Chemical Reactions and Equations

Pure water contains very few ions and therefore conducts electricity poorly. During electrolysis, adding a small amount of a suitable acid increases the concentration of mobile ions, allowing electric current to pass through the water more effectively. The acid is not added to colour oxygen, solidify hydrogen, or produce a precipitate. It supports conduction; the water is then decomposed at the electrodes into hydrogen and oxygen. Hence option A is correct.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The displaced copper deposits as a reddish-brown coating on the nail, while the iron sulphate formed makes the solution green. Unlike option A, copper cannot displace iron because copper is less reactive.

The blue colour of aqueous copper sulphate is associated mainly with hydrated Cu²⁺ ions. When an iron strip is placed in the solution, iron is more reactive and displaces copper: Fe + CuSO₄ → FeSO₄ + Cu. As Cu²⁺ ions are removed from the solution and copper metal is deposited, the intensity of the blue colour decreases; Fe²⁺ may also give the solution a pale green colour. Iron ions do not disappear, and sodium or barium ions are not involved. Therefore option C is correct.

Hydrogen removes oxygen from \(\mathrm{CuO}\), converting it to copper \(\mathrm{Cu}\); therefore, \(\mathrm{H_2}\) is the reducing agent. Hydrogen is itself oxidised to \(\mathrm{H_2O}\). In contrast, \(\mathrm{CuO}\) is reduced, so it is the oxidising agent rather than the reducing agent.

Barium chloride and sodium sulphate undergo a double-displacement reaction: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). Barium sulphate is a white insoluble precipitate, while sodium chloride remains dissolved. The other choices do not provide the required pair of products or do not represent two suitable aqueous salt solutions.

In option A, zinc, \(\mathrm{Zn}\), displaces copper from copper sulfate, so the reaction is a displacement reaction. The oxidation state of Zn changes from \(0\) to \(+2\), showing oxidation, while that of \(\mathrm{Cu^{2+}}\) changes from \(+2\) to \(0\), showing reduction. Thus, it is a redox displacement reaction. In contrast, option C also involves oxidation and reduction, but it is a photochemical decomposition reaction, not a displacement reaction.

CuO loses oxygen, so it is reduced; the oxidation number of copper changes from \\( +2 \\) to \\(0\\). H₂ gains oxygen to form H₂O, so it is oxidised; the oxidation number of hydrogen changes from \\(0\\) to \\( +1 \\). Therefore, this is a redox reaction, not merely a displacement reaction.

The reaction is CuO + H₂ → Cu + H₂O on heating. Oxygen is removed from copper(II) oxide and combines with hydrogen to form water. Removal of oxygen is the defining sign of reduction, so CuO is reduced to copper. Oxidation would involve gaining oxygen or losing electrons; precipitation and neutralisation do not describe this change.

In \(H_2O_2\), oxygen has an oxidation number of \(-1\). Some oxygen becomes \(-2\) in water, \(H_2O\), so it is reduced, while some becomes \(0\) in \(O_2\), so it is oxidised. Thus, oxygen undergoes both changes in the same reaction. In contrast, during the decomposition of \(KClO_3\), chlorine is reduced and oxygen is oxidised; the changes occur in different elements.

In the oxygen-based definitions of redox, loss of oxygen is reduction and gain of oxygen is oxidation. Therefore, substance A, which loses oxygen, is reduced, while substance B, which gains oxygen, is oxidised. Hence option B is correct. Option A reverses the definitions, and options C and D incorrectly assign the same process to both substances.

In the balanced equation, the mole ratio Al : CuCl₂ : Cu is 2 : 3 : 3. Thus, 2 mol Al reacts with 3 mol CuCl₂ to produce 3 mol Cu. Therefore, with sufficient CuCl₂, 1 mol Al produces 1.5 mol Cu. Option A incorrectly treats the coefficient ratio as 1 : 1.

In option B, zinc is more reactive than copper and displaces copper from \(\mathrm{CuSO_4}\). Heat is released during this reaction, so it is an exothermic displacement reaction. In contrast, option D involves exchange of ions and is a double-displacement reaction.

Rusting of iron requires both oxygen and water (moisture). Moist open air provides both, making it the most favourable environment for rusting. Oil prevents air and moisture from reaching the iron surface, whereas a dry airtight bottle lacks moisture.

Zinc is more reactive than copper. Therefore, zinc displaces copper from copper sulphate solution:

\(\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}\)

The reddish-brown coating is deposited copper, and the blue colour fades because blue \(\mathrm{Cu^{2+}}\) ions are removed from the solution. Unlike option B, copper cannot displace zinc because copper is less reactive than zinc.

Zinc is more reactive than iron. Therefore, when a zinc coating is scratched, zinc oxidises in preference to iron and provides sacrificial protection. As a result, the exposed iron can remain protected from rusting for some time. In contrast, a broken paint coating does not provide sacrificial protection, so exposed iron may rust on contact with air and moisture.

The spoilage of oils and fats is called rancidity. When these substances remain exposed to air, oxygen reacts with their components and produces unpleasant-smelling and unpleasant-tasting substances. This chemical deterioration is oxidation, so option D is correct. Precipitation forms an insoluble solid, electrolysis uses electric current, and galvanisation protects iron with zinc; none explains rancidity.

On heating in air, copper combines with oxygen: 2Cu + O₂ → 2CuO. Copper changes from elemental copper to copper(II) oxide, and the addition of oxygen is oxidation. The black colour indicates the new oxide, not merely a surface appearance. Thus option B is correct; no copper sulphate or displacement reaction is involved.

The oils present in chips can react with oxygen and become rancid through oxidation. Nitrogen is relatively inert and lowers the amount of oxygen in the packet, so oxidation of the oils slows down. The cushioning in option A may reduce chip breakage, but it does not prevent rancidity of the oils.

This is a redox reaction. Hydrogen has oxidation number 0 in \\(H_2\\), which becomes +1 in \\(H_2O\\); therefore, hydrogen is oxidised. Copper has oxidation number +2 in \\(CuO\\), which decreases to 0 in \\(Cu\\); therefore, copper(II) oxide is reduced. Option B states the opposite changes.

Balancing requires equal numbers of every atom on both sides. In option A, the left side contains 2 Pb, 4 N and 12 O atoms. The right side contains 2 Pb in 2PbO, 4 N in 4NO₂, and 2 + 8 + 2 = 12 O atoms. Therefore, 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂ is correct; the other options fail atom balance or change the product.