Chemical Reactions and Equations

The reaction is \(\mathrm{CuO + H_2 \rightarrow Cu + H_2O}\). Oxygen is removed from black \(\mathrm{CuO}\), producing reddish-brown copper. Removal of oxygen is reduction, so \(\mathrm{CuO}\) is reduced. In contrast, oxidation involves addition of oxygen.

The oxidation state of aluminium increases from \(0\) to \(+3\), so it loses electrons and is oxidised. Iron in \(Fe_2O_3\) has oxidation state \(+3\), which becomes \(0\) in elemental \(Fe\); thus \(Fe^{3+}\) gains electrons and is reduced. Option B reverses both changes.

On heating, the single compound lead nitrate, \(Pb(NO_3)_2\), breaks down to form \(PbO\), \(NO_2\), and \(O_2\). The breakdown of one compound into two or more substances by heat is a thermal decomposition reaction. In contrast, a combination reaction forms one product from two or more reactants.

A redox reaction involves transfer of electrons. The substance that loses electrons undergoes oxidation, while another substance gains those electrons and undergoes reduction. Therefore, oxidation and reduction occur together. Option B is incorrect because oxygen transfer is not required in every redox reaction.

In \(\mathrm{2Mg + O_2 \rightarrow 2MgO}\), the oxidation number of Mg changes from \(0\) to \(+2\), so Mg is oxidised. The oxidation number of O changes from \(0\) to \(-2\), so O is reduced. Therefore, this is a redox reaction. In contrast, option C is a precipitation reaction with no change in oxidation numbers.

The zinc coating acts as a barrier between iron and air and moisture. Rusting generally requires both oxygen and water/moisture, so preventing their contact reduces rusting of iron. Oxygen alone or moisture alone is an incomplete answer. In addition, zinc is more reactive than iron and can provide some protection even if the coating is scratched.

Zinc is more reactive than copper. Therefore, zinc displaces copper from aqueous \(\mathrm{CuSO_4}\), forming \(\mathrm{ZnSO_4}\) and \(\mathrm{Cu}\). Zinc changes from oxidation state 0 to \(+2\), while copper changes from \(+2\) to 0. Option B is a decomposition reaction, whereas option C is a double-displacement reaction.

Iron is more reactive than copper, so it displaces copper from copper sulphate. The reaction is \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The iron sulphate solution is green, while displaced copper forms the reddish-brown coating on the nail. This is not a double displacement reaction because iron replaces copper from its salt.

Zinc is more reactive than copper, so it displaces copper from copper sulphate solution: \(\mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu}\). As blue \(\mathrm{Cu^{2+}}\) ions are removed, the solution fades and reddish-brown copper is deposited on the zinc strip. The hydrogen bubbles in option B are typically produced when a metal reacts with an acid, not with copper sulphate solution.

In \(\mathrm{FeCl_3}\), Fe has oxidation number \(+3\), which becomes \(+2\) in \(\mathrm{FeCl_2}\). A decrease in oxidation number is reduction, so Fe is reduced. The oxidation number of Sn increases from \(+2\) in \(\mathrm{SnCl_2}\) to \(+4\) in \(\mathrm{SnCl_4}\); therefore, Sn is oxidised. Hence, option A is correct; both elements are not undergoing the same process.

This is a photodecomposition reaction. A single compound, \(AgCl\), decomposes in sunlight to form silver \(Ag\) and chlorine gas \(Cl_2\). The equation is balanced, with 2 Ag atoms and 2 Cl atoms on each side. In a displacement reaction, a free element replaces another element in a compound; no such free reactant is present here.

Heat on the reactant side indicates that the reaction absorbs heat, so it is endothermic. The breakdown of one compound into two or more products is a decomposition reaction. Therefore, the reaction is an endothermic decomposition reaction. Option D is also a decomposition reaction, but an exothermic reaction releases heat on the product side.

In option A, ions from two aqueous solutions exchange, producing insoluble \(\mathrm{AgCl}\) as a solid. Formation of an insoluble solid from aqueous solutions identifies a precipitation reaction. In option B, zinc displaces copper from its salt solution, so it is a displacement reaction rather than a precipitation reaction.

