SpecPoint Archives - Page 5 of 8 - TutorMyself Chemistry

1:24 understand why the noble gases (Group 0) do not readily react

The noble gases are inert (unreactive) because they have a full outer shell of electrons.

Hydr0Gen2023-01-10T12:26:58+00:00Categories: (d) The Periodic Table, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:24, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Ions, Topic: Metal Reactivity & Halogens, Topic: Metals (D), Topic: Simple Molecules & Covalent Bonding|

1:25 write word equations and balanced chemical equations (including state symbols): for reactions studied in this specification and for unfamiliar reactions where suitable information is provided

Example:

Sodium (Na) reacts with water (H₂O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).

Word equation:

sodium + water –> sodium hydroxide + hydrogen

Writing the chemical equation

A chemical equation represents what happens in terms of atoms in a chemical reaction.

Step 1: To write a chemical equation we need to know the chemical formulae of the substances.

Na + H₂O –> NaOH + H₂

Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.

During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.

2Na + 2H₂O –> 2NaOH + H₂

If asked for an equation, the chemical equation must be given.

State symbols are used to show what physical state the reactants and products are in.

State symbols	Physical state
(s)	Solid
(l)	Liquid
(g)	Gas
(aq)	Aqueous solution (dissolved in water)

Example:

A solid piece of sodium (Na) reacts with water (H₂O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).

2Na(s) + 2H₂O(l) –> 2NaOH(aq) + H₂(g)

Hydr0Gen2023-09-18T11:05:10+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:25, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil, Topic: Fundamentals, Topic: Ions, Topic: Key Reactions, Topic: Moles|

1:26 calculate relative formula masses (including relative molecular masses) (Mᵣ) from relative atomic masses (Aᵣ)

Relative formula mass (M_r) is mass of a molecule or compound (on a scale compared to carbon-12).

It is calculated by adding up the relative atomic masses (A_r) of all the atoms present in the formula.

Example:

The relative formula mass (M_r) for water (H₂O) is 18.

Water = H₂O

Atoms present = (2 x H) + (1 x O)

M_r = (2 x 1) + (1 x 16) = 18

Hydr0Gen2022-11-30T14:12:46+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:26, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Fundamentals, Topic: Key Calculations, Topic: Moles|

1:27 know that the mole (mol) is the unit for the amount of a substance

In Chemistry, the mole is a measure of amount of substance (n).

The abbreviation for mole is mol.

The mass of 1 mole of a substance is the relative formula mass (M_r) of the substance in grams.

Example:

The M_r of water is 18.

Therefore the mass of 1 mol of water equals 18 g.

Hydr0Gen2018-04-25T21:37:27+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:27, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Topic: Moles|

1:28 understand how to carry out calculations involving amount of substance, relative atomic mass (Aᵣ) and relative formula mass (Mᵣ)

The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:

$Amount(mol) = \frac{mass(g)}{M_r}$

Example 1:

Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO₂).

Step 1: Calculate the relative formula mass (M_r) of carbon dioxide (CO₂).

$Carbon\,dioxide = CO_2$

$Atoms\,present = (1 \times C) + (2 \times O)$

$M_r = (1 \times 12) + (2 \times 16) = 44$

Step 2: Use the formula to calculate the amount in moles.

$Amount = \frac{mass(g)}{M_r}$

$Amount = \frac{8.8}{44}$

$Amount = 0.2 mol$

Example 2:

Calculate the mass of 2 mol of copper(II) sulfate (CuSO₄).

Step 1: Calculate the relative formula mass (M_r) of copper(II) sulfate (CuSO₄).

$copper(II)\,sulfate = CuSO_4$

$Atoms\,present = (1 \times Cu) + (1 \times S) + (4 \times O)$

$M_r = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 159.5$

Step 2: Rearrange the formula to calculate the mass.

$mass = amount \times M_r$

$mass = 2 \times 159.5$

$mass = 319g$

Hydr0Gen2018-04-25T21:37:27+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:28, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Topic: Moles|

1:29 calculate reacting masses using experimental data and chemical equations

Example: When calcium carbonate (CaCO₃) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.

CaCO₃ –> CaO + CO₂

Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO₃)

$M_r(CaCO_3)=(1 \times 40) + (1\times12) + (3 \times 16)$

$M_r(CaCO_3)= 100$

$Amount = \frac{mass(g)}{M_r}$

$Amount = \frac{25}{100}$

$Amount = 0.25 mol$

Step 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO₃.

This step involves using the ratio of CaCO₃ to CaO from the chemical equation.

CaCO₃ –> CaO + CO₂

From the equation you can see that the ratio of CaCO₃ to CaO is 1:1.

Therefore if you have 0.25 mol of CaCO₃ this will produced 0.25 mol of CaO.

Step 3: Calculate the mass of 0.25 mol of CaO.

$M_r(CaO)=(1 \times 40) + (1 \times 16)$

$M_r(CaO)= 56$

$Mass(CaO)=amount \times M_r$

$Mass(CaO)= 0.25 \times 56$

$Mass(CaO)= 14g$

A simple format for laying out this method can be used.

Example: What mass of ammonia (NH₃) is formed when 7 g of nitrogen (N₂) is combined with hydrogen (H₂).

Hydr0Gen2018-04-25T21:37:27+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:29, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Topic: Moles|

1:30 calculate percentage yield

Yield is how much product you get from a chemical reaction.

The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.

In most chemical reactions, however, you rarely achieved your theoretical yield.

For example, in the following reaction:

CaCO₃ –> CaO + CO₂

You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO₃.

However, what if the actual yield is only 48 g of CaO.

