Explanation

(a)(i) Define the term standard electrode potential: It is the measure of the tendency of an element to form ions in solution relative-to the tendency of hydrogen atoms to form ions in solution at standard condition/ the potential difference that exists between an electrode and the hydrogen electrode under standard conditions (of I moldm\(^{-3}\) concentration at 298k and 1 atm).

(ii) State three factors that affect the discharge of ions during electrolysis: (1) Position of ions in the electrochemical series/standard electrode potential value. (2) Nature or reactivity of electrode used. (3) (relative) concentration of ions.

(iii) State two functions of a salt. bridge in an electrochemical cell: (1) Allows electrical contact between two solutions/complete the electric circuit (2) Maintains electrical neutrality in each half cell/allows ions flow in and out of salt bridge.

(b)(i) Describe briefly what happens when a solution of copper(II) tetraoxosulpahte(VI) is electrolysed using copper electrodes:

CuSO\(_4\) \(\rightleftharpoons\) Cu\(^{2+}\) + SO\(_4^{2-}\); Cu\(^{2+}\) discharged/Cu metal deposited/Cu cathode increases in mass.

H\(_2\)O \(\rightleftharpoons\) H\(+\) + OH\(-\) Copper anode dissolves/ionizes/decreases in size to give Cu\(^{2+}\) ions.

Solution remains blue/concentration of Cu\(^{2+}\) ions is constant.

(c) Calculate the mass of copper(II)tetraoxosulphate(V1) for 35 minutes using copper electrodes:. [H=1, O=16, S=32, Cu=64, IF=96,500]

Cu\(^{2+}\) + 2e ---> Cu

Q = It = 0.2 x 35 x 60 = 420C

2F = 2e = 1mole \(\frac{Cu}{2}\) x 96500C = 64g

420C = \(\frac{420}{ 2 x 96500}\) x 64gCu

= 0.14gCu.

(d)(i) State three characteristics of catalyst: (1) Remains unchanged in chemical nature and mass. (2) Does not affect the types of product formed. (3) Alters (increase or decrease) the rate of chemical reaction. (4) They are reactions specific. (5) Cannot initiate a reaction that is not feasible. (6) Has no effect on the equilibrium of a reversible reaction. (7) Required in small quantity.

(ii) Name one manufacturing process in which each of the following metals is used as a catalyst:

I. Iron – used in Haber process.

II. Nickel – hydrogenation of vegetable oil/alkenes.

III. Platinum – used in preparation of hydrogen iodido from hydrogen and iodine/contact process for H\(_2\)SO\(_4\) production/manufacture of HNO\(_3\) used in Hydrogenation.

Correct Answer: B. alkalinity

Correct Answer: C. Sucrose

Explanation

(a) Graham's Law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of it's vapour density or relative molecular mass.

\(\frac{RO_2}{R_y} = \sqrt{\frac{M_y}{MO_2}}; \frac{\frac{100}{4}}{{\frac{50}{3}}}\)

\(\frac{100}{4}\) x \(\frac{3}{50}\) = \(\sqrt{\frac{M_y}{32}}\)

= \(M_y = \frac{32 \times 9}{4}\) = 72

(b) X\(_{(g)}\) + 2Y\(_{(g)}\) \(\rightleftharpoons\) XY\(_{2(g)}\) \(\Delta\)H = – 52 KJ mol\(^{-1}\)

(i) The yield decreases

(ii) The forward reaction is accompanied by a decrease in the number of gaseous molecules. A decrease in pressure shifts the position of equilibrium to the left favouring the backward reaction.

(iii) i. The position of equilibrium is not affected by the presence of .a catalyst. ii. The presence of a catalyst lowers the activation energy or takes the reaction through another path of lower energy barrier.

(c)(i)

(ii) I. Boiling point II. Solubility III. Sublimation or property of solid.

(d) I. Neutralization II. Double decomposition or precipitation Ill. Displacement or reduction.

(iii) Hydrated salts contain water of crystallization while anhydrous salts do not.

Correct Answer: D. S

Correct Answer: D. depolarizer

Correct Answer: C) downward displacement of water

Correct Answer: A. 18 g/mol

Explanation

The molar mass of water (H2O) is calculated by adding the atomic masses of its constituent elements: 2 hydrogen atoms (each with a molar mass of approximately 1 g/mol) and 1 oxygen atom (with a molar mass of approximately 16 g/mol). Therefore, the molar mass of water is 2 x 1 g/mol + 1 x 16 g/mol = 18 g/mol.

Correct Answer: A. 0

Explanation

(a)(i) Heat the iron strongly in a combustion tube until it is red hot. Generate steam

by boiling water in a flask. Pass the steam over the red-hot iron. Hydrogen gas would be given off. Collect the hydrogen gas formed over water.

(ii) 3Fe(s) + 4H\(_2\)O\(_{(g)}\) Fe\(_3\)O\(_{4(s)}\) + 4H\(_{2(g)}\)

(iii) - from water gas / Bosch process

- from methane

- by electrolytic method

- from catalytic cracking of petroleum products

(b) (i) It favours the backward reaction / Equilibrium position shifts to the left

(ii) Equilibrium position shifts to the right / favours the forward reaction

(iii) Equilibrium position shifts to the right / favours the forward reaction.

(c)(i) Condensation polymerization is defined as the linking together of monomers (with two different functional groups) to form a polymer with the elimination of a small molecule.

(ii) - Nylon/polyamide

- Terylene

- Silk

- Wool

- Cellulose

- Proteins

- Starch

- Dacron/polyester

- Phenol methanal

- Urea methanal

- Melamine methanal

(iii) - CH\(_3\)COOC\(_2\)H\(_5\)

(d) (i) Allotropy is defined as the existence of two or more forms of an element in the same physical state.

(ii)- diamond

- graphite

(iii)

(e)