Oxidation & Reduction

1. 12 Chemistry Oxidation & Reduction

2. What is oxidation/reduction? Oxidation/Reduction reactions, collectively known as redox reactions, involve the transfer of electrons and always occur together An easy way to remember this is the mnemonic: ‘’OIL RIG’’ Oxidation Is Loss Reduction Is Gain Oxidation is the loss of electrons Reduction is the gain of electrons An example: CuO(aq) + Mg(s)  Cu + MgO(aq) As an ionic equation: Cu2+ + O2- + Mg  Cu + Mg2+ + O2- So if we look at the transfer of e- in the ‘net ionic equation’ Cu has gained e- Cu2+ + Mg  Cu + Mg2+ Mg has lost e-

3. Agents of Redox A substance that is oxidised loses electrons to another substance, so we call this substance a reducing agent A substance that is reduced, gains electrons from another substance, so we call this substance a oxidising agent

4. Exercise Considering the reaction between zinc and hydrochloric acid, write out: A balanced equation An ionic equation A net ionic equation (excludes the spectator ions) What are the oxidising and reducing agents

5. Exercise - Answers Considering the reaction between zinc and hydrochloric acid, write out: A balanced equation Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) An ionic equation Zn(s) + 2H+(aq) + 2Cl-(aq) Zn2+(aq) + 2Cl-(aq) + H2(g) A net ionic equation (excludes the spectator ions) Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Two half reactions to identify the oxidation and reduction Zn(s) Zn2+(aq) + 2e- (loss of e-) - oxidation 2H+(aq) + 2e-  H2(g) (gain of e-) - reduction 5. What are the oxidising and reducing agents r.a. – Zinc o.a. - Hydrogen

6. How do we know it’s redox? In every redox reaction there is an exchange of electrons, but since these transfers are not shown in overall reaction equations, we need a way to ‘see’ who is losing and who is gaining electrons. Oxidation States (aka Numbers) A set of rules applied to elements, ions and compounds will allow us to determine if we have a redox reaction or not. Oxidation – increase in number Reduction – decrease in number

7. How do we know it’s redox? Oxidation Number Rules • The oxidation state of a free element (i.e. not part of a compound) is zero (e.g. Zn(s), O2(g)) • The oxidation state of an element in an ionic compound is equal the electrical charge on its ion if monatomic. (e.g. Na+ = +1) • Oxidation states of elements in covalent compounds are calculated as if they were ionic. The most electronegative atom (closest to F in the periodic table) is assumed to gain electrons. (e.g. NH3; N = -3, H = +1) • The oxidation state of oxygen in a compound is normally -2, except for peroxides, when it is -1. (e.g. H2O2; H = 1, O = -1) • The oxidation state of hydrogen in a compound is normally +1, except for metal hydrides, when it is -1. (e.g. NaH; Na = 1, H = -1) • The sum of the oxidation states of all the elements in a neutral compound = zero. • The sum of the oxidation numbers of all atoms in a polyatomic ion = charge on the polyatomic ion.

8. Assigning Oxidation Numbers • When assigning oxidation numbers to the elements in a substance, take a systematic approach. Ask yourself the following questions: • Is the substance elemental? • Is the substance ionic? • If the substance is ionic, are there any monoatomic ions present? • Which elements have specific rules? • Which element(s) do(es) not have rules? • Use rule 6 or 7 from above to calculate these.

9. Assigning Oxidation Numbers • Example - Na2SO4 • Is the substance elemental? No, 3 elements are present • Is the substance ionic? Yes • If the substance is ionic, are there any monoatomic ions present? Yes, Na+, so OS of Na = +1 • Which elements have specific rules? Oxygen has a rule (#4) OS of O = -2 • Which element(s) do(es) not have rules? Sulfur does not have a rule • Use rule 6 or 7 from above to calculate these. • 2(Na) + 4(O) + (S) = 0  • (+2) + (-8) + S = 0  • S = +6