By using the following formula, the % yield can be calculated:

$yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product}$

$yield= \frac{48}{56}$

$yield=0.86$

$\% yield=86\%$

Hydr0Gen2023-01-10T11:13:31+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:30, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Equilibria, Topic: Key Calculations, Topic: Reversible Reactions (D)|

1:31a understand how the formulae of simple compounds can be obtained experimentally, including metal oxides

Finding the formula of a metal oxide experimentally

The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.

Example: When magnesium is burned in air, it reacts with oxygen (O₂) to form magnesium oxide (MgO).

Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)

Hydr0Gen2019-05-23T15:22:37+00:00Categories: Uncategorized|Tags: 1:31a, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, SubSpecPoint, Topic: Calculations|

1:31 understand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation

Finding the formula of a metal oxide experimentally

The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.

Example: When magnesium is burned in air, it reacts with oxygen (O₂) to form magnesium oxide (MgO).

Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)

Finding the formula of a salt containing water of crystallisation

When some substances crystallise from solution, water becomes chemically bound up with the salt. This is called water of crystallisation and the salt is said to be hydrated. For example, hydrated copper sulfate has the formula CuSO₄.5H₂O which formula indicates that for every CuSO₄ in a crystal there are five water (H₂O) molecules.

When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. If the hydrated copper sulfate (CuSO₄.5H₂O) are strongly heated in a crucible then they will break down and the water lost, leaving behind anhydrous copper sulfate (CuSO₄). The method followed is similar to that for metal oxides, as shown above. The difference of mass before and after heating is the mass of the water lost. These mass numbers can be used to obtain the formula of the salt.

Hydr0Gen2023-01-10T11:36:41+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:31, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D), Topic: Key Calculations, Topic: Specification experiments|

1:32 know what is meant by the terms empirical formula and molecular formula

The empirical formula shows the simplest whole-number ratio between atoms/ions in a compound.

The molecular formula shows the actual number of atoms of each type of element in a molecule.

For example, for ethane:

Molecular formula = C₂H₆

Empirical formula = CH₃

Here are some more examples:

Name	Molecular formula	Empirical Formula
pentene	C₅H₁₀	CH₂
butene	C₄H₈	CH₂
glucose	C₆H₁₂O₆	CH₂O
hydrogen peroxide	H₂O₂	HO
propane	C₃H₈	C₃H₈

Notice from the table that several different molecules can have the same empirical formula, which means that it is not possible to deduce the molecular formula from the empirical formula without some additional information. Also notice that sometimes it is not possible to simplify a molecular formula into simpler whole-number ratio, in which case the empirical formula is equal to the molecular formula.

Hydr0Gen2023-01-10T08:29:26+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:32, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Crude oil to plastics (D), Topic: Key Calculations|

1:33 calculate empirical and molecular formulae from experimental data

Calculating the Empirical Formula

Example: What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?

Step 1: Put the symbols for each element involved at the top of the page.

Step 2: Underneath, write down the masses of each element combining.

Step 3: Divide by their relative atomic mass (A_r).

Step 4: Divide all the numbers by the smallest of these numbers to give a whole number ratio.

Step 5: Use this to give the empirical formula.

(If your ratio is 1:1.5 then multiple each number by 2

If your ratio is 1:1.33 then x3. If your ratio is 1:1.25 x4)

Calculating the Molecular Formula

If you know the empirical formula and the relative formula mass you can work out the molecular formula of a compound.

Example: A compound has the empirical formula CH₂, and its relative formula mass is 56. Calculate the molecular formula.

Step 1: Calculate the relative mass of the empirical formula.

Step 2: Find out the number of times the relative mass of the empirical formula goes into the M_r of the compound.

Step 3: This tells how many times bigger the molecule formula is compared to the empirical formula.

Empirical formula calculations involving water of crystallisation

When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. For example, barium chloride crystals contain water of crystallisation, and therefore would have the formula BaCl₂.nH₂O where the symbol ‘n’ indicates the number of molecules of water of crystallisation. This value can be calculated using the following method.

Example: If you heated hydrated barium chloride (BaCl₂.nH₂O) in a crucible you might end up with the following results.

Step 1: Calculate the mass of the anhydrous barium chloride (BaCl₂) and the water (H₂O) driven off.

Step 2: Use the empirical formula method to find the value of n in the formula.

Hydr0Gen2023-01-10T08:30:04+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:33, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Crude oil to plastics (D), Topic: Key Calculations|

1:34 (Triple only) understand how to carry out calculations involving amount of substance, volume and concentration (in mol/dm³) of solution

Concentration is a measurement of the amount of substance per unit volume.

In Chemistry, concentration is measured in mol/dm³ (read as moles per cubic decimetre).

The following formula allows for the interconversion between a concentration (in mol/dm³), the amount (in moles) of a substance in a solution, and the volume of the solution (in dm³).

$amount(mol) = concentration(mol/dm^3) \times volume(dm^3)$

or

$concentration(mol/dm^3) = \frac{amount(mol)}{volume(dm^3)}$

Note: 1 dm³ = 1000 cm³

For example, to convert 250 cm³ into dm³:

$250cm^3 = \frac{250}{1000}$

$250cm^3 = 0.25dm^3$

Example 1: 0.03 mol of sodium carbonate (Na₂CO₃) is dissolved in 300 cm³ of water. Calculate the concentration of the solution.

Step 1: Convert the volume of water from cm³ into dm³.

$volume = 300cm^3$

$volume = \frac{300}{1000}dm^3$

$volume = 0.300dm^3$

Step 2: Use the molar concentration formula to calculate of the concentration of the solution.

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.03}{0.300}$

$concentration = 0.1 mol/dm^3$

Example 2: Calculate the amount, in moles, of 25cm³ of hydrochloric acid (HCl) with a concentration of 2 mol/dm³.

Step 1: Convert the volume of HCl from cm³ into dm³.

