Oxidation and Reduction Topic 9 Oxidation and Reduction

IB Core Objective 9.1.1 Define oxidation and reduction in terms of electron loss and gain. Define: Give the precise meaning of a word, phrase or physical quantity.

9.1.1 Define oxidation and reduction in terms of electron loss and gain. Oxidation: The loss of electrons Fe2+(aq) → Fe3+(aq) + e- Reduction: The gain of electrons 2H+(aq) + 2e- → H2(g)

9.1.1 Define oxidation and reduction in terms of electron loss and gain. Helpful Mnemonic This is Leo the Lion LEO goes GER Loss of Electrons is Oxidation Gain of Electrons is Reduction

9.1.1 Define oxidation and reduction in terms of electron loss and gain. Or another if you prefer… OIL RIG Oxidation Is Loss of electrons. Reduction Is Gain of electrons.

IB Core Objective 9.1.2 Deduce the oxidation number of an element in a compound. Deduce: Reach a conclusion from the information given.

9.1.2 Deduce the oxidation number of an element in a compound. In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

9.1.2 Deduce the oxidation number of an element in a compound. A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

9.1.2 Deduce the oxidation number of an element in a compound. A species is reduced when it gains electrons. Here, each of the H+ gains an electron and they combine to form H2.

9.1.2 Deduce the oxidation number of an element in a compound. It may be easier to find what is being reduced and oxidized by splitting the equation into “half equations”. For example, with Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) It can be split up as: Zn(s) → Zn2+(aq) + 2e- and 2H+(aq) + 2e- → H2(g)

9.1.2 Deduce the oxidation number of an element in a compound. It is not always easy to split equations into half equations. Consider the following reaction: Can you tell which is being oxidized? If not, then we need to use oxidation numbers. N2(g) + 3H2(g)  2NH3(g)

9.1.2 Deduce the oxidation number of an element in a compound. Oxidation Number The charge that an atom would have if all covalent bonds were broken so that the more electronegative element kept all the electrons.

9.1.2 Deduce the oxidation number of an element in a compound. Oxidation Number Rules Elements in elemental state = 0 F = -1 (always) O = -2 (except in H2O2 where its +1) H = +1 (except in hydrides H-) Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1) The sum of the oxidation numbers in a neutral compound is 0. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

9.1.2 Deduce the oxidation number of an element in a compound. Elements in elemental state = 0 F = -1 (always) O = -2 (except in H2O2 where its +1) H = +1 (except in hydrides H-) Halides = -1 except when bonded to oxygen or other halides higher in the group (more reactive one will be -1) Find the oxidation number for the following: Nitrogen in N2 = Carbon in CH4 = Sulfur in H2SO4 = Phosphorous in PCl4+ = Iodine in IO4- = Answers: 0, -4, +6, +5, +7

IB Core Objective 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. Deduce: Reach a conclusion from the information given. (Obj 3)

9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. Let’s go back to the equation: What is the oxidation number for nitrogen on both sides? Has it been oxidized or reduced? Answer: Oxidation number goes from 0 to -3. It has gained electrons, therefore it has been reduced. N2(g) + 3H2(g)  2NH3(g)

MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq) 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. Consider the reaction between MnO4− and C2O42− : MnO4−(aq) + C2O42−(aq)  Mn2+(aq) + CO2(aq)

MnO4− + C2O42-  Mn2+ + CO2 First, assign oxidation numbers. 9.1.4 Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers. +7 +3 +4 +2 MnO4− + C2O42-  Mn2+ + CO2 First, assign oxidation numbers. Next, find out if carbon and manganese are being oxidized or reduced. Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

IB Core Objective 9.1.3 State the names of compounds using oxidation numbers. State: Give a specific name, value or other brief answer without explanation or calculation. (Obj 1)

9.1.3 State the names of compounds using oxidation numbers. For elements that have a variable oxidation number, the oxidation state is signified by Roman numerals. Example: Fe+3 would be written as Iron(III) How would you write the following? FeCl2 FeCl3 MnO4- Cr2O3 Answers: iron(II) chloride, iron(III) chloride, permanganate (VII), chromium(III) oxide Challenge: How would you write the formula for ammonium dichromate? Answer: (NH4)2Cr2O7

Ammonium dichromate volcano (NH4)2Cr2O7 --> Cr2O3 + 4 H2O + N2 Is chromium oxidized or reduced in this reaction? Is nitrogen oxidized or reduced in this reaction? Answer: Chromium is reduced from +6 to +3 Nitrogen is oxidized from +3 to 0

IB Core Objective 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Deduce: Reach a conclusion from the information given.

