Question 1: Explain the principle of aqueous electrolysis and how it differs from molten electrolysis.

Answer: Aqueous electrolysis involves passing an electric current through an aqueous solution to cause decomposition, with water and dissolved ions participating. Unlike molten electrolysis, which involves only ionic compounds in a liquid state, aqueous electrolysis involves water as a reactant and can produce different products at electrodes due to water's dissociation.

Question 1: Define electrolysis and describe the roles of the anode and cathode.

Answer: Electrolysis is a process that uses electrical energy to drive a non-spontaneous chemical reaction. The anode is the positive electrode where oxidation occurs, and the cathode is the negative electrode where reduction takes place.

Question 2: Describe what occurs at the electrodes during the electrolysis of aqueous sodium chloride solution.

Answer: At the anode, chloride ions (Cl-) are oxidised to produce chlorine gas (Cl₂). At the cathode, water is reduced to produce hydrogen gas (H₂) and hydroxide ions (OH-).

Question 1: Calculate the volume of hydrogen gas produced at the cathode when 0.5 mol of electrons are supplied during electrolysis at room temperature (25°C) and pressure (1 atm). Use the ideal gas law: PV = nRT. (R = 8.314 J/mol·K)

Answer: The number of moles of electrons (n) is 0.5 mol. Since 2 electrons produce 1 molecule of H₂, moles of H₂ = 0.25 mol. Using PV = nRT, V = nRT/P. V = (0.25 mol)(8.314 J/mol·K)(298 K) / (101325 Pa) ≈ 0.061 L or 61 mL.

Question 1: Design a simple experiment to electrolyse dilute copper sulfate solution. Identify the variables you would control and safety precautions to take.

Answer: The experiment involves setting up electrodes in a dilute copper sulfate solution connected to a power supply. Control variables include the voltage, electrode material, and concentration. Safety precautions include wearing goggles and gloves, ensuring proper insulation of electrical connections, and handling chemicals carefully to prevent spills and burns.

Question 1: During electrolysis of a potassium iodide solution, the volume of iodine produced at the anode was measured as 45 mL over 30 minutes, with a current of 0.5 A. Calculate the number of moles of iodine produced. (Use F = 96485 C/mol, 1 mol of electrons produces 1 mol of iodine.)

Answer: Current (I) = 0.5 A, time (t) = 1800 seconds. Total charge Q = I × t = 0.5 × 1800 = 900 C. Number of moles of electrons = Q / F = 900 / 96485 ≈ 0.00933 mol. Since 1 mol of electrons produces 1 mol of iodine, moles of iodine = 0.00933 mol.

Question 1: Explain why water undergoes reduction at the cathode during electrolysis of an aqueous sodium chloride solution. (6 marks)

Answer: In aqueous solutions, water molecules are present at the cathode and can be reduced because they have a higher reduction potential compared to some metal cations. When sodium chloride is electrolysed, chloride ions are oxidised at the anode, but at the cathode, water molecules are more readily reduced to produce hydrogen gas and hydroxide ions. The reduction of water occurs because it has a lower activation energy for reduction under these conditions, making it more favourable than sodium cations for gaining electrons.

Question 1: Describe one industrial application of electrolysis involving aqueous solutions and explain its significance.

Answer: One example is the electrolysis of brine (saltwater) to produce chlorine gas, hydrogen gas, and sodium hydroxide. This process is vital for producing chemicals used in disinfectants, plastics, and cleaning products, making it an important industrial process.