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

Step 2: Rearrange the molar concentration formula to calculate of the amount, in moles of HCl.

$amount= concentration \times volume$

$amount = 2 \times 0.025$

$amount = 0.050 mol$

Titration calculations, example 1

The titration method that is used to prepare a soluble salt is also used to determine the concentration of an unknown solution.

For example, a titration problem will look like this:

A pupil carried out a titration to find the concentration of a solution of hydrochloric acid (HCl). She found that 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide solution (NaOH) required 23.50 cm³ of dilute hydrochloric acid for neutralisation. Calculate the concentration, in mol/dm³ of the acid.

The chemical equation for this reaction is:

NaOH(aq) + HCl(aq) –> NaCl(aq) + H₂O(l)

volume 25.0cm³ 23.50cm³

concentration 0.100mol/dm³

It can be useful, as shown above, to write the values of the volumes & concentration underneath the equation.

Step 1: Calculate the amount, in moles, of the solution that you know the values for both volume and concentration. In this case sodium hydroxide (NaOH).

First convert the volume of NaOH from cm³ into dm³.

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

Then rearrange the molar concentration formula to calculate of the amount, in moles of NaOH.

$amount= concentration \times volume$

$amount = 0.100 \times 0.025$

$amount = 0.0025 mol$

Step 2: Deduce the amount, in moles of the solution with the unknown concentration. In this case hydrochloric acid (HCl).

From the chemical equation the ratio of NaOH to HCl is 1:1

Therefore if you have 0.0025 mol of NaOH, this will react with 0.0025 mol of HCl.

Step 3: Calculate the concentration of the hydrochloric acid (HCl).

Convert the volume of HCl from cm³ into dm³.

$volume = 23.5cm^3$

$volume = \frac{23.5}{1000}dm^3$

$volume = 0.0235dm^3$

Use the molar concentration formula to calculate of the concentration of the HCl.

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.0025}{0.0235}$

$concentration = 0.106 mol/dm^3$

Titration calculations, example 2

25.0 cm³ of sodium carbonate solution (Na₂CO₃) of unknown concentration was neutralised by 30.0 cm³ of 0.100 mol/dm³ nitric acid (HNO₃).

Na₂CO₃(aq) + 2HNO₃(aq) –> 2NaNO₃(aq) + CO₂(g) + H₂O(l)

Calculate the concentration, in mol/dm³ of the sodium carbonate solution.

Step 1: Calculate the amount, in moles, of nitric acid (HNO₃).

$volume = 30cm^3$

$volume = \frac{30}{1000}dm^3$

$volume = 0.030dm^3$

$amount= concentration \times volume$

$amount = 0.100 \times 0.030$

$amount = 0.0030 mol$

Step 2: Deduce the amount, in moles of sodium carbonate (Na₂CO₃).

Using the equation:

Na₂CO₃(aq) + 2HNO₃(aq) –> 2NaNO₃(aq) + CO₂(g) + H₂O(l)

Ratio Na₂CO₃:HNO₃ = 1:2

$amount=\frac{0.0030}{2} = 0.0015mol$

Step 3: Calculate the concentration of sodium carbonate (Na₂CO₃).

$volume = 25cm^3$

$volume = \frac{25}{1000}dm^3$

$volume = 0.025dm^3$

$concentration = \frac{amount}{volume}$

$concentration = \frac{0.0015}{0.025}$

$concentration = 0.06 mol/dm^3$

Converting mol/dm³ into g/dm³

Concentration can also be expressed in g/dm³ (grams per cubic decimetre).

Therefore mol/dm³ can be converted into g/dm³.

Example: Convert 0.06 mol/dm³ of sodium carbonate (Na₂CO₃) into g/dm³

Step 1: calculate the relative formula mass (M_r) of sodium carbonate (Na₂CO₃).

$M_r= (2 \times 23) + (1 \times 12) (3 \times 16) = 106$

Step 2: Recall the formula giving the relationship between mass, amount and formula mass.

$mass= amount(in\,moles) \times M_r$

therefore

$mass/dm^3= moles/dm^3 \times M_r$

$mass/dm^3= 0.06 \times 106$

$mass/dm^3= 6.36 g/dm^3$

Hydr0Gen2019-05-13T15:41:32+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:34, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Triple|

1:35 (Triple only) understand how to carry out calculations involving gas volumes and the molar volume of a gas (24dm³ and 24,000cm³ at room temperature and pressure (rtp))

The molar volume of a gas is the volume that one mole of any gas will occupy.

1 mole of gas, at room temperature and pressure (rtp), will always occupy 24 dm³ or 24,000 cm³.

Note: 1 dm³ = 1000 cm³

The following formulae allows for the interconversion between a volume in dm³ or cm³ and a number of moles for a given gas:

$Amount(mol) = \frac{volume(dm^3)}{24}$

or

$Amount(mol) = \frac{volume(cm^3)}{24000}$

Example 1:

Calculate the amount, in moles, of 12 dm³ of carbon dioxide (CO₂).

$Amount = \frac{volume(dm^3)}{24}$

$Amount = \frac{12}{24}$

$Amount = 0.5mol$

Example 2:

Calculate the volume at rtp in cubic centimetres (cm³), of 3 mol of oxygen, (O₂).

$Volume = amount \times 24000$

$Volume = 3 \times 24000$

$Volume = 72000cm^3$

Hydr0Gen2019-06-21T14:38:23+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:35, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Key Calculations, Triple|

1:36 practical: know how to determine the formula of a metal oxide by combustion (e.g. magnesium oxide) or by reduction (e.g. copper(II) oxide)

To determine the formula of a metal oxide by combustion.

Example: When magnesium is burned in air, it reacts with oxygen (O₂) to form magnesium oxide (MgO).