9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Let’s look at an equation that we worked with before…. What is wrong with this equation? Answer: It is not balanced! We have worked with half equations before (zinc and hydrogen). Now we’ll dig deeper. MnO4− + C2O42-  Mn2+ + CO2

General rules for balancing half equations 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. General rules for balancing half equations 1) Balance atoms being oxidized or reduced 2) Add H20 to balance Oxygen atoms 3) Add H+(aq) to balance Hydrogen atoms 4) Add e- to balance charge

Oxidation Half-Reaction 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Oxidation Half-Reaction C2O42−  CO2 To balance the carbon, we add a coefficient of 2: C2O42−  2 CO2

Oxidation Half-Reaction 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Oxidation Half-Reaction C2O42−  2 CO2 The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side. C2O42−  2 CO2 + 2 e−

Reduction Half-Reaction 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Reduction Half-Reaction MnO4−  Mn2+ The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side. MnO4−  Mn2+ + 4 H2O

Reduction Half-Reaction 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Reduction Half-Reaction MnO4−  Mn2+ + 4 H2O To balance the hydrogen, we add 8 H+ to the left side. 8 H+ + MnO4−  Mn2+ + 4 H2O

Reduction Half-Reaction 9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. Reduction Half-Reaction 8 H+ + MnO4−  Mn2+ + 4 H2O To balance the charge, we add 5 e− to the left side. 5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O

IB Core Objective 9.2.2 Deduce redox equations using half-equations. Deduce: Reach a conclusion from the information given.

Combining the Half-Reactions 9.2.2 Deduce redox equations using half-equations. Combining the Half-Reactions Now we evaluate the two half-reactions together: C2O42−  2 CO2 + 2 e− 5 e− + 8 H+ + MnO4−  Mn2+ + 4 H2O To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

Combining the Half-Reactions 5 C2O42−  10 CO2 + 10 e− 9.2.2 Deduce redox equations using half-equations. Combining the Half-Reactions 5 C2O42−  10 CO2 + 10 e− 10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O When we add these together, we get: 10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2 +10 e−

Combining the Half-Reactions 10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  9.2.2 Deduce redox equations using half-equations. Combining the Half-Reactions 10 e− + 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2 +10 e− The only thing that appears on both sides are the electrons. Subtracting them, we are left with: 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2

9.2.1 Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction. 9.2.2 Deduce redox equations using half-equations. Practice Given two half-equations: Cr2O72-(aq) → Cr3+(aq) Fe2+ → Fe3+ Deduce the half-equations for each, then deduce the redox equation.

Answer Cr2O72-(aq) + 14H+(aq) + 6Fe2+(aq) → 2Cr3+(aq) + 7H2O(l) + 6Fe3+(aq)

IB Core Objective 9.2.3 Define the terms oxidizing agent and reducing agent. Define: Give the precise meaning of a word, phrase or physical quantity. (Obj 1)

9.2.3 Define the terms oxidizing agent and reducing agent. Oxidizing agent: Substance that is reduced and causes the oxidation of another substance in a redox reaction. Reducing agent: Substance that is oxidized and causes the reduction of another substance in a redox reaction. I am oxidizing agent man. I am here to take your electrons.

IB Core Objective 9.2.4 Identify the oxidizing and reducing agents in redox equations. Identify: Find an answer from a given number of possibilities. (Obj 2)

9.2.4 Identify the oxidizing and reducing agents in redox equations. Identify the oxidizing and reducing agents in the following equations: Sn2+(aq) + 2Fe3+(aq) → Sn4+(aq) Fe2+(aq) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

9.2.4 Identify the oxidizing and reducing agents in redox equations. Deduce the following half equations, deduce the redox equation, and identify the oxidizing agent and the reducing agent. MnO4-(aq) → Mn-2(aq) SO2(aq) → SO42-(aq)

IB Core Objective 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. Deduce: Reach a conclusion from the information given.