10. Assigning Oxidation Numbers • Example - K2C2O4 • Is the substance elemental? No, 3 elements are present • Is the substance ionic? Yes , metal + non-metal • If the substance is ionic, are there any monoatomic ions present? Yes, K+, so OS of K = +1 • Which elements have specific rules? Oxygen has a rule (#4) OS of O = -2 • Which element(s) do(es) not have rules? Carbon does not have a rule • Use rule 6 or 7 from above to calculate these. • 2(K) + 4(O) + (C) = 0  • (+2) + (-8) + 2(C) = 0  • C = +3

11. Exercises Determine the oxidation number of each element in the following compounds: Ba(NO3)2 NF3 (NH4)2SO4

12. Exercises Answers • 1. Ba(NO3)2 • Is the substance elemental? • No, three elements are present. • Is the substance ionic? • Yes, metal + non-metal. • Are there any monoatomic ions? • Yes, barium ion is monoatomic. • Barium ion = Ba2+ • Oxidation # for Ba = +2 • Which elements have specific rules? • Oxygen has a rule....-2 in most compounds • Oxidation # for O = -2 • Which element does not have a specific rule? • N does not have a specific rule. • Use rule 6 to find the oxidation # of N • Let N = Oxidation # for nitrogen • (# Ba) (Oxid. # of Ba) + (# N) (Oxid. # N) + (# of O) (Oxid. # of O) = 0 • 1(+2) + 2(N) + 6(-2) = 0 • N = +5

13. Exercises Answers • 2. NF3 • Is the substance elemental? • No, two elements are present. • Is the substance ionic? • No, two non-metals. • Are there any monoatomic ions? • Since it is molecular, there are no ions present. • Which elements have specific rules? • F = -1 • Which element does not have a specific rule? • N does not have a specific rule. • Use rule 6 to find the oxidation # of N • Let N = oxidation # of N • (# N) (Oxid. # N) + (# F) (Oxid. # F) = 0 • 1(N) + 3(-1) = 0 • N = +3

14. Exercises Answers • 3. (NH4)2SO4 • Is the substance elemental? • No, four elements are present. • Is the substance ionic? • Yes, even though there are no metals present, the ammonium ion is a common polyatomic cation. • Are there any monoatomic ions? • No, the cation and anion are both polyatomic. • Which elements have specific rules? • H = +1 because it is attached to a non-metal (N) • O = -2 • Which elements do not have a specific rule? • Neither N nor S has a specific rule. • You must break the compound into the individual ions that are present and then use rule 7 to find the oxidation numbers of N and S. Notice that if you try to use rule 6, you end up with one equation with two unknowns: 2N + 8(+1) + 1S + 4(-2) = 0 • The two ions present are NH4+ and SO42-. • N + 4(+1) = +1 so N = -3 • S + 4(-2) = -2 so S = +6

15. Naming compounds Recall, that many elements have multiple valencies, so they are able to form a variety of compounds. We name these compounds using their oxidation numbers. Iron has two common oxidation states: +2 and +3 This means that iron can form two different compounds with oxygen which has an oxidation state = -2 FeO – we call this iron(II) oxide (say, “iron 2 oxide”) Fe2O3 – we call this iron(III) oxide (say, “iron 3 oxide”) Fe2+ Fe3+

16. Using oxidation numbers to identify redox in a reaction Not all reactions are redox reactions • To determine if a redox reaction has occurred, assign oxidation numbers to each element on both sides of the reaction and if: • Oxidation # increases – that element has been oxidised • Oxidation # decreases – that element has been reduced • Oxidation # does not change – no oxidation has occured Example: CuO + H2 Cu + H20 Cu: +2  0 (reduced) O: -2  -2 (no change) H: 0  +1 (oxidised) Redox has occurred You try: 2CrO42- + 2H+ Cr2O72- + H20 Cr: +6  +6 O: -2  -2 H: +1  +1 No changes means no redox

