxygen. — This says the reaction is not a redox reaction at all, and that cannot be true of any reaction in which free elements combine. An element in its free state is at oxidation state zero, and the moment it forms a compound with a different element the shared electrons are assigned unequally, so at least one oxidation state must move. Here both move: hydrogen from 0 to +1 and oxygen from 0 to −2. The reasoning that leads a candidate to this option is usually one of two things — treating a combination or synthesis reaction as a category separate from redox, or assuming that a covalent product with no ions cannot involve oxidation and reduction. Neither holds. The categories overlap freely: this reaction is simultaneously a combination reaction, the combustion of hydrogen, a strongly exothermic reaction and a redox reaction, and being one of those does not stop it being the others.
• (d)During the reaction, hydrogen is reduced while oxygen is oxidized. — This states the correct chemistry backwards, and it is the most dangerous option in the set because it is written in the vocabulary a candidate is most confident about. Hydrogen is oxidised in this reaction, not reduced, and oxygen is reduced, not oxidised. The safest way to keep the directions straight is to define them by electrons rather than by the names of the elements: oxidation is loss of electrons and reduction is gain, so the element whose oxidation state rises has lost electron density and has been oxidised. Hydrogen goes from 0 to +1, so it has lost; oxygen goes from 0 to −2, so it has gained. The confusion is helped along by the everyday sense of the word oxidation, which suggests that oxygen is the thing being oxidised when in fact oxygen is what does the oxidising. This option and option (b) assert the same thing in two different vocabularies, one in terms of oxidation states and one in terms of oxidation and reduction, which is itself a signal that neither can be the intended answer.

Oxidation and reduction were first defined by the gain and loss of oxygen, then generalised to the loss and gain of electrons, and are most usefully handled through oxidation states, which extend the idea to covalent compounds where no electron is fully transferred. An oxidation state is assigned by a set of conventions: an element in its free state is zero; a monatomic ion takes the charge it carries; fluorine is always −1; oxygen is normally −2, but −1 in peroxides and positive when combined with fluorine; hydrogen is normally +1, but −1 in metal hydrides; and the oxidation states in a species sum to its overall charge, which is zero for a neutral molecule. The underlying rule behind all of these is that the electrons of each bond are counted to the more electronegative atom. Once the states are assigned, oxidation is an increase in oxidation state and reduction a decrease, and the two always occur together, which is why such reactions are called redox reactions. The species that is oxidised acts as the reducing agent, since it supplies the electrons; the species that is reduced acts as the oxidising agent. Combustion in oxygen is the archetypal redox process, and the combustion of hydrogen in this question is its simplest instance, releasing a great deal of energy and forming only water — which is the basis of interest in hydrogen as a fuel. A few related patterns are worth carrying: displacement of a less reactive metal from its salt by a more reactive one is a redox reaction, corrosion and rusting are redox processes, respiration and photosynthesis are redox sequences in biology, and in disproportionation a single element is both oxidised and reduced in the same reaction.

This is a two-line calculation dressed as a statement question, and its design is worth examining because the same design recurs. The chemistry needed is minimal: two conventions — free elements are zero, oxygen in water is −2 — settle every number in the problem. What the item actually tests is whether the candidate can move between two vocabularies for the same fact. Two options are phrased in terms of oxidation states rising and falling, and two in terms of substances being oxidised and reduced, and a candidate who has learnt only one of the two phrasings has to translate under time pressure. That is also where the structural shortcut lives. Option (b) and option (d) say the same thing in the two different vocabularies, so in a question with one correct answer neither can be it, and the choice reduces to the two remaining options before any oxidation state is worked out. Spotting a pair of options that are logically equivalent is a genuinely transferable examination skill and it is worth practising deliberately in the science block, where the same claim can so easily be dressed in two sets of words. The third distractor tests something different again: whether the candidate believes that a combination reaction producing a covalent molecule can be a redox reaction. It can, and the fact that the product has no ions in it changes nothing, because oxidation states are an accounting convention rather than a claim about real charges.