Strong oxidizing agent 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. Recall in acids and bases that a strong acid had a weak conjugate base. Same in redox reactions. The conjugate of a powerful oxidizing agent is a weak reducing agent. F2 + 2e- ↔ 2F- Weak reducing agent Strong oxidizing agent

Mr. F can really attract the electrons (more electronegative). 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. Mr. F can really attract the electrons (more electronegative). When Mr. F has the electrons, he doesn’t want to let them go. So although he is a good oxidizing agent, he is a poor reducing agent. (He doesn’t like to reduce the number of his electrons!)

Think back to Topic 3 on Periodicity. 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. Think back to Topic 3 on Periodicity. What are the trends in electronegativity?

Compare What exception do you see? 9.3.1 Deduce a reactivity series based upon the chemical behaviour of a group of oxidizing and reducing agents. Compare What exception do you see? Hydrogen (Lithium is another exception)

IB Core Objective 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. Deduce: Reach a conclusion from the information given.

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq) Feasible? A: Yes 9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq) Feasible? A: Yes I2(aq) + 2Cl-(aq) → Cl2(aq) + 2I-(aq) A: No Chlorine attracts electrons more strongly than iodine, so chlorine is a better oxidizing agent.

9.3.2 Deduce the feasibility of a redox reaction from a given reactivity series. Zn(s) + Cu2+(aq) → Cu(s) + Zn2+(aq) Feasible? A: Yes Cu(s) + Zn2+(aq) → Zn(s) + Cu2+(aq) A: No These examples are all displacement reactions, because they involve a more reactive metal or non-metal displacing the reactive one from its salt.

IB Core Objective 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. Explain: Give a detailed account of causes, reasons or mechanisms.

9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. A Voltaic cell is a device for converting chemical energy into electrical energy using a redox reaction.

Anode(-): Oxidation, forms a negative charge 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. Anode(-): Oxidation, forms a negative charge Cathode(+): Reduction, forms a positive charge e- e- e- e- 2+ e- e- e- 2+ e- 2+ 2+

Balance (-) charge by replacing it with some more negative ions 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. Lets harness some Energy!! Problem, the highly negative charge on electrode causes (+) ions to be attracted back Zn(s) Cu(s) Solution Balance (-) charge by replacing it with some more negative ions 2+ 2+ e- e- e- 2+ 2+ 2+ e- 2+ 2+ 2+ 2+ 2+

Lets harness some Energy!! 9.4.1 Explain how a redox reaction is used to produce electricity in a Voltaic cell. Lets harness some Energy!! 2+ Zn(s) + + + - - Cu(s) - - + + - 2+ 2+ e- e- e- 2+ 2+ http://www.dynamicscience.com.au/tester/solutions/chemistry/redox/galvan5.swf 2+ e- 2+ 2+ 2+ 2+ 2+

IB Core Objective 9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). State: Give a specific name, value or other brief answer without explanation or calculation.

9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode. Which of the metals is being reduced? So which is the cathode?

Lead and zinc are set up in a voltaic cell. 9.4.2 State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode). Lead and zinc are set up in a voltaic cell. Which one would be oxidized? Which one is being reduced? A: Zinc is being oxidized. Lead is being reduced. Which one would be the cathode and which would be the anode? Zinc would be the anode, lead is the cathode.

IB Core Objective 9.5.1 Describe, using a diagram, the essential components of an electrolytic cell. Describe: Give a detailed account.

Draw a diagram of an electrolytic cell. 9.5.1 Describe, using a diagram, the essential components of an electrolytic cell. Homework: Draw a diagram of an electrolytic cell. Provide a brief description what is happening at each step, including the components, where oxidation and reduction is occurring, how current is conducted, and the products of a molten salt. If you do this effectively, you will have down objectives 9.5.1 – 9.5.4

9.5.1 Describe, using a diagram, the essential components of an electrolytic cell. Need to have a liquid containing ions, which is called an electrolyte.

IB Core Objective 9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode). State: Give a specific name, value or other brief answer without explanation or calculation.

The anode attracts anions. 9.5.2 State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode). The anode attracts anions. When the anions reach they anode, they lose electrons. So are they oxidized or reduced? A: oxidized When cations reach the cathode they gain electrons and they are reduced.

IB Core Objective 9.5.3 Describe how current is conducted in an electrolytic cell. Describe: Give a detailed account.