2Mg + O₂ –> 2MgO

Method:

Weigh a crucible and lid
Place the magnesium ribbon in the crucible, replace the lid, and reweigh
Calculate the mass of magnesium
Heat the crucible with lid on until the magnesium burns (lid prevents magnesium oxide escaping therefore ensuring accurate results)
Lift the lid from time to time (this allows air to enter)
Stop heating when there is no sign of further reaction
Allow to cool and reweigh
Repeat the heating, cooling and reweigh until two consecutive masses are the same (this ensures all Mg has reacted and therefore the results will be accurate)
Calculate the mass of magnesium oxide formed (mass after heating – mass of crucible)

It is then possible to use this data to calculate the empirical formula of the metal oxide.

To determine the formula of a metal oxide by reduction

Example: When copper (II) oxide is heated in a stream of methane, the oxygen is removed from the copper (II) oxide, producing copper, carbon dioxide and water:

4CuO + CH₄ –> 4Cu + CO₂ + 2H₂O

By comparing the mass of copper produced with the mass of copper oxide used it is possible to determine the formula of the copper oxide.

As an alternative, hydrogen gas could be used instead of methane:

CuO + H₂ –> Cu + H₂O

It is then possible to use this data to calculate the empirical formula of the metal oxide.

There is an important safety point to note with both versions of this experiment. Both methane and hydrogen are explosive if ignited with oxygen. It is important that a good supply of the the gas is allowed to fill the tube before the gas it lit. This flushes out any oxygen from the tube, so the gas will only burn when it exits the tube and comes into contact with oxygen in the air.

Hydr0Gen2019-03-16T10:58:55+00:00Categories: (e) Chemical formulae, equations and calculations, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:36, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Calculations, Topic: Key Calculations, Topic: Specification experiments|

1:37 understand how ions are formed by electron loss or gain

Ions are electrically charged particles formed when atoms lose or gain electrons.

They have the same electronic structures as noble gases.

Metal atoms form positive ions (cations).

Non-metal atoms form negative ions (anions).

Hydr0Gen2022-11-30T14:13:00+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:37, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Ions|

1:38a understand how to use the charges of these ions in ionic formulae: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)

When given this information of the following ions, it is possible to work out the formulae of ionic compounds which include these ions.

Name of Ion	Formula	Charge
Sulfate	SO₄^2-	-2
Carbonate	CO₃^2-	-2
Nitrate	NO₃^-	-1
Hydroxide	OH^-	-1
Ammonium	NH₄⁺	+1
Silver ion	Ag⁺	+1
Zinc ion	Zn²⁺	+2
Hydrogen ion	H⁺	+1
Copper (II) ion	Cu²⁺	+2
Iron (II) ion	Fe²⁺	+2
Iron (III) ion	Fe³⁺	+3
Lead (II) ion	Pb²⁺	+2

Ion charges on the periodic table

Hydr0Gen2022-12-08T10:17:11+00:00Categories: Uncategorized|Tags: 1:38a, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, SubSpecPoint, Topic: Ions, Topic: Key Reactions|

1:38b know the charges of these ions: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)

Name of Ion	Formula	Charge
Sulfate	SO₄^2-	-2
Carbonate	CO₃^2-	-2
Nitrate	NO₃^-	-1
Hydroxide	OH^-	-1
Ammonium	NH₄⁺	+1
Hydrogen ion	H⁺	+1

Ion charges on the periodic table

Sam Mahal2022-12-08T10:46:44+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:38b, GCSE, GCSE_Chemistry_Single, SpecPoint, Topic: Acids & Salts|

1:38 know the charges of these ions: metals in Groups 1, 2 and 3, non-metals in Groups 5, 6 and 7, Ag⁺, Cu²⁺, Fe²⁺, Fe³⁺, Pb²⁺, Zn²⁺, hydrogen (H⁺), hydroxide (OH⁻), ammonium (NH₄⁺), carbonate (CO₃²⁻), nitrate (NO₃⁻), sulfate (SO₄²⁻)

Name of Ion	Formula	Charge
Sulfate	SO₄^2-	-2
Carbonate	CO₃^2-	-2
Nitrate	NO₃^-	-1
Hydroxide	OH^-	-1
Ammonium	NH₄⁺	+1
Silver ion	Ag⁺	+1
Zinc ion	Zn²⁺	+2
Hydrogen ion	H⁺	+1
Copper (II) ion	Cu²⁺	+2
Iron (II) ion	Fe²⁺	+2
Iron (III) ion	Fe³⁺	+3
Lead (II) ion	Pb²⁺	+2

Ion charges on the periodic table

Hydr0Gen2023-01-10T11:37:49+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:38, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Acids & Salts, Topic: Acids & Salts (D)|

1:39 write formulae for compounds formed between the ions listed in 1:38

Writing the electron configuration of an atom allows you to work out the electron configuration of the ion and therefore the charge on the ion.

Examples:

Atom = Mg

Electron configuration = 2,8,2

remove the two electrons from the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)

Ion = Mg²⁺

[2,8]²⁺

Atom = O

Electron configuration = 2,6

add two electrons to the outer shell to achieve the same electron configuration as the nearest noble gas, Neon (Ne 2,8)

Ion = O^2- [2,8]^2-

Hydr0Gen2022-11-30T14:13:29+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:39, Chemistry, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Ions, Topic: Key Reactions|

1:40 draw dot-and-cross diagrams to show the formation of ionic compounds by electron transfer, limited to combinations of elements from Groups 1, 2, 3 and 5, 6, 7 only outer electrons need be shown

Sodium chloride, NaCl

Magnesium chloride, MgCl₂

Potassium oxide, K₂O

Calcium oxide, CaO

Aluminium oxide, Al₂O₃

Magnesium nitride, Mg₃N₂

Hydr0Gen2023-01-10T11:24:32+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:40, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Ions, Topic: Structure & Bonding|

1:41 understand ionic bonding in terms of electrostatic attractions

Ionic bonding: a strong electrostatic attraction between oppositely charged ions.