• An element in its free state has an oxidation state of zero, so both H₂ and O₂ enter this reaction at zero. In water, oxygen is assigned −2 and each hydrogen +1, because the electrons of each bond are counted to the more electronegative atom and the states must sum to zero for a neutral molecule.
• Hydrogen therefore moves from 0 to +1, an increase, and oxygen from 0 to −2, a decrease. An increase in oxidation state is oxidation and a decrease is reduction, so hydrogen is oxidised and oxygen is reduced in the combustion of hydrogen.
• The species that is oxidised is the reducing agent and the species that is reduced is the oxidising agent. Here hydrogen is the reducing agent and oxygen the oxidising agent, which is the reverse of what the everyday sense of the word oxidation suggests.
• The standard exceptions to the oxidation state conventions are worth memorising: oxygen is −1 in peroxides such as hydrogen peroxide and positive when combined with fluorine, as in oxygen difluoride, and hydrogen is −1 in metal hydrides such as sodium hydride.
• Oxidation states in a covalent molecule are a formal bookkeeping device, not real ionic charges. Water contains no H⁺ or O²⁻ ions; the convention exists so that electron redistribution in covalent reactions can be tracked and recognised as redox at all.
• Categories of reaction overlap. This reaction is simultaneously a combination reaction, the combustion of hydrogen, a strongly exothermic reaction and a redox reaction, and belonging to one category does not exclude it from the others.

• Assuming oxygen must be the element oxidised because the word oxidation contains its name. Oxygen is the oxidising agent here and is itself reduced, moving from 0 to −2.
• Treating a combination reaction between two elements as something other than a redox reaction. Free elements are at zero, so the moment they form a compound at least one oxidation state must change.
• Believing that a covalent product rules out oxidation and reduction. Oxidation states are an accounting convention that assigns bonding electrons to the more electronegative atom; they do not assert that ions exist.
• Confusing a decrease in oxidation state with a decrease in electrons. Oxygen's state falls to −2 precisely because it has gained electron density, which is reduction.
• Failing to notice that two options in this set assert the same thing in different vocabularies. In a single-answer question, a logically equivalent pair can both be discarded before any chemistry is attempted.

Redox appears in this exam in four shapes. The commonest is the one here: a balanced equation is printed and the candidate is asked which statement about the oxidation states, or about which species is oxidised and which reduced, is correct. The second asks for the oxidation state of a named element in a specific compound, where the marks turn entirely on the exceptions — oxygen in hydrogen peroxide, hydrogen in sodium hydride, chlorine in a chlorate or a perchlorate. The third names the oxidising or reducing agent in a reaction, which requires the extra step of remembering that the oxidised species is the reducing agent. The fourth is applied, asking why iron rusts, what happens in a dry cell, or why a more reactive metal displaces a less reactive one from its salt. The preparation that covers all four is short: the assignment rules with their exceptions, the two definitions in terms of electrons, the agent terminology, and practice at reading a printed equation and writing the oxidation state of every element above its symbol before looking at the options at all.

No directly related past PYQ was found.

• practice — not a real PYQ

  In which one of the following compounds does oxygen exhibit an oxidation state of −1?

  • (a)H₂O
  • (b)H₂O₂
  • (c)OF₂
  • (d)CO₂

  Answer(b) H₂O₂ — in hydrogen peroxide the two oxygen atoms are bonded to each other as well as to a hydrogen each, and a bond between identical atoms contributes nothing to either atom's oxidation state, so each oxygen is left at −1. Oxygen is −2 in water and in carbon dioxide, and it is +2 in oxygen difluoride, where the partner is fluorine, the only element more electronegative than oxygen.

• practice — not a real PYQ

  In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, which species is oxidised and which acts as the oxidising agent, respectively?

  • (a)Zn and CuSO₄
  • (b)CuSO₄ and Zn
  • (c)Zn and ZnSO₄
  • (d)Cu and ZnSO₄

  Answer(a) Zn and CuSO₄ — zinc goes from 0 in the free metal to +2 in zinc sulphate, an increase, so it is oxidised and is the reducing agent. Copper goes from +2 in copper sulphate to 0 in the free metal, a decrease, so it is reduced, which makes copper sulphate the oxidising agent. The sulphate ion is a spectator and its own oxidation states do not change.