Alkali Metals Reaction with Water and Common Ions

• Group 1 metals (alkali metals) react with water to produce a metal hydroxide and hydrogen.
• The general word equation for the reaction is: alkali metal + water → metal hydroxide + hydrogen.
• An example using sodium is: sodium + water → sodium hydroxide + hydrogen.
• The corresponding chemical equation is: 2Na + 2H2O → 2NaOH + H2.
• Observations during the reaction include the metal floating and moving on the water surface, along with fizzing.
• Sodium (Na) forms a ball shape during the reaction, while potassium (K) produces a lilac flame.
• Because metal hydroxides are alkaline, universal indicator (UI) turns blue/purple in the solution.

Compound Ions to Memorise

• Sulfate: SO₄²⁻
• Carbonate: CO₃²⁻
• Nitrate: NO₃⁻
• Hydroxide: OH⁻

Ion Charges

• Group 1 metals form +1 ions (e.g., Na⁺).
• Group 2 metals form +2 ions (e.g., Ca²⁺).
• Group 3 metals form +3 ions (e.g., Al³⁺).
• Group 5 non-metals form -3 ions (e.g., N³⁻).
• Group 6 non-metals form -2 ions (e.g., O²⁻).
• Group 7 non-metals form -1 ions (e.g., Cl⁻).

Displacement Reactions

• Displacement reactions involve a more reactive metal displacing a less reactive metal from a compound.
• These reactions are examples of redox reactions, where both reduction and oxidation occur.
• Copper(II) compounds, such as CuSO₄, form blue solutions.
• When a more reactive metal (e.g., calcium) is added to copper sulfate, calcium sulfate and copper are formed.
• The blue color of the solution fades, and pink copper metal is observed.

Redox Reactions

• Reduction is the loss of oxygen or the gain of electrons.
• Oxidation is the gain of oxygen or the loss of electrons.
• A reducing agent reduces another substance and is itself oxidized.
• An oxidizing agent oxidizes another substance and is itself reduced.

Corrosion of Iron/Steel

• Both water and oxygen are necessary for iron to rust.
• Rust is chemically known as hydrated iron(III) oxide.
• Experiments to demonstrate the requirements for rusting include using boiled water with an oil layer (water, no oxygen) and a drying agent with a stopper (oxygen, no water).

Preventing Rust

• Barrier method: Coating iron with paint, oil, plastic, grease, or other metals to prevent contact with water and oxygen.
• The barrier method fails if the coating is scratched.
• Sacrificial method: Adding a more reactive metal (e.g., magnesium or zinc) that corrodes instead of the iron.
• Galvanisation: Coating iron with a thin layer of zinc which protects iron, even if scratched, because zinc is more reactive.

Reactivity Series

• The reactivity series lists metals in order of reactivity.
• K, Na, Li, Ca, Mg, Al, Zn, Fe, Cu, Ag, Au is an example of a reactivity series.
• K, Na, Li, and Ca react violently with water and explosively with acids.
• Mg, Al, Zn, and Fe react very slowly with water and fairly quickly with acids.
• Cu, Ag, and Au do not react with water or dilute acids.

Metal Reactions with Acid

• The general equation is: metal + acid → salt + hydrogen.
• Observations include bubbles and an increase in temperature (exothermic reaction).
• The most reactive metal produces the greatest temperature increase.
• Exothermic reactions release heat energy.

Flame and Cation Tests

• These are chemical tests used to identify metal ions or compounds.

Ionic Compounds

• Metal salts are ionic compounds with high melting points.
• This is due to their giant ionic lattice structure and the strong electrostatic forces of attraction between positive and negative ions.
• Overcoming these forces requires a lot of energy.

Determining Oxygen Percentage in Air

• Metals like copper can be used to determine the percentage of oxygen in dry air.
• Copper reacts with oxygen to form copper oxide (solid), effectively removing oxygen from the air.
• The reaction is: copper + oxygen → copper oxide.

Calculating Percentage of Oxygen

• The formula to calculate the percentage of oxygen in air is: % oxygen = (change in air volume / total starting volume) × 100.
• For example: If the start volume of air is 90cm³ and the end volume is 72cm³, the change in air volume is 18cm³ (90-72).
• Therefore, the % oxygen in air is (18/90) × 100 = 20%.