17. Redox half-reactions An ionic equation can be written as two half equations that show each process. Cu2+ + Mg  Cu + Mg2+ Oxidation Half-reaction Mg  Mg2+ + 2e- Reduction Half-reaction Cu2+ + 2e-  Cu Note: the number of atoms and charge must balance in half-reactions just as in full reactions

18. More complex half-reactions With simple monatomic ions, half-reactions are simple as in the previous example. However, with polyatomic ions, the process is a bit more complex. Some require acidic conditions to proceed Consider the reduction of dichromate (VI) ion to chromium (III) Cr2O72- Cr3+

19. More complex half-reactions Now you try… Consider the oxidation of sulfur dioxide to sulfate SO2 SO42-

20. More complex half-reactions Answer Consider the oxidation of sulfur dioxide to sulfate SO2 SO42-

21. Adding redox half-reactions Consider the two half reactions from the previous example in acidic solution: SO2(aq) + 2H2O(l) SO42-(aq)+ 4H+(aq)+ 2e- (oxidation) Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq)+7H2O(l)(reduction) We can add the two half reactions to get the net ionic reaction, but notice that adding these two half reactions results in an imbalance in the number of electrons. In this case, we must multiply the first by 3 to get the electrons to cancel. In addition some of the waters and H ions also cancel. 3SO2(aq) + Cr2O72-(aq)+ 2H+(aq)3SO42-(aq)+ 2Cr3+(aq)+H2O(l) What is the oxidising agent? Reducing agent?

22. Adding redox half-reactions Consider the two half reactions from the previous example in acidic solution: SO2(aq) + 2H2O(l) SO42-(aq)+ 4H+(aq)+ 2e- (oxidation) Cr2O72-(aq) + 14H+(aq) + 6e-  2Cr3+(aq)+7H2O(l)(reduction) We can add the two half reactions to get the net ionic reaction, but notice that adding these two half reactions results in an imbalance in the number of electrons. In this case, we must multiply the first by 3 to get the electrons to cancel. In addition some of the waters and H ions also cancel. 3SO2(aq) + Cr2O72-(aq)+ 2H+(aq)3SO42-(aq)+ 2Cr3+(aq)+H2O(l) What is the oxidising agent? Reducing agent? Cr2O72-, SO2

23. Reactions of metals • Reactions with oxygen (combustion) • All metals form oxides except Ag, Au and Pt • Metal + oxygen  metal oxide • e.g. 2Mg + O2 2MgO • Tendency to form metal oxides: • Li, Na, K, Ca, Ba (react at room temp) • Mg, Al, Fe, Zn (react slowly at room temp, vigorously when heated) • Sn, Pb, Cu (react slowly and only when heated) heat

24. Reactions of metals • Reactions with water • Reactive metals react with water or steam • Metal + water  metal hydroxide + hydrogen gas • e.g. Na + 2H2O  2NaOH + H2 • Metal + steam  metal oxide + hydrogen gas • e.g. Zn + H2O  ZnO + H2 • Relative reactivity: • Li, Na, K, Ca, Ba (react with water at room temp) • Mg, Al, Zn, Fe (react with steam at high temp) • Sn, Pb, Cu, Ag, Au, Pt (do not react)

25. Reactions of metals • Reactions with dilute acid • More metals react with acid than water • Metal + acid  salt + hydrogen gas • Zn + 2HCl  ZnCl2 + H2 • Relative reactivity: • Li, Na, K, Ca, Mg, Al, Zn, Fe, Co, Ni (react readily) • Sn, Pb (slow to react without heat) • Ag, Hg, Pt, Au (do not react)

26. Reactivity Series Based on the ease of reactions with oxygen, water and acids, metals can be organised in order of reactivity, known as an activity series. Activity series for metals: K>Na>Ba>Ca>Mg>Al>Zn>Fe>Sn>Pb>Cu>Ag>Hg>Pt>Au most reactiveleast reactive Grp 1>Grp 2> Grp 3>some TM (Zn, Fe)>Grp 4>more TM N.B. TM = transition metals

