Chemistry 2013 Jamb Examination

In the given reaction, fluorine gas \(F_2\) acts as an oxidising agent. Each fluorine atom in \(F_2\) has an oxidation state of 0. During the reaction, it gains electrons to form hydrogen fluoride (HF), where the oxidation state of fluorine becomes -1. The gain of electrons indicates reduction. The oxygen in water, on the other hand, goes from -2 in \(H_2O\) to 0 in \(O_2\), showing oxidation. Thus, \(F_2\) is the species that is reduced because it accepts electrons from water molecules.

Polarization occurs in an electrochemical cell when hydrogen gas forms and accumulates on the surface of the electrode, usually the cathode, creating a barrier that hinders the flow of current. This build-up of hydrogen bubbles increases the internal resistance of the cell and decreases its efficiency. The phenomenon is reduced by using depolarisers such as manganese(IV) oxide, which react with hydrogen to remove it from the electrode surface.

The electrolysis of acidified water produces oxygen at the anode and hydrogen at the cathode according to the equation:\[ 2H_2O(l) \rightarrow 2H_2(g) + O_2(g) \]At the anode, 4 moles of electrons produce 1 mole of \(O_2\):\[ 4e^- \rightarrow O_2 \]Calculate total charge passed:\[ Q = I \times t = 5 \times 193 = 965 \, C \]Find moles of electrons:\[ n(e^-) = \frac{Q}{F} = \frac{965}{96500} = 0.01 \, mol \]Find moles of \(O_2\):\[ n(O_2) = \frac{n(e^-)}{4} = \frac{0.01}{4} = 0.0025 \, mol \]Find volume of \(O_2\) at s.t.p:\[ V = n \times 22.4 \]\[= 0.0025 \times 22.4 = 0.056 \, dm^3 \]\[ V = 0.056 \, dm^3 = 56 \, cm^3 \]Hence, the volume of oxygen evolved is 0.056 dm³ (or 56 cm³).

Solubility in \(g \, dm^{-3}\) is the mass of solute that dissolves in 1 dm³ (1000 g) of solvent at a given temperature.
Given:
Mass of solute = 24.4 g
Mass of solvent = 42 g\[ \text{Solubility} = \dfrac{24.4}{42} \times 1000 \]\[= 581 \, gdm^{-3} \]This means 581 g of lead(II) trioxonitrate(V) would dissolve in 1 dm³ of water at 20°C.

Grease stains are organic and non-polar, and thus dissolve best in organic, non-polar solvents. Turpentine is a non-polar organic solvent that dissolves oils, fats, and grease effectively. It works by breaking the intermolecular forces binding the grease to the surface, allowing easy removal. Ammonia, ethanol, and borax solutions are more effective for polar or ionic stains, not greasy materials.

Sewage introduces large quantities of organic matter into water. Bacteria decompose this matter, consuming dissolved oxygen in the process. The oxygen depletion leads to hypoxic conditions that can suffocate fish and other aquatic organisms. This phenomenon is known as eutrophication. As bacterial population increases, oxygen levels decrease, upsetting the ecosystem balance and causing the death of aquatic life.

Dilution of a solution follows the relation:\[ C_1V_1 = C_2V_2 \]Where:
\( C_1 = 2.0 \; mol\, dm^{-3} \), \( V_1 = 2.5 \; dm^3 \)
Final volume, \( V_2 = 2.5 + 10 = 12.5 \; dm^3 \)\[ C_2 = \dfrac{C_1V_1}{V_2} \]\[= \dfrac{2.0 \times 2.5}{12.5} = 0.4 \; mol\, dm^{-3} \]Hence, the final concentration is 0.4 mol dm⁻³.

Adding a strong acid such as hydrochloric acid to any solution increases the concentration of hydrogen ions \(H^+\), thereby lowering the pH. The original solution with pH 6.4 is weakly acidic, and the addition of more \(H^+\) ions from HCl makes it more acidic. Consequently, the pH decreases and becomes less than 6.4.

Hydrogen bonding occurs when a hydrogen atom is directly bonded to a highly electronegative atom such as fluorine, oxygen, or nitrogen. In hydrogen fluoride (HF), the large difference in electronegativity between hydrogen and fluorine creates strong dipole–dipole attractions between neighbouring molecules, forming hydrogen bonds. These bonds account for HF’s unusually high boiling point compared to similar-sized molecules like CH₄ or HBr.

The element with configuration 1s² 2s² 2p⁶ 3s² is magnesium (Mg), a metal, while 1s² 2s² 2p⁴ is oxygen (O), a non-metal. When these two elements react, magnesium loses two electrons to form \(Mg^{2+}\), and oxygen gains two electrons to form \(O^{2-}\). The electrostatic attraction between these oppositely charged ions forms an ionic bond, resulting in the compound magnesium oxide (MgO).