Chemical Reactions and Equations

In electrolysis of water, hydrogen gas is formed at the cathode and oxygen gas at the anode. From the balanced equation \(2H_2O \rightarrow 2H_2 + O_2\), the volume ratio of hydrogen to oxygen is \(2:1\). Therefore, the greater volume of hydrogen is obtained at the cathode. Oxygen forms at the anode, but its volume is half that of hydrogen. Exam tip: Remember: \(H_2\) at the cathode and \(O_2\) at the anode.

Pure water is a very poor conductor because it contains only a small concentration of ions. Adding a small amount of a suitable acid, commonly dilute sulphuric acid in the school experiment, increases the concentration of mobile ions. These ions carry charge through the solution and allow electrolysis to proceed effectively. The acid is not added to colour the water, stop oxygen, or increase hydrogen’s mass, so A is correct.

Decomposition is the breaking of one compound into simpler substances, but the two named types differ in how the reaction receives energy. Thermal decomposition is driven by heating, as in the breakdown of calcium carbonate on strong heating. Electrolytic decomposition is driven by an electric current, as in water electrolysis. Product colour, vessel shape, and reactant mass do not define the classification, so A is correct.

Iron is above copper in the reactivity series, so it can displace copper from copper sulphate solution. The reaction is Fe + CuSO₄ → FeSO₄ + Cu. Aqueous iron(II) sulphate is pale green, while copper metal deposits on the nail. Therefore formation of iron sulphate explains the green colour. The other proposed compounds contain elements not involved in this reaction, so A is correct.

The blue colour of copper sulphate solution is mainly associated with hydrated Cu²⁺ ions. When an iron nail is placed in the solution, iron displaces copper: Fe + Cu²⁺ → Fe²⁺ + Cu. Copper ions are removed from the solution and deposited as copper metal, while the pale-green Fe²⁺ solution forms. Consequently, the blue colour fades because the number of Cu²⁺ ions decreases, so A is correct.

This is a double-displacement precipitation reaction. The \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) ions combine to form insoluble barium sulphate as a white precipitate: \(\mathrm{BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)}\). Hence, the products are barium sulphate and sodium chloride. Option B contains sulphite rather than sulphate, and sulphite ions are not present in the reactants. Exam tip: In such questions, identify the insoluble salt to find the precipitate product.

The governing concept is a double-displacement precipitation reaction. In a reaction such as BaCl₂ + Na₂SO₄ → BaSO₄ + 2NaCl, the ions exchange partners, so it is double displacement. BaSO₄ is insoluble and appears as a white solid, so the reaction also involves precipitation. Therefore option A identifies both features; the other choices describe unrelated reaction types.

In chemical equations, a downward arrow is used to mark a precipitate. A precipitate is a solid product that forms from a solution and does not dissolve appreciably in the solvent, so it separates and settles. The symbol therefore communicates the physical state and separation of the product. A gas would be indicated by an upward arrow, while heat and light are not represented by this notation.

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.

Removal of oxygen from a substance is reduction, so substance A undergoes reduction. Removal of hydrogen is oxidation, so substance B undergoes oxidation. Therefore, the correct sequence is reduction and oxidation. Option B gives the two processes in reverse order. Exam tip: Addition of oxygen or removal of hydrogen indicates oxidation.

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.

The zinc coating forms a barrier between iron and air and moisture, preventing the oxygen and water needed for rusting from reaching the iron. If the coating is scratched, zinc, being more reactive than iron, corrodes first and provides sacrificial protection. Option B is incorrect because zinc does not convert iron into zinc. Exam tip: Remember that galvanisation gives both barrier and sacrificial protection.

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.

Rancidity is the unpleasant smell and taste produced when fats and oils undergo oxidation. Filling a packet with nitrogen removes or greatly reduces the oxygen available around the food, so oxidation occurs more slowly and the food remains acceptable for longer. Nitrogen is not increasing oxygen, forming water, or producing a precipitate. Therefore, option A correctly explains the preservation method.

Antioxidants slow the oxidation of food, especially oils and fats. Oxidation of oils and fats causes rancidity, producing an unpleasant smell and taste. Reduction is the opposite of oxidation, so it is not correct. Exam tip: Associate the spoilage of oily and fatty foods with rancidity caused by oxidation.