27. Metal Displacement The reactivity series of metals gives us a guide as to how readily reactions will take place among metals www.bbc.co.uk For example, Mg is more reactive than Cu, so Mg will displace Cu from solution. Cu will precipitate and Mg will dissolve into solution

28. Redox titrations Titration is an analytical method to determine the unknown concentration of a substance in solution. The most common titrations are acid-base and redox. Unlike most acid-base titrations, many redox titrations equivalence points require no indicator solution Iron determination:The concentration of Fe2+ can be determined by the purple oxidizing agent, KMnO4, which is reduced to colourless Mn2+. The overall redox reaction is below. 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O (See Pearson p. 420) Winkler method for dissolved oxygen: 2 Mn2+ (aq) + O2 (aq) + 4OH– → 2 MnO2(s) + H20            (1:2) MnO2(s) + 2 I–(aq) + 4H+ → Mn2+(aq) + I2(aq) + 3H2O   (1:1) 2 S2O32-(aq) + I2 (aq) → S4O62-(aq) + 2 I– (aq)    (2:1) The Winkler titration is a more complex series of redox reactions. Notice the stoichiometric relationship/ratio of O2 to S2O32- is 1:4 overall. (See Pearson p.423)

29. Voltaic Cell – metal equilibrium Recall the half-equations that we used to show oxidation and reduction reactions. Consider a solid metal in a solution of it’s component ions (see diagram). This shows there is an equilibrium between the metal and its ion. If we connect these two metals together, something interesting happens….

30. Voltaic cell- Danielle Cell Because there is a difference in reactivity between these two metals, e- will flow from the more reactive (anode) where oxidation occurs, to the less reactive (cathode) where reduction occurs through an external wire. Voltmeter A salt bridge filled with an electrolyte such as KCl allows the ions in solution to flow, completing the circuit. + - The cathode is the site of reduction and is positive and attracts electrons The anode is the site of oxidation and is negative due to the production of electrons Recall that Zn was found to be more reactive than Cu. This means Zn is more likely to be oxidised. So if we connect these two metals together, a spontaneous reaction occurs and we can get electrons to flow in a closed circuit called a voltaic cell (aka Galvanic Cell) Reduction: e- gain Oxidation: e- loss Overall reaction Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) Another mnemonic: An Ox Red Cat

31. Quantifying the Reactivity Series The reactivity series of metals can be estimated using the reactions previously described. However, a more precise method can allow us to quantify the differences in reactivity and can include more than just metals. If we connect a half-cell to the standard hydrogen half-cell, there are 2 possibilities: The substance is a stronger oxidising agent than H+ (gets reduced) – see Cu above The substance is a weaker oxidising agent than H+ (gets oxidised) – see Zn below • A hydrogen half-cell (left) is used as a standard to which all other substances are compared. • This cell consists of a platinum electrode over which hydrogen gas is bubbled under standard conditions: • 1 mol dm-3 [H+] • 298 K • 1 atm (H2) Standard H2 half-cell

32. Standard Reduction Potentials Notice these reactions are all written as reductions , by convention • With the standard hydrogen half-cell as a reference, we are able to measure standard electrode potentials (E0) measured in volts (V). • E0 values are either: • Negative – if the substance is a stronger reducing agent or; • Positive – if the substance is a stronger oxidising agent Notice hydrogen is set to 0.00V. All above are + and all below are – compared to this reference cell

33. Cell Potential Considering the example containing Cu and Zn half-cells, we can measure the potential difference by placing a voltmeter across the external wire connecting the two. Voltmeter (1.10V) The voltmeter for this pair of half-cells “shows” that 1.10V is produced. - + Since e- flow from one half-cell to the other, we can measure the energy difference between the two. Potential difference is the measure of the energy difference between the two half-cells in volts (V) and is also called the cell potential or emf. Overall reaction Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq) The potential difference can be estimated using the Standard Reduction Potentials Table. The substance higher on the table (more negative E0) will be oxidised (Zn), while the lower (Cu) is reduced