Hydr0Gen2023-01-10T11:24:42+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:41, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Ions, Topic: Structure & Bonding|

1:42 understand why compounds with giant ionic lattices have high melting and boiling points

Ionic compounds have high melting and boiling points because they have a giant structure with strong electrostatic forces between oppositely charged ions that require a lot of energy to break.

Giant 3D lattice of sodium chloride (NaCl)

Hydr0Gen2023-01-10T11:24:55+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:42, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Ions, Topic: Structure & Bonding|

1:43 Know that ionic compounds do not conduct electricity when solid, but do conduct electricity when molten and in aqueous solution

Ionic compounds do not conduct electricity when solid.

However, ionic compounds do conduct electricity if molten or in solution.

Hydr0Gen2023-01-10T11:25:05+00:00Categories: (f) Ionic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:43, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Electrolysis, Topic: Structure & Bonding|

1:44 know that a covalent bond is formed between atoms by the sharing of a pair of electrons

A covalent bond is formed between two non-metal atoms by sharing a pair of electrons in order to fill the outer shell.

Hydr0Gen2023-01-10T11:26:44+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:44, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:45 understand covalent bonds in terms of electrostatic attractions

Covalent bonding: a strong attraction between a shared pair of electrons and two nuclei.

Hydr0Gen2023-01-10T11:26:54+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:45, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:46 understand how to use dot-and-cross diagrams to represent covalent bonds in: diatomic molecules, including hydrogen, oxygen, nitrogen, halogens and hydrogen halides, inorganic molecules including water, ammonia and carbon dioxide, organic molecules containing up to two carbon atoms, including methane, ethane, ethene and those containing halogen atoms

Hydr0Gen2023-01-10T11:28:29+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:46, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Crude oil to plastics (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:47 explain why substances with a simple molecular structures are gases or liquids, or solids with low melting and boiling points. The term intermolecular forces of attraction can be used to represent all forces between molecules

Carbon dioxide (CO₂) has a simple molecular structure. This just means that it is made up of molecules.

Within each molecule are atoms bonded to each other covalently. These covalent bonds inside the molecules are strong.

However, between the molecules are weak forces of attraction that require little energy to break. These forces are not covalent bonds. This is why simple molecular substances such as carbon dioxide have a low boiling point.

So when carbon dioxide changes from a solid to a gas, for example, that process can be represented as:

CO₂ (s) → CO₂ (g)

Notice that even though there has been a dramatic change of state from solid to gas, the substance before and after the change is always made up of carbon dioxide molecules. During the change of the state the covalent bonds within each molecule remain unbroken. Instead it is the weak forces of attraction between the molecules which have been overcome.

Hydr0Gen2023-01-10T11:29:01+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:47, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil, Topic: Bonding & Structure (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:48 explain why the melting and boiling points of substances with simple molecular structures increase, in general, with increasing relative molecular mass

Larger molecules tend to have higher boiling points.

This is because larger molecules (molecules with more mass) have more forces of attraction between them. These forces, although weak, must be overcome if the substance is to boil, and larger molecules have more attractions which must be overcome.

Hydr0Gen2023-01-10T11:29:19+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:48, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Alkanes & Crude Oil, Topic: Bonding & Structure (D), Topic: Simple Molecules & Covalent Bonding, Topic: Structure & Bonding|

1:49 explain why substances with giant covalent structures are solids with high melting and boiling points

Diamond has a high melting point because it is a giant covalent structure with many strong covalent bonds that require a lot of energy to break.

Hydr0Gen2023-01-10T11:30:25+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:49, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Structure & Bonding|

1:50 explain how the structures of diamond, graphite and C₆₀ fullerene influence their physical properties, including electrical conductivity and hardness

Allotropes are different forms of the same element. Three different allotropes of carbon are shown here as examples: diamond, graphite and C₆₀ fullerene.

Diamond is made up of only carbon atoms, in a giant 3D lattice, where each of those atoms has a strong covalent bonds to 4 other carbon. Every one of carbon’s 4 outer electrons is involved in one of these strong covalent bonds.

Diamond is extremely hard because it is a giant covalent structure with many strong covalent bonds.

Because it is hard, diamond is used in high speed cutting tools, eg diamond-tipped saws.

Graphite is also made of only carbon atoms, and is also a giant structure, but it is formed of layers where each carbon atom has a strong covalent bond to 3 other carbons. This means each carbon atom has one electron not involved in a covalent bond, and these electrons form a sea of delocalised electrons between the layers.

Even though it is a non-metal, graphite can conduct electricity because the delocalised electrons are free to move.

Each layer is a giant structure, with weak forces of attraction between the layers. These layers can easily slide over each other.

Graphite is soft and slippery because it has weak forces of attraction between layers. It is used as a lubricant and in pencils because it is soft and slippery.

C₆₀ fullerene which is a simple molecular structure (also known as a buckyball) is also made of only carbon atoms, but it forms molecules of 60 carbon atoms. The molecule has weak intermolecular forces of attraction between them which take little energy to overcome. Hence C₆₀ fullerene has a low melting point, and it is soft.

C₆₀ fullerene cannot conduct electricity. Although in each molecule every carbon is only covalently bonded to 3 others and the other electrons are delocalised, these electrons cannot jump between different molecules.

Hydr0Gen2023-01-10T11:30:51+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:50, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Structure & Bonding|

1:51 know that covalent compounds do not usually conduct electricity

Electric current is a flow of charged particles that can move.

Covalent compounds do not conduct electricity.

