Chemical Reactions and Equations

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 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.

Start with \(CuO + H_2 \rightarrow Cu + H_2O\). Copper is already balanced with coefficient \(1\) for \(CuO\) giving \(1\) for \(Cu\), but oxygen on the left is \(1\) while oxygen in \(H_2O\) is \(1\) only if there is \(1\) water molecule; hydrogen then becomes \(2\) on both sides only for \(H_2\) coefficient \(1\). However, the oxidation-reduction balance requires matching atoms on both sides: overall, the simplest integer-balanced form is obtained by doubling coefficients to keep consistency with the full stoichiometric form, giving \(2CuO(s)+2H_2(g)\rightarrow 2Cu(s)+2H_2O(l)\).

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.

When heat is written on the products side, it means that heat is released during the reaction. A reaction that releases heat is called an exothermic reaction. In contrast, an endothermic reaction absorbs heat, so heat is written on the reactants side.

Absorption of heat with the reactants shows that the reaction is endothermic. The breakdown of a single compound into two or more products is the characteristic of a decomposition reaction. Therefore, it is an endothermic decomposition reaction. In contrast, in a combination reaction, two or more reactants generally form one product.

In combustion, oxidation of a fuel releases energy, mainly as heat and often as light. In respiration, glucose undergoes controlled oxidation and releases energy for life processes. Therefore, both are exothermic processes. In contrast, a requirement for light is associated with processes such as photosynthesis, not with both combustion and respiration.

During photosynthesis, plants absorb sunlight and store it as chemical energy in glucose. In contrast, respiration releases energy during the oxidation of glucose, and part of this energy is captured in ATP. Therefore, option D is correct. Option B is incorrect because respiration does not require absorption of external light energy. Exam tip: remember photosynthesis as energy storage and respiration as energy release.

During mechanical chewing, food is broken into smaller pieces; its size and surface area change, but no new substance is formed. In chemical digestion, enzymes convert complex food substances into simpler substances, so new substances are formed. Therefore, option A is correct; option C is incorrect because chemical digestion is not merely a physical change.

In option A, the ions of two ionic compounds exchange partners. \(\mathrm{Pb^{2+}}\) ions combine with \(\mathrm{I^-}\) ions to form yellow, insoluble \(\mathrm{PbI_2}\) as a precipitate, while \(\mathrm{KNO_3}\) remains in solution. In contrast, option B is a single displacement reaction because zinc displaces hydrogen.

Hydrogen has oxidation number 0 in \(\mathrm{H_2}\) and +1 in \(\mathrm{H_2O}\), so hydrogen is oxidised. Copper has oxidation number +2 in \(\mathrm{CuO}\) and 0 in \(\mathrm{Cu}\), so copper(II) oxide is reduced. Thus, this is a redox reaction because oxidation and reduction occur together.

In option A, oxygen in \(H_2O_2\) has an oxidation number of \(-1\). One part is reduced to \(-2\) in \(H_2O\), while another part is oxidised to \(0\) in \(O_2\). Thus, \(H_2O_2\) undergoes both oxidation and reduction; this is a disproportionation reaction. Option B is also a redox reaction, but zinc is oxidised and copper ions are reduced, so different substances undergo the two changes.

In a chemical equation, a downward arrow indicates the formation of an insoluble solid that settles down, called a precipitate. An upward arrow indicates that a gas is evolved during the reaction. Option B reverses these two meanings. Exam tip: remember ↓ for precipitate and ↑ for evolved gas.

In \(H_2O_2\), oxygen has an oxidation number of \(-1\). In the products, it becomes \(-2\) in \(H_2O\) and \(0\) in \(O_2\). Thus, one portion of oxygen is reduced while another is oxidised, making this a disproportionation reaction. In contrast, in option B, zinc is only oxidised and copper ions are only reduced.

Iron is more reactive than copper, so it displaces copper from copper sulphate: \(Fe + CuSO_4 \rightarrow FeSO_4 + Cu\). The displaced copper forms a reddish-brown coating on the nail, and the iron sulphate formed makes the solution green. In contrast, an unchanged blue solution would not indicate displacement.

In option A, zinc, which is more reactive than copper, displaces copper from copper sulphate. The oxidation state of zinc changes from \(0\) to \(+2\), while that of copper changes from \(+2\) to \(0\). Thus, zinc is oxidised and copper is reduced, making this both a displacement and a redox reaction. Option B is a double-displacement (precipitation) reaction, not a displacement reaction.

The formation of a substance with new properties shows that a new substance was produced; therefore, the process is a chemical reaction. The equality of total mass before and after the reaction in a closed container is consistent with the law of conservation of mass. Options A and D describe physical changes, which do not produce a new substance, while option B contradicts the observation.

The reaction is \(\mathrm{CaCO_3 \rightarrow CaO + CO_2}\). In a sealed container, the \(\mathrm{CO_2}\) produced remains inside, so the total mass of the products includes both \(\mathrm{CaO}\) and \(\mathrm{CO_2}\). By the law of conservation of mass, this total equals the mass of the original \(\mathrm{CaCO_3}\). Formation of a gas does not reduce total mass; it would decrease only if the gas escaped from the container.

Subscripts in a chemical formula show the number of atoms of each element in a compound. Changing a subscript changes the compound’s composition and chemical identity; for example, changing \(\mathrm{H_2O}\) to \(\mathrm{H_2O_2}\) changes water into hydrogen peroxide. To balance an equation, coefficients placed before formulae are changed, not subscripts.

One \(\mathrm{H_2O}\) molecule contains 2 hydrogen atoms. Therefore, \(2\mathrm{H_2O}\) contains \(2 \times 2 = 4\) hydrogen atoms. In a balanced equation, the number of atoms of each element is the same on both sides, so the reactant side must also contain 4 hydrogen atoms. The option 2 applies to only one water molecule.

The balanced equation is \(2\mathrm{Mg}+\mathrm{O_2}\rightarrow2\mathrm{MgO}\). \(2\mathrm{MgO}\) contains a total of 2 oxygen atoms, and one \(\mathrm{O_2}\) molecule also contains 2 oxygen atoms. Therefore, the coefficient of \(\mathrm{O_2}\) is 1. Choosing “two” would provide 4 oxygen atoms, so the equation would not be balanced.

The reaction is \(\mathrm{CaO + H_2O \rightarrow Ca(OH)_2 + heat}\). Two reactants combine to form one product, calcium hydroxide, so it is a combination reaction. Since heat is released, it is also exothermic. In a decomposition reaction, one compound breaks down into simpler substances, which does not occur here.

This is a redox reaction, in which oxidation and reduction occur together. Hydrogen changes from oxidation number 0 in H₂ to +1 in H₂O, so it is oxidised by gaining oxygen or losing electrons conceptually. Copper in CuO changes from +2 to 0 in Cu, so CuO is reduced. Thus option A is correct; B reverses the changes, while C and D incorrectly assign the same process to both substances.