The diagram shows an electrolytic cell - HSC - SSCE Chemistry - Question 32 - 2018 - Paper 1

To calculate the minimum voltage required to drive the electrolytic cell, we need to examine the half-reactions occurring at the electrodes.

At the anode, bromide ions are oxidized: $2Br^-(aq) \rightarrow Br_2(g) + 2e^- \quad E^\circ = -1.10 \text{ V}$

At the cathode, water is reduced: $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq) \quad E^\circ = -0.83 \text{ V}$

The total potential for the cell can be found by summing the standard reduction potentials: $E_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.83) - (-1.10) = 0.27 \text{ V}$

Therefore, the minimum voltage required to drive the cell rounds up to a total potential, leading us to conclude: A voltage greater than 1.93 V is required to effectively drive the reaction.

To compare the rates of corrosion of iron and steel, the following procedure could be used:

1. Sample Preparation: Cut equal-sized samples of iron and steel, ensuring they have the same surface area to maintain uniform conditions.

2. Environment Setup: Prepare a test tube filled with distilled water and place the samples in it. Optionally, introduce a mild salt solution to increase the rate of corrosion, as saltwater can accelerate it.

3. Observation Period: Allow the samples to sit in the solution undisturbed for a period of three weeks.

4. Data Collection: After the period, remove the samples, rinsing them gently with distilled water. Observe and record any visual signs of corrosion, such as rust formation or weight loss.

5. Analysis: Compare the extent of corrosion between the two metals by measuring factors such as changes in mass or the presence of rust.

Rusting is a specific type of corrosion that primarily affects iron, leading to the formation of iron(III) oxide. It occurs when iron reacts with oxygen and moisture in the environment.

1. Oxidation Reaction: Iron acts as the anode and is oxidized: $Fe(s) \rightarrow Fe^{2+}(aq) + 2e^-$

2. Reduction Reaction: At the cathode, oxygen is reduced, often in the presence of water: $O_2(g) + 4e^- + 2H_2O(l) \rightarrow 4OH^-(aq)$

These reactions combine to form iron(III) oxide, or rust: $4Fe^{2+}(aq) + O_2(g) + 6H_2O(l) \rightarrow 4Fe(OH)_3(s) \rightarrow 2Fe_2O_3\cdot 3H_2O(s)$

Rusting typically occurs in the presence of impurities such as salts, which can accelerate this reaction. It not only hampers the structural integrity of iron but results in a visible reddish-brown coating, indicative of corrosion.