Hydr0Gen2023-01-10T11:31:03+00:00Categories: (g) Covalent bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:51, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Bonding & Structure (D), Topic: Electrolysis, Topic: Structure & Bonding|

1:52 (Triple only) know how to represent a metallic lattice by a 2-D diagram

When metal atoms join together the outer electrons become ‘delocalised’ which means they are free to move throughout the whole structure.

Metals have a giant regular arrangement of layers of positive ions surrounded by a sea of delocalised electrons.

Hydr0Gen2019-06-21T14:39:25+00:00Categories: (h) Metallic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:52, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Structure & Bonding, Triple|

1:53 (Triple only) understand metallic bonding in terms of electrostatic attractions

Metallic bonding is the strong electrostatic attraction between positive metal ions and a sea of delocalised electrons.

Hydr0Gen2019-06-21T14:39:36+00:00Categories: (h) Metallic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:53, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Structure & Bonding, Triple|

1:54 (Triple only) explain typical physical properties of metals, including electrical conductivity and malleability

Metals are good conductors because they have delocalised electrons which are free to move.

Metals are malleable (can be hammered into shape) because they have layers of ions that can slide over each other.

Hydr0Gen2019-06-21T14:39:47+00:00Categories: (h) Metallic bonding, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:54, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Structure & Bonding, Triple|

1:55 (Triple only) understand why covalent compounds do not conduct electricity

Electrical conductivity is the movement of charged particles.

In this case, charged particles means either delocalised electrons or ions.

These particles need to be free to move in a substance for that substance to be conductive.

Covalent compounds do not conduct electricity because there are no charged particles that are free to move.

Hydr0Gen2019-05-13T15:40:12+00:00Categories: (i) Electrolysis, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:55, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Structure & Bonding, Triple|

1:56 (Triple only) understand why ionic compounds conduct electricity only when molten or in aqueous solution

Ionic compounds only conduct electricity only when molten or in solution.

When solid the ions are not free to move.

When molten or in solution the ions are free to move.

Hydr0Gen2019-05-13T15:39:56+00:00Categories: (i) Electrolysis, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:56, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Structure & Bonding, Triple|

1:57 (Triple only) know that anion and cation are terms used to refer to negative and positive ions respectively

A negative ion is called an anion. Examples are the bromide ion (Br⁻) and the oxide ion (O²⁻).

A positive ion is called a cation. Examples are the sodium ion (Na⁺) and the aluminium ion (Al³⁺).

A trick to remember this is to write the ‘t’ as a ‘+’ in the word cation: ca+ion

Hydr0Gen2019-05-13T15:39:45+00:00Categories: (i) Electrolysis, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:57, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Triple|

1:58 (Triple only) describe experiments to investigate electrolysis, using inert electrodes, of molten compounds (including lead(II) bromide) and aqueous solutions (including sodium chloride, dilute sulfuric acid and copper(II) sulfate) and to predict the products

Electrolysis: The breaking down of a substance caused by passing an electric current through an ionic compound which is molten or in solution. New substances are formed.

ELECTROLYSIS OF MOLTEN IONIC COMPOUNDS

Example: The electrolysis of molten lead bromide (PbBr₂)

Solid lead bromide is heated and becomes molten. Explanation: ions become free to move.

Electrodes attached to a power source are placed in the molten lead bromide. Explanation: these electrodes are made of either graphite or platinum because both conduct electricity and are fairly unreactive.

From the diagram, the left-hand electrode becomes positively charged, this is called the anode. The right-hand becomes negatively charged, this is called the cathode. Explanation: delocalised electrons flow from the anode to the cathode.

At the anode a brown gas is given off. This is bromine gas (Br₂(g)). Explanation: Negatively charged bromide ions are attracted to the anode (positive electrode). At the anode, bromide ions lose electrons (oxidation) and become bromine molecules.

At the cathode a shiny substance is formed. This is molten lead (Pb(l) ). Explanation: Positively charged lead ions are attracted to the cathode (negative electrode). At the cathode, lead ions gain electrons (reduction) and become lead atoms.

Overall Reaction

word equation: lead bromide –> lead + bromine

chemical equation: PbBr₂(l) –> Pb(l) + Br₂(g)

Remember

OILRIG : Oxidation Is the Loss of electrons and Reduction Is the Gain of electrons

PANCAKE : Positive Anode, Negative Cathode

ELECTROLYSIS OF IONIC SOLUTIONS

Rules for working out elements formed from electrolysis of solutions

Follow these rules to decide which ions in solution will react at the electrodes:

At the cathode

Metal ions and hydrogen ions are positively charged. Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series:

The metal will be produced if it is less reactive than hydrogen
Hydrogen gas (H₂) will be produced if the metal is more reactive than hydrogen

At the anode

At the anode, the product of electrolysis is always oxygen gas (O₂) unless the solution contains a high concentration of Cl^–, Br^- or I^– ions, in which case a halogen is produced, e.g. chlorine gas (Cl₂), bromine gas (Br₂), and iodine gas (I₂).

The electrolysis of sodium chloride solution (NaCl(aq))

Solid sodium chloride is dissolved in water. Explanation: The sodium ions and chloride ions become free to move.

The solution also contains hydrogen ions (H⁺) and hydroxide ions (OH^–). Explanation: Water is a very weak electrolyte. It ionises very slightly to give hydrogen ions and hydroxide ions:

H₂O(l) ⇋ H⁺(aq) + OH^–(aq)

Chloride ions (Cl^–) and hydroxide ions (OH^–) are attracted to the anode.

Sodium ions (Na⁺) and hydrogen ions (H⁺) are attracted to the cathode.

