Chemical Reactions and Equations

Some silver salts, such as silver chloride, are light-sensitive. On exposure to light, they may undergo photochemical decomposition and form metallic silver; therefore, they are stored in dark-coloured bottles to reduce exposure to light. Dark bottles do not increase their solubility in water.

The electrolysis of water is represented by \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). At the same temperature and pressure, gas volumes are proportional to their mole ratio, so \(H_2:O_2 = 2:1\). Therefore, for 12 mL of oxygen, the volume of hydrogen is \(2 \times 12 = 24\) mL. The 12 mL option is incorrect because hydrogen and oxygen are not produced in equal volumes.

During electrolysis of water, reduction occurs at the cathode, producing hydrogen gas, whereas oxidation at the anode produces oxygen gas. The overall reaction is \(2H_2O(l) \rightarrow 2H_2(g) + O_2(g)\). Therefore, oxygen is produced at the anode and hydrogen at the cathode. Option B reverses the gases at the two electrodes.

Iron is more reactive than copper, so it displaces copper from copper sulphate: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The ferrous sulphate formed makes the solution light green, while displaced copper deposits as a reddish-brown coating on the nail. A temperature rise alone, as in option D, is not sufficient evidence of a displacement reaction.

Pure water contains very few ions, so it is a poor conductor of electricity. Dilute sulfuric acid supplies ions, increasing the electrical conductivity of the solution and allowing current to pass more readily. The acid does not replace the external electrical source. During electrolysis, hydrogen is formed at the cathode and oxygen at the anode, with a volume ratio of 2:1, not 1:1.

Iron is more reactive than copper, so it displaces copper from copper sulphate solution: \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). The iron(II) sulphate formed gives the solution a pale green colour. Copper metal is also produced, but it deposits as a reddish-brown coating on the nail; it does not make the solution green.

Iron is more reactive than copper in the reactivity series. Therefore, iron displaces copper from copper sulphate solution. The liberated copper forms a reddish-brown deposit on the nail surface. The reaction is \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\). Option B is incorrect because the less reactive metal, copper, cannot displace iron from its compound.

The blue colour of copper sulphate solution is due to \(\mathrm{Cu^{2+}}\) ions. Since iron is more reactive than copper, it displaces copper in the reaction \(\mathrm{Fe + CuSO_4 \rightarrow FeSO_4 + Cu}\), depositing copper metal. Thus, the concentration of \(\mathrm{Cu^{2+}}\) ions decreases and the blue colour fades. In contrast, \(\mathrm{Fe^{2+}}\) ions are produced rather than decreased.

In a displacement reaction, a more reactive metal displaces a less reactive metal from its salt solution. Since metal A displaces metal B and B cannot displace A, A is more reactive than B. If option B were correct, B would be able to displace A from its salt solution.

This is a double-displacement precipitation reaction: \(\mathrm{BaCl_2(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2NaCl(aq)}\). The \(\mathrm{Ba^{2+}}\) and \(\mathrm{SO_4^{2-}}\) ions form insoluble white barium sulphate, while sodium chloride remains in solution. Option B is wrong because it incorrectly replaces the sulphate ion with a sulphite ion. Exam tip: After exchanging ions in a double-displacement reaction, check the solubility of the products.

The oxidation number of \(Al\) increases from 0 to \(+3\), so aluminium is oxidised. At the same time, it reduces \(Fe^{3+}\) in \(Fe_2O_3\) to \(Fe\). A substance that causes reduction of another substance while itself being oxidised is the reducing agent; therefore, \(Al\) is the reducing agent. In contrast, \(Fe_2O_3\) is the oxidising agent because its \(Fe^{3+}\) ions are reduced.

A precipitate is an insoluble solid produced when two aqueous solutions react and one product cannot remain dissolved. It separates from the solution and settles or appears as a solid, so a downward arrow is used to indicate its formation. The symbol does not mean that the substance releases heat or light. A gas escaping is represented by an upward arrow, making option B the opposite notation and option A correct.

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 Al, the oxidation number is 0, while in Al₂O₃ it becomes +3; aluminium loses electrons and is oxidised. Iron in Fe₂O₃ has oxidation number +3 and changes to elemental iron with oxidation number 0, so it gains electrons and is reduced. This is a redox reaction in which aluminium acts as the reducing agent and iron(III) oxide as the oxidising agent. Therefore option A is correct.

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.

Rusting is an electrochemical corrosion process that needs iron, oxygen and water or moisture. The oil coating acts as a physical barrier, so air and moisture cannot easily reach the iron surface; consequently, the corrosion reaction becomes slower. Oil does not change iron into another metal, dissolve rust completely, or increase oxidation. Therefore, option A gives the correct prevention mechanism.