Question 1: Define a primary cell and explain how it differs from a secondary cell.

Answer: A primary cell is a chemical cell that cannot be recharged once the chemical reactions are complete, whereas a secondary cell can be recharged by reversing the chemical reactions using an external electrical source.

Question 2: Describe the role of the electrolyte in a chemical cell.

Answer: The electrolyte conducts ions between the electrodes, enabling the flow of charge and maintaining the chemical reactions inside the cell.

Question 3: Explain why zinc is often used as the anode in galvanic cells.

Answer: Zinc is used because it has a high tendency to oxidise (it is a good reducing agent), which makes it suitable as the anode in galvanic cells.

Question 1: Calculate the standard cell potential (E°) for a cell consisting of a zinc electrode (E° = -0.76 V) and a copper electrode (E° = +0.34 V).

Answer: E° = E°(cathode) - E°(anode) = 0.34 V - (-0.76 V) = 1.10 V

Question 2: A secondary lithium-ion battery has a capacity of 3.0 Ah and operates at a voltage of 3.7 V. Calculate the total energy stored in the battery in joules.

Answer: Energy (J) = Voltage (V) × Capacity (Ah) × 3600 (seconds in an hour) = 3.7 V × 3.0 Ah × 3600 = 40,080 Joules

Question 1: Describe a simple experiment to test the voltage of a typical dry cell and identify the variables involved.

Answer: Connect the dry cell to a voltmeter and record the reading. Variables include the type of cell, temperature, and the condition of the voltmeter electrodes.

Question 2: List three safety precautions to take when handling batteries in a laboratory setting.

Answer: Avoid short circuits by connecting correctly, handle batteries with dry hands, and do not expose batteries to excessive heat or flames.

Question 1: A student measures the voltage of a rechargeable nickel-metal hydride (NiMH) cell and records a voltage of 1.25 V. The standard potential for this cell is 1.30 V. Suggest possible reasons for the discrepancy and how it could affect the performance.

Answer: Possible reasons include partial discharge, internal resistance, or temperature effects. This lower voltage may reduce the efficiency and capacity of the battery during use.

Question 1: Explain how the chemical reactions in a lead-acid car battery produce electrical energy. Include the overall cell reaction and the roles of the electrodes and electrolyte. (6 marks)

Answer: In a lead-acid battery, the anode is made of lead (Pb) and the cathode of lead dioxide (PbO₂). The electrolyte is sulphuric acid (H₂SO₄). During discharge, Pb reacts with H₂SO₄ to produce PbSO₄ and releases electrons at the anode; PbO₂ reacts with H₂SO₄ and electrons at the cathode to form PbSO₄ and water. The overall reaction is: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. The flow of electrons from the anode to the cathode produces electrical energy, which powers the vehicle.