This is a photochemical decomposition reaction because the single compound \(AgCl\) breaks down in sunlight to form silver and chlorine gas. In a displacement reaction, a second, more reactive element removes another element from its compound; no such reactant is present here.

\(\mathrm{aq}\) stands for “aqueous”, meaning the species is dissolved in water (so it exists in solution). \(\mathrm{l}\) stands for “liquid”, meaning the species is in the liquid state as a separate phase (it is not indicating “in water”). Therefore, \(\mathrm{aq}\) specifies water as the solvent, whereas \(\mathrm{l}\) specifies only that the substance is a liquid.

In \(\mathrm{CuO + H_2 \rightarrow Cu + H_2O}\), copper in \(\mathrm{CuO}\) changes from oxidation state \(+2\) to \(0\), so copper is reduced. Hydrogen in \(\mathrm{H_2}\) changes from \(0\) to \(+1\), so hydrogen is oxidised. Therefore, this is a redox reaction. In contrast, oxidation numbers do not change in option B.

According to the law of conservation of mass, total mass is neither created nor destroyed in a chemical reaction. The gas formed remains inside the closed container, so the combined mass of the container and all its contents is the same before and after the reaction. A rise in temperature indicates an energy change, not an increase in mass.

Subscripts in a chemical formula show the fixed ratio of atoms in a compound. For example, \(\mathrm{H_2O}\) contains two hydrogen atoms and one oxygen atom. Changing the subscript to make \(\mathrm{H_2O_2}\) gives hydrogen peroxide, not water. Therefore, only coefficients are changed to balance an equation; subscripts must remain unchanged.

The chemical formula of water is \(H_2O\). One water molecule has 2 hydrogen atoms and 1 oxygen atom, so it contains 3 atoms in total. Therefore, two water molecules contain \(2 \times (2+1)=6\) atoms. Option 5 is incorrect because it does not represent the total number of hydrogen and oxygen atoms in two complete water molecules.

Magnesium reacts slowly with oxygen in air to form a thin magnesium oxide layer on its surface. This coating reduces direct contact between magnesium and oxygen. Rubbing with sandpaper removes the coating, so magnesium burns more readily with a dazzling white flame. The rubbing does not change magnesium’s melting point or remove oxygen from the air.

Calcium oxide, \(\mathrm{CaO}\) (quicklime), reacts with water to form calcium hydroxide, \(\mathrm{Ca(OH)_2}\) (slaked lime):
\[\mathrm{CaO + H_2O \rightarrow Ca(OH)_2}\]
This is an exothermic reaction. Calcium carbonate requires \(\mathrm{CO_2}\) for its formation, so it is not produced by adding water alone.

Calcium carbonate undergoes thermal decomposition: \(\mathrm{CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)}\). Therefore, calcium oxide is the solid product and carbon dioxide is the gaseous product. Calcium hydroxide would require water, which is not a reactant in this reaction.

Thermal decomposition of calcium carbonate produces carbon dioxide: \(\mathrm{CaCO_3 \xrightarrow{\Delta} CaO + CO_2}\). When this gas is passed through lime water, \(\mathrm{Ca(OH)_2}\), it forms an insoluble precipitate of calcium carbonate, making the lime water milky: \(\mathrm{Ca(OH)_2 + CO_2 \rightarrow CaCO_3 \downarrow + H_2O}\). Oxygen does not form a calcium carbonate precipitate under these conditions.

Lead nitrate undergoes thermal decomposition:
\[
2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}
\]
\(\mathrm{NO_2}\) is the brown gas, while \(\mathrm{O_2}\) is colourless. Option B incorrectly names nitrogen gas; nitrogen is released in this reaction as \(\mathrm{NO_2}\), not as \(\mathrm{N_2}\).

On heating, lead nitrate undergoes thermal decomposition: \(2\mathrm{Pb(NO_3)_2} \xrightarrow{\Delta} 2\mathrm{PbO} + 4\mathrm{NO_2} + \mathrm{O_2}\). The yellow solid residue is lead(II) oxide, \(\mathrm{PbO}\). In contrast, lead(II) chloride contains chloride ions, which cannot be formed by merely heating lead nitrate.