At the anode a green gas is given off. This is chlorine gas (Cl₂(g)). Explanation: chloride ions lose electrons (oxidation) and form molecules of chlorine. The chloride ions react at the anode instead of the hydroxide ions because the chloride ions are in higher concentration. The amount of chlorine gas produced might be lower than expected because chlorine is slightly soluble in water.

Electron half equation: 2Cl^–(aq) –> Cl₂ (g) + 2e^–

At the cathode a colourless gas is given off. This is hydrogen gas (H₂(g)). Explanation: hydrogen ions gain electrons (reduction) and form molecules of hydrogen. The hydrogen ions react at the cathode because hydrogen is below sodium in the reactivity series.

Electron half equation: 2H⁺(aq) + 2e^– –> H₂ (g)

The solution at the end is sodium hydroxide (NaOH(aq)).

The electrolysis of copper sulfate solution (CuSO₄(aq))

Copper sulfate solution is composed of copper ions (Cu²⁺), sulfate ions (SO₄^2-), hydrogen ions (H⁺) and hydroxide ions (OH^–).

At the cathode a brown layer is formed. This is copper. Explanation: copper ions gain electrons (reduction) and form atoms of copper. The copper ions react at the cathode instead of hydrogen ions because copper is below hydrogen in the reactivity series.

Electron half-equation: Cu²⁺(aq) + 2e^– –> Cu (s)

At the anode, bubbles of gas are given off. This is oxygen gas (O₂(g)). Explanation: hydroxide ions lose electrons (oxidation) and form molecules of oxygen and water. The hydroxide ions react at the anode instead of the sulfate ions because the hydroxide ions are less stable.

Electron half-equation: 4OH^–(aq) –> O₂ (g) + 2H₂O(l) + 4e^–

The electrolysis of sulfuric acid (H₂SO₄(aq))

Sulfuric acid is composed of sulfate ions (SO₄^2-), hydrogen ions (H⁺) and hydroxide ions (OH^–).

At the cathode bubbles of gas are formed. This is hydrogen gas (H₂(g)). Explanation: hydrogen ions gain electrons (reduction) and form molecules of hydrogen.

Electron half-equation: 2H⁺(aq) + 2e^– –> H₂(g)

At the anode, bubbles of gas are given off. This is oxygen gas (O₂(g)). Explanation: hydroxide ions lose electrons (oxidation) and form molecules of oxygen and water. The hydroxide ions react at the anode instead of the sulfate ions because the hydroxide ions are less stable.

Electron half-equation: 4OH^–(aq) –> O₂ (g) + 2H₂O(l) + 4e^–

Twice the volume of hydrogen gas is produce compared to oxygen gas. Explanation: from the two half equations, O₂ needs 4e^–but H₂ only needs 2e^–as can be seen from the equation

2H₂O(l) –> 2H₂(g) + O₂(g)

There are twice the amount (in moles) of H₂ compared to O₂

Hydr0Gen2022-09-16T13:20:34+00:00Categories: (i) Electrolysis, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:58, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Specification experiments, Triple|

1:59 (Triple only) write ionic half-equations representing the reactions at the electrodes during electrolysis and understand why these reactions are classified as oxidation or reduction

Oxidation: the loss of electrons or the gain of oxygen

Reduction: the gain of electrons or the loss of oxygen

Example: The electrolysis of lead (II) bromide, PbBr₂

At the cathode (negative electrode): Pb²⁺(l) + 2e^– → Pb (l) (reduction)

At the anode (positive electrode): 2Br^–(l) → Br₂ (g) + 2e^– (oxidation)

Example: The electrolysis of aluminium oxide, Al₂O₃

At the cathode: Al³⁺ + 3e^– → Al (reduction)

At the anode: 2O^2- → O₂ + 4e^– (oxidation)

Example: The electrolysis of sodium chloride solution (NaCl (aq))

At the cathode: 2H⁺(aq) + 2e^– → H₂ (g) (reduction)

At the anode: 2Cl^–(aq) → Cl₂ (g) + 2e^– (oxidation)

Example: The electrolysis of copper sulfate solution (CuSO₄(aq))

At the cathode: Cu²⁺(aq) + 2e^– → Cu (s) (reduction)

At the anode: 4OH^–(aq) → O₂ (g) + 2H₂O (l) + 4e^– (oxidation)

Hydr0Gen2019-05-13T15:39:26+00:00Categories: (i) Electrolysis, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:59, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Triple|

1:60 (Triple only) practical: investigate the electrolysis of aqueous solutions

The diagram shows an electrolytic cell.

The electrolyte is an aqueous solution. For example it might be concentrated sodium chloride, NaCl (aq).

The test tubes over the electrodes must not completely cover them to make sure the ions are free to move throughout the solution.

In the case of NaCl (aq) bubbles of gas will be seen forming at the electrodes. These float up and collect in the test tubes when each gas can be tested to assess its identity.

Hydr0Gen2019-05-13T15:39:17+00:00Categories: (i) Electrolysis, 1 Principles of chemistry, Edexcel iGCSE Chemistry|Tags: 1:60, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Electrolysis, Topic: Specification experiments, Triple|

2:01 understand how the similarities in the reactions of lithium, sodium and potassium with water provide evidence for their recognition as a family of elements

Group 1 metals such as potassium, sodium and lithium, react with water to produce a metal hydroxide and hydrogen. For example:

lithium + water → lithium hydroxide + hydrogen

2Li (s) + 2H₂O (l) → 2LiOH (aq) + H₂ (g)

The observations for the reaction of water with either potassium or sodium or lithium have the following similarities:

fizzing (hydrogen is produced)
metal floats and moves around on the water
metal disappears

In each case a metal hydroxide solution is produced.

These similarities in the reactions provide evidence that the 3 metals are in the same group of the Periodic Table (i.e. have the same number of electrons in their outer shell).