34. Predicting redox reactions Oxidising agents get stronger as we move down the left side of the Standard Reduction Potentials Table This means a substance such as Cl2 will oxidise any of the reductants above it and itself be reduced Reducing agents get stronger as we move up the right side of the table This means a substance such as Zn will reduce any oxidants below it and itself be oxidised. This means the reaction will be reversed (oxidation) Example: Cl2 will oxidise both Pb(s) or Fe(s) which are higher in the table This means for example: Pb(s)  Pb2+ + 2e- Notice the reaction has been reversed for oxidation Example: Zn(s) will reduce Cu2+ which is lower in the table. This means: Zn(s)  Zn2+ + 2e- (ox) Cu2+ + 2e-  Cu(s)(red)

35. Predicting redox reactions Predicting a redox reaction using the table: Will Cl2 react with Zn(s)? If so, what is the cell potential? Cl2 will oxidise Zn(s) which is higher in the table, so yes, a reaction will occur. This means: Zn(s)  Zn2+ + 2e- (ox) E0 = +0.76V (rvrsd) 1/2Cl2 + e-  Cl- (red) E0 = +1.36V Added together: Zn(s) +Cl2  Zn2+ + 2Cl- E0cell = E0 (red) - E0 (ox) E0cell = 1.36 – (-0.76) E0cell = 2.12V (this is the predicted cell potential) Note: spontaneous reactions always have a positive E0 value.

36. Exercises Using the table to the right, predict if the following substances will result in a reaction. If there is a reaction, write out the half-equations and determine the cell potential. Cu2+ , Ba(s) F2, Pb(s) Na+ , Fe(s)

37. Exercises - Answers Using the table to the right, predict if the following substances will result in a reaction. If there is a reaction, write out the half-equations and determine the cell potential. Cu2+ , Ba(s) F2, Pb(s) Na+ , Fe(s) Cu2+ + 2e-  Cu(s) 0.34V Ba(s)  Ba2+ + 2e- 2.90V Ecell = 2.90 + 0.34 = 3.24V 1/2F2 + e-  F- 2.87V Pb(s)  Pb2+ + 2e- 0.13VEcell = 2.87 + 0.13 = 3.00V No rxn. (Na+ is a weaker oxidising agent than water (E0 = -0.83V) and has a less positive E0 value than the reducing agent Fe(s)

38. Exercise In your notebook, draw the voltaic cell to the right and: Label the anode and cathode including signs Label the salt bridge Indicate the direction of e-flow in the cell Write out the oxidation and reduction half reactions Predict the cell potential using the Standard Reduction Potentials Table

39. Exercise - Answers In your notebook, draw the voltaic cell to the right and: Label the anode and cathode including signs Label the salt bridge Indicate the direction of e-flow in the cell Write out the oxidation and reduction half reactions Predict the cell potential using the Standard Reduction Potentials Table - + Reduction Ag+ + e-  Ag(s) (E0 = +0.80V) Oxidation Cu(s)  Cu2+ + 2e- (E0 = -0.34V) Ecell = Eox + Ered = -0.34V + 0.80V = +0.46V

40. Electrolytic Cells Previously, we have seen that voltaic cells generate electric current in an external circuit by spontaneous redox reactions. Voltaic cell: Chemical energy  Electrical energy • There is a second type of electrochemical cell where we input electrical energy into a system to cause a non-spontaneous reaction to occur. • These reactions are called electrolysis reactions. Some examples: • Electrolysis of water • Electrolysis of molten NaCl • Electroplating Electrolytic cell: Electrical energy  Chemical energy

41. Electrolytic Cell Components • The electrolytic cell is different to the voltaic cell in operation and structure. The electrolytic cell contains: • Power source • One container (instead of two) • Electrolyte to allow conduction of current in the solution • Reduction at the Cathode (-) • Oxidation at the Anode (+) • Inert electrodes (e.g. carbon) Note the polarities of the electrodes have changed from the voltaic cell and are determined by which terminal of the power supply they are connected to. Reduction still occurs at the cathode and oxidation at the anode. At the cathode, e- are supplied from the power source, so a reduction of the strongest oxidising agent in solution occurs. (positive ions or water) At the anode, e- are removed, so oxidation of the strongest reducing agent occurs here. (negative ions or water)

