Chemical Reactions and Equations

The governing concept is reduction as removal of oxygen. Black copper(II) oxide reacts with hydrogen: CuO + H₂ → Cu + H₂O. Oxygen is removed from CuO, and reddish-brown copper is produced, so CuO is reduced. Hydrogen is oxidised because it gains oxygen and forms water. Thus option A is correct; the other options reverse the change or name unrelated processes.

Oxidation can be defined as gain of oxygen, loss of hydrogen, or loss of electrons. In CuO + H₂ → Cu + H₂O, hydrogen combines with oxygen removed from copper oxide and forms water. Because hydrogen gains oxygen, it is oxidised. Option B describes the opposite trend, and options C and D do not occur in this reaction, so option A is the only correct explanation.

The governing concept is electron transfer in a redox reaction. When one substance loses electrons, it is oxidised; those electrons cannot disappear, so another substance accepts them and is reduced. The same linked transfer can also be described through oxygen gain and loss. Therefore oxidation and reduction necessarily occur together, making option A correct; the other choices describe neither the defining mechanism nor a necessary condition.

Rusting is a slow electrochemical oxidation of iron. Iron loses electrons while oxygen is reduced, and a thin water layer allows ions to move between the reacting regions. In ordinary conditions, removing either oxygen or moisture greatly prevents or slows rust formation. Thus option A gives the correct roles. The other choices incorrectly describe melting, gas formation, blocking, or inactivity.

Paint protects iron by acting as a barrier against oxygen and water. A scratch breaks that continuous protective layer and exposes fresh iron to air and moisture. Electrochemical oxidation can then begin at the exposed region, producing rust that may spread beneath the coating. Therefore option A is correct. Paint does not generate oxygen, and the other options describe impossible transformations.

Rusting requires electrochemical oxidation of iron in the presence of oxygen and water. A continuous oil film covers the iron and acts as a barrier, reducing the access of humid air to the metal surface. With less oxygen and moisture reaching the iron, the oxidation current and rust formation decrease. Hence option A is correct; oil does not transmute iron, convert rust to water, or increase oxygen supply.

Rancidity is mainly caused by oxidation of unsaturated oils and fats when they react with oxygen in air. The oxidation produces unpleasant-smelling and bad-tasting compounds, so the food becomes stale or rancid. Antioxidants, refrigeration, and airtight packaging slow this process by reducing oxidation. Therefore option A is correct; precipitation, electrolysis, and galvanisation are different processes and do not explain the spoilage of fats.

The governing concept is that a displacement reaction occurs when a more reactive element replaces a less reactive element from its compound, while a redox reaction involves simultaneous oxidation and reduction. In Zn + CuSO₄ → ZnSO₄ + Cu, zinc displaces copper and is oxidised to Zn²⁺; Cu²⁺ is reduced to copper. B is decomposition, C is double displacement, and D is combination plus redox, not displacement.

A careful scientific conclusion must be based only on the observations. Gas evolution suggests that a gaseous product may have formed, often indicating a chemical reaction, while cooling of the container suggests that heat was absorbed from the surroundings. These observations support option A, but they do not identify the exact reaction or prove rusting. The other options either ignore an observation or make an unjustified claim.

Iron is above copper in the reactivity series, so it displaces copper from copper sulphate: Fe + CuSO₄ → FeSO₄ + Cu. The newly formed iron sulphate solution is pale green, while copper appears as the reddish-brown coating on the nail. Thus A correctly explains both observations. The reaction is not combination, decomposition by heating, or neutralisation because no acid-base reaction is involved.

The reactivity series shows that iron is more reactive than copper. Therefore iron displaces copper from copper sulphate according to Fe + CuSO₄ → FeSO₄ + Cu. Ferrous sulphate gives the solution a light green colour, and copper forms the reddish-brown coating. Hence B is correct. A reverses the reactivity relationship, while C and D ignore the formation of new substances.

On heating, sodium hydrogen carbonate, NaHCO₃, undergoes thermal decomposition: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂. The carbon dioxide produced forms bubbles in the batter, and these bubbles expand during heating, making the cake rise and become porous. Therefore option A is correct. Hydrogen is not produced in this decomposition, while chlorine and nitrogen are not products of sodium hydrogen carbonate heating.

The governing concept is the use of characteristic gas tests. Hydrogen burns rapidly and produces a ‘pop’ when a burning splint is brought near the gas. Carbon dioxide does not give that test; it turns limewater milky because calcium carbonate forms. Thus option A correctly pairs each gas with its diagnostic test. Smell and colour are unreliable here, and reversing the tests gives option B, which is incorrect.

The governing concept is the general basic character of many metal oxides. A basic oxide reacts with an acid in a neutralisation reaction to form salt and water; for example, CuO + 2HCl → CuCl2 + H2O. Some metal oxides also react with water to form hydroxides. Therefore option A gives the defining reason. Option B describes acidity, while C and D are factually incorrect generalisations.

The electrolysis of concentrated sodium chloride solution, called the chlor-alkali process, produces sodium hydroxide, hydrogen, and chlorine. At the cathode, water is reduced to hydrogen and hydroxide ions; at the anode, chloride ions are oxidised to chlorine. The sodium ions combine with hydroxide ions in solution to form NaOH. Therefore option A is correct, while the other gas pairs are not products of brine electrolysis.

The chlor-alkali process is the electrolysis of concentrated brine, or aqueous sodium chloride. At the cathode, water molecules gain electrons in a reduction reaction and form hydrogen gas and hydroxide ions: 2H2O + 2e− → H2 + 2OH−. Chlorine is formed at the anode, not the cathode, while carbon dioxide and ammonia are not products. Therefore option A is correct.