Hydr0Gen2023-01-10T12:29:10+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:01, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:02 understand how the differences between the reactions of lithium, sodium and potassium with air and water provide evidence for the trend in reactivity in Group 1

Lithium is the first element in group 1 of the Periodic Table. The observations for the reaction of lithium and water are:

fizzing (hydrogen gas is released)
lithium floats and moves around on the water
lithium disappears

Sodium is the second alkali metal in the group. The reaction of sodium and water is more vigorous than lithium’s:

fizzing (hydrogen gas is released)
sodium floats and moves around on the water
sodium melts into a silver-coloured ball
sodium disappears

Potassium is the third alkali metal in the group. The reaction of potassium and water is more vigorous than sodium’s:

fizzing (hydrogen gas is released)
potassium floats and moves around on the water
catches fire with a LILAC flame
potassium disappears

When the group 1 metals react with air they oxidise, showing a similar trend in reactivity as we go down the group of the Periodic Table.

Therefore, as we go down group 1 (increasing atomic number), the elements become more reactive: Li<Na<K<Rb<Cs<Fr

Hydr0Gen2023-01-10T12:29:20+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:02, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:03 use knowledge of trends in Group 1 to predict the properties of other alkali metals

From the data in the table, it is possible to deduce the properties of francium from the trends in the other group 1 metals.

For example, we can predict that francium will have a melting point around 20⁰C and a density of just over 2g/cm³.

We can also predict that francium will react violently with water, producing francium hydroxide and hydrogen.

Alkali metal	Melting point (⁰C)	Density (g/cm³)	Reaction with water	Products
lithium (Li)	181	0.53	fizzing	lithium hydroxide + hydrogen
sodium (Na)	98	0.97	rapid fizzing	sodium hydroxide + hydrogen
potassium (K)	63	0.86	vigorous fizzing and lilac flame	potassium hydroxide + hydrogen
rubidium (Rb)	39	1.53	?	rubidium hydroxide + hydrogen
caesium(Cs)	29	1.88	?	caesium hydroxide + hydrogen
francium (Fr)	?	?	?	?

Hydr0Gen2023-01-10T12:29:32+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:03, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Topic: Metals (D)|

2:04 (Triple only) explain the trend in reactivity in Group 1 in terms of electronic configurations

As you go down the group the outer electron lost from the group 1 metal is further from the nucleus therefore the electron is less attracted by the nucleus and therefore more easily lost.

Hydr0Gen2019-06-21T14:03:07+00:00Categories: (a) Group 1 (alkali metals) – lithium, sodium and potassium, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:04, Chemistry, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Metal Reactivity & Halogens, Triple|

2:05 know the colours, physical states (at room temperature) and trends in physical properties of chlorine, bromine and iodine

Element	Colour	State at room temp
Chlorine (Cl₂)	Green	Gas
Bromine (Br₂)	Red-brown	Liquid
Iodine (l₂)	Grey	Solid

Chlorine is a toxic gas, so should be handled in a fume cupboard.

Hydr0Gen2023-01-10T11:50:40+00:00Categories: (b) Group 7 (halogens) – chlorine, bromine and iodine, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:05, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Halogens (D), Topic: Metal Reactivity & Halogens|

2:06 use knowledge of trends in Group 7 to predict the properties of other halogens

If you look at the trends in the physical properties of the halogens, Cl₂, Br₂, I₂ you can make predictions about the properties of the other halogens.

Element	Colour	State at room temp
Fluorine (F₂)	Yellow	Gas
Astatine (At₂)	Black	Solid

Hydr0Gen2023-01-10T11:51:01+00:00Categories: (b) Group 7 (halogens) – chlorine, bromine and iodine, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:06, Chemistry, Double, GCSE, GCSE_Chemistry_Single, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Halogens (D), Topic: Metal Reactivity & Halogens|

2:07 understand how displacement reactions involving halogens and halides provide evidence for the trend in reactivity in Group 7

Group 7 elements are called the Halogens. As you go up group 7 (decreasing atomic number), the elements become more reactive. For example, fluorine is the most reactive and astatine is the least reactive.

A more reactive halogen will displace a less reactive halogen, e.g. chlorine will displace bromine:

By reacting a halogen solution with a potassium halide solution and making observations, the order of their reactivity can be deduced:

	Potassium chloride, KCl(aq)	Potassium bromide, KBr(aq)	Potassium iodide, KI(aq)
Chlorine, Cl₂(aq)	No change	Colourless to orange	Colourless to brown
Bromine, Br₂(aq)	No change	No change	Colourless to brown
Iodine, I₂(aq)	No change	No change	No change

From the above results, chlorine displaces both bromine and iodine, and bromine displaces iodine. Therefore the order of reactivity is: chlorine is more reactive than bromine, which in turn is more reactive than iodine.

Hydr0Gen2023-01-10T11:51:13+00:00Categories: (b) Group 7 (halogens) – chlorine, bromine and iodine, 2 Inorganic chemistry, Edexcel iGCSE Chemistry|Tags: 2:07, Chemistry, Double, GCSE, GCSE_Chemistry_SpecPoint, SpecPoint, Topic: Halogens (D), Topic: Metal Reactivity & Halogens, Topic: Specification experiments|

2:08 (Triple only) explain the trend in reactivity in Group 7 in terms of electronic configurations

The higher up we go in group 7 (halogens) of the periodic table, the more reactive the element. The explanation concerns how readily these elements form ions, by attracting a passing electron to fill the outer shell.

In fluorine the outer electron shell is very close to the positively charged nucleus, so the attraction between this nucleus and the negatively charged electrons is very strong. This means fluorine is very reactive indeed.

However, for iodine the outer electron shell is much further from the nucleus so the attraction is weaker. This means iodine is less reactive.