42. Electrolysis of molten NaCl Molten NaCl consists of Na+ and Cl- ions that migrate towards the electrodes of opposite charge in an electrolytic cell completing the circuit The following equation represents the breaking apart of NaCl(l): 2NaCl(l) → 2Na(l) + Cl2 (g) The half-reactions involved in this process are: E° reduction 2Na+(l) + 2e- → Na(s) -2.71 V oxidation Cl-(l) → Cl2(g) + 2e- -1.36V ________________________________________ net voltage required - 4.07V The negative voltage (-4.07V) that results when we add up the half-reactions indicates that the overall reaction will not be spontaneous. An EMF of more than 4.07 volts will be required for this reaction to occur.

43. Electrolysis of Water This is one of the most common demonstrations of a simple electrolysis. It involves the oxidation and reduction of water. Overall: 2 H2O(l)  2 H2(g) + O2(g) The half reactions are: 2H2O + 2e-  H2 + 2OH- Eored = -0.83 V 2H2O  O2 + 4H+ + 4e- Eoox = -1.23 V Note: in any aqueous solution, these two reactions are possible during electrolysis

44. Electrolysis of Aqueous NaCl – competing reactions At the anode the possible reactions are: Cl-(l) → Cl2 (g) + 2 e- E0oxid = -1.36V 2H2O(l) → O2 (g) +4H+(aq) + 4e- E0oxid = -1.23V At the cathode the possible reactions are: 2H2O(l) + 2e-→ +H2(g) + 2OH- (aq) E0red = - 0.83V Na+(l) + e- → Na(s)  E0red = - 2.71V • How to predict? • Oxidation: most - E0 on the reduction potential table (E0oxid most +) • Reduction: most + E0 on the reduction potential table (E0red most +) Electrolysis of a concentrated salt solution (brine) Source: www.answers.com/topic/electrolysis These are the products unless the solution is highly concentrated. In this case, we will start to see the evolution of chlorine gas as in the diagram. So, which two reactions are predicted? See further examples in Derry, “Chemistry Higher Level”, pp 257-262

45. Electroplating Electroplating is an application of electrolysis where a thin layer of a metal such as silver or nickel is plated over another metal such as iron or copper to provide a bright shiny finish to the object. Some examples: food cans (Sn), tools (Ni, Cr), galvanising (Zn). Silver plating Objects such as flatware are sometimes electroplated using silver. The reactions are: Ag(s)  Ag+(aq) + e- Ag+(aq) + e-  Ag(s) Identify the anode and cathode in this example.

46. Factors affecting amount of products formed 1. Charge on the ion – 2. Current – 3. Duration of Electrolysis – How do you think these factors will affect the products formed in an electrolysis reaction?

47. Factors affecting amount of products formed 1. Charge on the ion – assuming the same amount of applied current a M+ ion will produce 2x the amount of atoms as M2+ which requires twice the number of e-. 2. Current – if the current in an electrolytic cell increases, then the amount of electrons increases and so does the amount of product 3. Duration of Electrolysis – assuming there are enough reactants, the longer current is applied, the more products will be produced Note: Each e- carries a charge of 1.602 x 10-19 C (coulombs)

48. Inert vs Active electrodes Inert Inert electrodes (C and Pt are the most common) do not take part in the reactions at the surface of the electrode. They simply provide a surface for the electron transfer reactions to take place. Active Active electrodes are ones that may take part in a reaction at their surface. Fe and Ag are more reactive substances and are more likely to be oxidised than water. See values below. Ag(s) Ag2+(aq) + 2e- E0 = -0.80V Fe(s)  Fe2+(aq) + 2e- E0 = +0.44V 2H2O(l) O2(g) + 4H+(aq) + 4e- E0 = -1.23V

49. R. Slider 2012 Oxidation & Reduction