Corrosion:

Corrosion is the weakening of a metal over time through rusting, oxidation, or chemical reactions.

The more reactive a metal, the more quickly it corrodes. This is because they lose their electrons faster.  Some metals, such as gold, do not corrode at all.

Some metals, such as aluminium, will form a layer of tarnish when they oxidise – stopping oxygen from getting to the surface of the metal.

The statue of liberty is another example of tarnish protecting the metal from corroding further.

Rusting:

Rusting only occurs in iron. Iron reacting with oxygen is the same as corrosion.

For iron to rust it needs oxygen and water.

As you can see on the right, without oxygen and water, the nail does not rust.

iron + oxygen + water → hydrated iron oxide

Electroplating uses electrolysis to put a thin layer of a less reactive metal onto the object you want to protect.

• The cathode (negative electrode) should be the metal you want to protect or improve the appearance of.
• The anode (positive electrode) should be the metal you are using to protect it.
• The electrolyte (ionic liquid) should contain ions of the protecting metal.

Example: Explain how electroplating can be used to coat a spoon in silver. (3)

Part 1: Attach the pure silver to the anode. The silver atoms will turn into positive ions by losing electrons (oxidised) and move through the electrolyte to the cathode.

Ag → Ag⁺ + e^-

Part 2: The electrolyte will contain silver ions as well so that the ions are free to move.

Part 3: The silver ions will move to the cathode, where they will gain their electrons back (reduced) and turn back into metal atoms - coating the metal you want to protect.

Ag⁺ + e^- → Ag

Key word definitions:

• Actual Yield: The mass of a desired product actually produced in a chemical reaction.
• Theoretical Yield: The maximum calculated mass that could be produced.
• Percentage Yield: A measure of how much product you actually make in a chemical reaction compared to how much you should make in theory, expressed as a percentage.

Example: In a chemical reaction, 5.3g of sodium carbonate is reacted with excess hydrochloric acid. It was calculated that 5.85g of sodium chloride should be produced, but only 4.2g of sodium chloride was actually formed. Calculate the percentage yield for the reaction.

• Percentage yield = Actual Yield ÷ Theoretical Yield x 100.
• Percentage yield = 4.2g ÷ 5.85g x 100 = 71.8%

You might also be asked to calculate the theoretical yield - this is the same as a maximum mass calculation (see CH83)

Atom Economy is the percentage of mass converted into useful products and is calculated using the equation on the right.

Example: Ammonium chloride reacts with calcium hydroxide to form ammonia, calcium chloride and water. Work out the atom economy for the formation of calcium chloride in this reaction. (A_r: N = 14, H = 1, Cl = 35.5, Ca = 40, O = 16)

2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O

Step 1: Find out the formula mass (CH78) for calcium chloride, CaCl₂.

• 1 x Ca = 1 x 40 = 40
• 2 x Cl = 2 x 35.5 = 71
• Relative Formula Mass = 40 + 71 = 111.

Step 2: Find out the total formula mass of all of the products combined:

• 2NH₃: 2 x (14 + 3) = 34
• 2H₂O: 2 x (2 + 16) = 36
• Total formula mass of products = 34 + 36 + 111 = 181

Step 3: Use these two values to calculate the percentage of useful products:

• Atom Economy = formula mass of useful product (calcium chloride) ÷ total formula mass of all products x 100
• Atom economy = 111 ÷ 181 x 100 = 61.3%

The molar volume is the volume of a gas when you have one mole of molecules.

At room temperature (20^oC) and pressure (1atm), the

molar volume for 1 mole is always 24dm³mol^-1.

You can use this to work out masses and volumes for any balanced equation:

Example 1: Calculate the volume of 0.50 mol of oxygen at room temperature & pressure. (Molar volume = 24dm³mol^-1)

• Volume = moles x molar volume
• = 0.5 x 24 = 12dm³

Example 2: In an experiment, hydrogen gas is made by reacting 20.0g of sodium with excess water. Calculate the maximum volume of hydrogen that could be formed.

2Na _(s) + 2H₂O_(l) → 2NaOH_(s) + H_{2 (g)}

• Step 1: Find out the formula mass (CH78) for one sodium = 23
• Step 2: Find out the moles (CH86) for sodium = mass ÷ M_r = 20 / 23 = 0.870 moles
• Step 3: The ratio of sodium to hydrogen is 2:1, so you have 0.87/2 = 0.435 moles of hydrogen.
• Step 4: Volume = moles of H₂ x molar volume = 0.435 moles x 24 = 10.4dm³

Example: Sulfur trioxide is made by reacting sulfur dioxide with oxygen. It is a reversible reaction. The forward reaction is exothermic.

2SO₂ + O₂ ⇌ 2SO₃

Explain the effects of temperature, concentration, pressure and a catalyst on the position of equilibrium in the formation of sulfur trioxide.  

Temperature:

• Increasing the temperature favours the endothermic reaction.
• In the reaction above, the backwards reaction is endothermic so equilibrium will shift to the left.
• This means increasing the temperature will give a lower yield of sulfur trioxide, SO₃.
• The yield of the reactants (SO₂ & O₂) would increase.
• The reason we increase the temperature is because we get that (smaller) yield much faster than at lower temperatures.

Concentration:

• If you increase the concentration of reactants, there will be more frequent collisions.
• This will move the equilibrium to the right and give you a larger yield of ammonia.

Pressure:

• Increasing the pressure favours the side with less molecules.
• In the reaction above, there are three molecules on the left (one O₂ molecule and two SO₂ molecules) and two molecules on the right (two SO₃ molecules).
• Therefore, increasing the pressure will move equilibrium to the right (the side with the least molecules) and increase the yield of sulfur trioxide.

Catalyst:

• Using a catalyst speeds up the rate of attainment of equilibrium, but does not affect the position of equilibrium.