Topic: Electrolysis

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.

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.

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.

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

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

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 (SO42-), hydrogen ions (H+) and hydroxide ions (OH). 
  • At the cathode a shiny orange 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 (SO42-), 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 4ebut 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

 

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).

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:      2O2-    →    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 (CuSO(aq))

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

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

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.

 

 

2:16 understand how metals can be arranged in a reactivity series based on their displacement reactions between: metals and metal oxides, metals and aqueous solutions of metal salts

A metal will displace another metal from its oxide that is lower in the reactivity series. For example, a reaction with magnesium and copper (II) oxide will result in the magnesium displacing the copper from its oxide:

A metal will also displace another metal from its salt that is lower in the reactivity series. For example, the reaction between zinc and copper (II) sulfate solution will result in zinc displacing the copper from its salt:

The blue colour of the copper (II) sulfate solution fades as colourless zinc sulfate solution is formed.

2:20 in terms of gain or loss of oxygen and loss or gain of electrons, understand the terms: oxidation, reduction, redox, oxidising agent, reducing agent, in terms of gain or loss of oxygen and loss or gain of electrons

Oxidation

  • Oxidation is the loss of electrons. For example a sodium atom (Na) loses an electron to become a sodium ion (Na⁺). Another example is a chloride ion (Cl⁻) losing an electron to become a chlorine atom (Cl).
  • Another definition of oxidation is the gain of oxygen. For example if carbon combines with oxygen to form carbon dioxide, the carbon is being oxidised.

 

Reduction

  • Reduction is the gain of electrons. For example a sodium ion (Na⁺) gains an electron to become a sodium ion (Na). Another example is a chlorine atom (Cl) gaining an electron to become a chloride atom (Cl⁻).
  • Another definition of reduction is the loss of oxygen. For example when aluminium oxide is broken down to produce aluminium and oxygen, the aluminium is being reduced.

 

Redox: A reaction involving oxidation and reduction.

A good way to remember the definitions of oxidation and reduction in terms of electrons is:

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

 

Oxidising agent: A substance that gives oxygen or removes electrons (it is itself reduced).

 

Reducing agent: A substance that takes oxygen or gives electrons (it is itself oxidised).

 

2:22 (Triple only) know that most metals are extracted from ores found in the Earth’s crust and that unreactive metals are often found as the uncombined element

Most metals are found in the Earth’s crust combined with other elements. Such compounds are found in rocks called ore, rocks from which it is worthwhile to extract a metal.

A few very unreactive metals, such as gold, are found native which means they are found in the Earth’s crust as the uncombined element.

 

2:23 (Triple only) explain how the method of extraction of a metal is related to its position in the reactivity series, illustrated by carbon extraction for iron and electrolysis for aluminium

Extraction of a metal from its ore typically involves removing oxygen from metal oxides.

 

If the ore contains a metal which is below carbon in the reactivity series then the metal is extracted by reaction with carbon in a displacement reaction.

 

If the ore contains a metal which is above carbon in the reactivity series then electrolysis (or reaction with a more reactive metal) is used to extract the metal.

2:24 (Triple only) be able to comment on a metal extraction process, given appropriate information

Extraction of a metal from its ore typically involves removing oxygen from metal oxides.

 

If the ore contains a metal which is below carbon in the reactivity series then the metal is extracted by reaction with carbon in a displacement reaction.

 

If the ore contains a metal which is above carbon in the reactivity series then electrolysis (or reaction with a more reactive metal) is used to extract the metal.

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Section 1: Principles of chemistry

      a) States of matter

      b) Atoms

      c) Atomic structure

     d) Relative formula masses and molar volumes of gases

     e) Chemical formulae and chemical equations

     f) Ionic compounds

     g) Covalent substances

     h) Metallic crystals

     i) Electrolysis

 Section 2: Chemistry of the elements

     a) The Periodic Table

     b) Group 1 elements: lithium, sodium and potassium

     c) Group 7 elements: chlorine, bromine and iodine

     d) Oxygen and oxides

     e) Hydrogen and water

     f) Reactivity series

     g) Tests for ions and gases

Section 3: Organic chemistry

     a) Introduction

     b) Alkanes

     c) Alkenes

     d) Ethanol

Section 4: Physical chemistry

     a) Acids, alkalis and salts

     b) Energetics

     c) Rates of reaction

     d) Equilibria

Section 5: Chemistry in industry

     a) Extraction and uses of metals

     b) Crude oil

     c) Synthetic polymers

     d) The industrial manufacture of chemicals