9.5.3 Describe how current is conducted in an electrolytic cell. Electricity is supplied from an external source and used to create a non-spontaneous reaction. Electrolyte solution can conduct electricity because the ions move towards oppositely charged electrodes.

IB Core Objective 9.5.4 Deduce the products of the electrolysis of a molten salt. Deduce: Reach a conclusion from the information given.

9.5.4 Deduce the products of the electrolysis of a molten salt. Sodium chloride Negative chloride ions are attracted to the positive ions. There they lose electrons and are oxidized to chlorine gas: 2Cl-(l) → Cl2(g) + 2e- Positive sodium ions are attracted to the negative cathode. They gain electrons and are reduced to sodium metal: Na+(l) + e- → Na(l)

9.5.4 Deduce the products of the electrolysis of a molten salt. Question For every 2 mol of electrons that flow through the circuit, how many mol of chlorine gas and sodium metal will be produced? A: 1 mol of chlorine gas and 2 mol of sodium.

IB HL Objective 19.1.1 Describe the standard hydrogen electrode. Describe: Give a detailed account. (Obj 2)

19.1.1 Describe the standard hydrogen electrode. Electrode: An electrical conductor through which electric current leaves or enters Anode: Negative electrode where oxidation takes place. Cathode: Positive electrode where reduction takes place.

19.1.1 Describe the standard hydrogen electrode. The potential of any two electrodes can be compared using this apparatus You will learn more about this voltaic cell later (SL topic).

19.1.1 Describe the standard hydrogen electrode. Electrode potentials from the voltaic cell are measured relative to the standard hydrogen electrode (SHE).

19.1.1 Describe the standard hydrogen electrode. The reference half-reaction is the reduction of H+(aq) to H2(g): 2H+(aq) + 2e- ↔ H2(g)

IB HL Objective 19.1.2 Define the term standard electrode potential (Eѳ). Define: Give the precise meaning of a word, phrase or physical quantity. (Obj 1)

Electromotive Force (emf) Water only spontaneously flows one way in a waterfall. Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy.

Electromotive Force (emf) 19.1.2 Define the term standard electrode potential (Eѳ). Electromotive Force (emf) The potential difference between the anode and cathode in a cell is called the electromotive force (emf). It is also called the cell potential, and is designated Ecell.

19.1.2 Define the term standard electrode potential (Eѳ). The standard hydrogen electrode (SHE) is defined as having a potential of zero. Standard electrode potentials also refer to the conditions. In the SHE, the platinum electrode is surrounded by H2 gas at 1 atm (1.01 x 105 Pa), electrode is immersed in strong acid at 1.00 mol dm-3, and is kept at 298 K. in a standard hydrogen electrode is equal to 0 V. is the standard reduction potential.

19.1.2 Define the term standard electrode potential (Eѳ). The electrode potential at standard conditions can be found through this equation: Ecell  = Ered (cathode) − Ered (anode) There is also another way….If the half-equation is being oxidized instead of reversed, just flip the values . Example: K ↔ K+ +e- Eѳ = +2.92 V. Then add the two values from the two half-reactions together!

IB HL Objective 19.1.3 Calculate cell potentials using standard electrode potentials. Calculate: Find a numerical answer showing the relevant stages in the working (unless instructed not to do so). (Obj 2)

19.1.3 Calculate cell potentials using standard electrode potentials. Reduction potentials for many electrodes have been measured and tabulated.

19.1.3 Calculate cell potentials using standard electrode potentials. For the oxidation in this cell, For the reduction, Ered = −0.76 V  Ered = +0.34 V 

19.1.3 Calculate cell potentials using standard electrode potentials.  = Ered (cathode) − (anode) = +0.34 V − (−0.76 V) = +1.10 V

IB HL Objective 19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values. Predict: Give an expected result. (Obj 3)

19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values. Standard electrode potentials allow predictions to be made about which reactions could theoretically occur. If the cell potential is negative, a spontaneous reaction cannot occur. If the cell potential is positive, then the reaction could occur spontaneously.

