Review the key concepts, formulae, and examples before starting your quiz.
🔑Concepts
Primary Cells (Dry Cell): These are cells where the redox reaction occurs only once and the cell becomes dead after a period of use as the reaction cannot be reversed. The Leclanché cell is the most common example.
Dry Cell Components: The anode consists of a Zinc () container. The cathode is a carbon (graphite) rod surrounded by powdered manganese dioxide () and carbon. The electrolyte is a moist paste of ammonium chloride () and zinc chloride ().
Dry Cell Reactions: At the anode: . At the cathode: .
Role of : The gas produced at the cathode combines with ions from to form a complex ion , which prevents the buildup of gas pressure inside the cell.
Secondary Cells (Lead Accumulator): These cells can be recharged by passing current through them in the opposite direction. They can be used through several cycles of discharging and charging.
Lead Accumulator Components: The anode is a lead () plate. The cathode is a grid of lead packed with lead dioxide (). The electrolyte is a solution of sulphuric acid () with a density of .
Discharging of Lead Accumulator: During discharge, chemical energy is converted to electrical energy. is consumed, and is deposited on both electrodes.
Recharging of Lead Accumulator: An external source of is applied. The cell reactions are reversed: .
📐Formulae
💡Examples
Problem 1:
Write the overall cell reaction for a lead storage battery during the discharging process and state what happens to the density of the electrolyte.
Solution:
The overall reaction is: . The density of the electrolyte decreases.
Explanation:
During discharging, is consumed to form water and solid . Since is denser than water, its consumption and the production of lead to a decrease in the overall density/specific gravity of the electrolyte.
Problem 2:
Why is the dry cell not truly 'dry'?
Solution:
It contains a moist paste of and .
Explanation:
For ions to move and conduct electricity between the electrodes, a liquid or moisture-rich medium is necessary. If the cell were completely dry, the internal resistance would be infinite, and no current would flow.
Problem 3:
Calculate the amount of produced if Faradays of electricity is drawn from a lead accumulator.
Solution:
are produced at each electrode, totaling for the whole cell.
Explanation:
From the half-reactions, mole of produces mole of by releasing . Thus, of charge produces mole of at the anode and mole at the cathode. Therefore, will produce moles at each electrode based on the stoichiometry of the -electron process.