19.1.4 Predict whether a reaction will be spontaneous using standard electrode potential values. Can copper metal reduce hydrogen ions to hydrogen gas? Cu ↔ Cu2+ + 2e- Eѳ = -0.34 V 2H+ + 2e- ↔ H2 Eѳ = 0.00 V Eѳcell = -0.34 V Answer= No

IB HL Objective 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. Predict: Give an expected result. Explain: Give a detailed account of causes, reasons or mechanisms.

19.2.1 Predict and explain the products of electrolysis of aqueous solutions. Question How would you know if a solution has electrolytes in it? A: It conducts electricity. Electrolytes are easy to determine the products, since there is one positive ion and one negative ion. So sodium chloride may dissociate in solution. Anything else?

H2O(l) ↔ H+(aq) + OH-(aq) 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. Water is a poor conductor of electricity, but it still dissociates into ions: H2O(l) ↔ H+(aq) + OH-(aq) In order to electrolyze water, need to add something more to conduct the current that easily produces ions, but won’t be oxidized/reduced. So a small amount of sulfuric acid is added.

4OH-(aq) → O2(g) + 2H2O(l) + 4e- 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. If water is electrolyzed, what would the products be? Half-equations for each ion? A: for the hydrogen ions: 2H+(aq) + 2e- → H2(g) for the hydroxide ions: 4OH-(aq) → O2(g) + 2H2O(l) + 4e-

You will also need to know aqueous sodium chloride for this objective. 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. You will also need to know aqueous sodium chloride for this objective. We will be going over this more with the SL students, since they will need to know this as well.

A: CuSO4(aq) → Cu2+(aq) + SO42-(aq) H2O(l) ↔ H+(aq) + OH-(aq) 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. What if we were to electrolyze copper(II) sulfate solution? What are equations for ions being formed in solution? A: CuSO4(aq) → Cu2+(aq) + SO42-(aq) H2O(l) ↔ H+(aq) + OH-(aq) Inert (non-reactive) platinum or graphite electrodes can be used.

(-) electrode: Cu2+(aq) + 2e- → Cu(s) 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. Based on the dissociation equations, which would be attracted to the positive electrode? Which would be attracted to the negative electrode? What are the half-reactions for these? Be sure to take into account Eѳ potentials. (-) electrode: Cu2+(aq) + 2e- → Cu(s) (+) electrode: 4OH-(aq) → 2H2O(l) + O2(g) + 4e- Because copper is below hydrogen in the reactivity series, it will gain electrons instead of the hydrogen. Because the sulfate would have a more positive value, it would keep its electrons over the hydroxide.

(-) electrode: Cu2+(aq) + 2e- → Cu(s) 19.2.1 Predict and explain the products of electrolysis of aqueous solutions. What if a copper electrode is used? What would the half equations be at each electrode? (-) electrode: Cu2+(aq) + 2e- → Cu(s) (+) electrode: Cu(s) → Cu2+(aq) + 2e-

IB HL Objective 19.2.2 Determine the relative amounts of the products formed during electrolysis. Determine: Find the only possible answer.

4OH-(aq) → O2(g) + 2H2O(l) + 4e- 19.2.2 Determine the relative amounts of the products formed during electrolysis. In the electrolysis of water, what would the mol ratio be for products? For oxygen gas: 4OH-(aq) → O2(g) + 2H2O(l) + 4e- We need four mol of electrons to produce one mol of oxygen. For hydrogen gas: 2H+(aq) + 2e- → H2(g) Four mol of electrons would produce two mol of hydrogen gas. Therefore the ratio would be 2 mol H2: 1 mol O2

19.2.2 Determine the relative amounts of the products formed during electrolysis. What factors will influence the amount of products produced and the rate? The magnitude of current (increasing the flow of electrons). The stronger the current, the faster the reaction will take place. Time: More time that current is allowed to pass, more products will be formed. Charge on the ions: Look at the half-equations, and you can determine the mol ratios.

IB HL Objective 19.2.3 Describe the use of electrolysis in electroplating. Describe: Give a detailed account.

19.2.3 Describe the use of electrolysis in electroplating. Electroplating: Cathode becomes coated in a layer of metal (negative electrode). Cathode is where reduction takes place.

19.2.3 Describe the use of electrolysis in electroplating. Electroplating can be used to purify substances. For example, copper is used for electrical wiring, and it needs to be pure otherwise the resistance increases. So the positive electrode is impure copper and the negative electrode becomes pure copper.