1. GazeSJ Version 3.1 2013 Chemistry NCEA L3 3.7 Redox

2. This achievement standard involves describing oxidation-reduction processes. 1 Processes include reactions in electrochemical and electrolytic cells, This includes the use of reduction potentials and spontaneity of reactions. Knowledge of preferential discharge in electrolytic cells is not required. 2 Calculations are limited to include the use of oxidation numbers, mole ratios and electrode potentials. 3 Knowledge of appearance of the following reactants and their products is required. Oxidants will be limited to: O 2, Cl 2, I 2, Fe 3+, dilute acid (with metals), H 2 O 2, MnO 4 – (reacting in acidic, basic or neutral conditions), Cu 2+, Cr 2 O 7 2– / H +, OCl –, concentrated HNO 3, IO 3 –, MnO 2. Reductants will be limited to: metals, C, H 2, Fe 2+, Br –, I –, H 2 S, SO 2, (SO 3 2–, HSO 3 - ), S 2 O 3 2–, H 2 O 2, H 2 C 2 O 4. Appropriate information relating to other oxidants or reductants will be provided. Standard reduction potentials will be included where required. Achievement Criteria

3.  A redox reaction is where one substance is oxidised and the other substance is reduced. OxidationReduction >loss of electrons >loss of hydrogen >gain of oxygen >gain of electrons >gain of hydrogen >loss of oxygen Oxidation numbers are used to determine what is oxidised and what is reduced in a reaction. Redox Terms

4. An Iron nail left in copper sulfate Fe (s) + Cu (aq) Fe (aq) + Cu (s) 2+ Copper is reduced – gained electrons Oxidising agent (oxidant) Iron is oxidised – lost electrons Reducing Agent (reductant) Electron transfer

5. Iron Ore smelting 2 Fe 2 O 3(s) + 3C (s) 4Fe (s) + 3CO 2(g) Iron oxide is reduced – lost oxygen Oxidising agent (oxidant) carbon is oxidised – gained oxygen Reducing Agent (reductant) Iron Ore Oxygen transfer

6. Sulfur production 2 H 2 S (g) + O 2(g) 2S (s) + 2H 2 O (l) Hydrogen sulphide is oxidised – lost hydrogen Reducing Agent (reductant) Oxygen gas is reduced – gained hydrogen Oxidising agent (oxidant) Hydrogen transfer

7. A balanced redox equation is broken into two half-equations, to show how electrons are transferred. Fe (s) + Cu (aq) Fe (aq) + Cu (s) 2+ Reduction half equation - oxidant is reduced to a product Fe Fe + 2e Oxidation half equation – reductant is oxidised to a product Cu + 2e Cu 2+ - - Half Equations

8. ReductantOxidant >reducing agent >is oxidised >loses electrons + hydrogen >gains oxygen >oxidising agent >is reduced >gains electrons + hydrogen >loses oxygen Summary of terms

9. Oxidants – L3 triiodideI 3 - → I- + I 2 iodide Purple brown (-1) colourless (aq)(-1) ion NameConditions Half equation / colour change /ON Name permanganate (manganate(VII)) acidifiedMnO 4 - + 8H + + 5e - → Mn 2+ + 4H 2 O manganese (II) purple (+7) colourless (+2) ion permanganate (manganate(VII)) neutralMnO 4 - + 4H + + 3e - → MnO 2 + 2H 2 O manganese purple (+7) brown (s) (+4) dioxide permanganate (manganate(VII)) alkalineMnO 4 - + e - → MnO 4 2- manganate purple (+7) green (+6) Dichromate (dichromate(VI)) acidifiedCr 2 O 7 2- + 14H + + 6e - → 2Cr 3+ + 7H 2 O chromic (III) ion orange (+6) green (blue) (+3) Hydrogen peroxide acidifiedH 2 O 2 + 2H + + 2e - → 2H 2 O water colourless (-1) colourless (-2) chlorineCl 2 + 2e - → C l- chloride ion pale yellow/green (0) colourless (-1) Hypochlorite (chlorate (I)) alkalineClO - + H 2 O + 2e - → Cl - + 2OH - chloride ion colourless (+1) colourless (-1) iodineI 2 + 2e - → 2I - iodide grey (s) (0) colourless (aq)(-1) ion

10. NameConditions Half equation / colour change /ON Name chlorate (chlorate(V)) acidifiedClO 3 - + 6H + + 6e - → Cl - + 3H 2 O chloride colourless (+5) colourless (-1) ion Nitric acid (nitrate ion) concentratedNO 3 - + 2H + + e - → NO 2 + H 2 O nitrogen colourless (+5) brown (g) (+4) dioxide Iron (III) ionFe 3+ + e - → Fe 2+ Iron (II) ions orange (+3) pale green (+3) Copper (II) ionCu 2+ + e - → Cu + copper (I) ion blue (+2) white (+1) oxygenO 2 + 4e - → 2O 2- oxide ion colourless (0) colourless (-2) Dilute acid (H + ions) 2H + +2e - → H 2 hydrogen gas Colourless (+1) colourless (0) Manganese dioxide acidifiedMnO 2 + 2H + + 2e - → Mn 2+ + 2H 2 O manganese colourless (+4) colourless (+2) ion iodateIO 3 - + 2e - → I 2 iodine Colourless (aq) (+5) grey (s)(0) Oxidants – L3 bromateBrO 3 - + 2e- → Br - bromide ion Colourless (aq) (+5) colourless (-1)

11. Name Half equation / colour change /ONName hydrogenH 2(g) → 2H + (aq) + 2e - colourless (0) colourless (+1) hydrogen ion zincZn (s) → Zn 2+ + 2e - grey (0) colourless (+2) zinc (II) ion Iron (II) ionFe 2+ (aq) → Fe 3+ (aq) + e - pale green (+2) orange (+3) Iron (III) ion oxalate ion (ethanediotae (COO - ) 2 ) C 2 O 4 2- (aq) → 2CO 2(aq or g) + 2e - colourless (+3) colourless (+4) carbon dioxide oxalic acid (ethanedioic acid) H 2 C 2 O 4(aq) → 2CO 2(aq or g) + 2H + + 2e - colourless (+3) colourless (+4) carbon dioxide sulfite ionsSO 3 2- (aq) + H 2 O → SO 4 2- (aq) + 2H + + 2e - colourless (+4) colourless (+6) sulphate ion thiosulfate2S 2 O 3 2- (aq) → S 4 O 6 2- aq) + 2e - colourless (+2) colourless solution (+2.5) tetrathionate magnesiumMg (s) → Mg 2+ (aq) + 2e - grey (0) colourless (+2) magnesium ion Reductants – L3 carbonC (s) + 2H 2 O → CO 2(g) + 4H + + 4e - carbon black (0) colourless (+4) dioxide

12. Name Half equation / colour change /ONName carbon monoxideCO (g) + H 2 O → CO 2(g) + 4H + + 2e - colourless (+2) colourless (+4) carbon dioxide iron metalFe (s) → Fe 2+ + 2e - silver (0) pale green (+2) iron (II) ion copper metalCu (s) → Cu 2+ (aq) + 2e - Red/brown (0) blue (+2) copper (II) ion hydrogen sulphideH 2 S (g) → S (s) + 2H + + 2e - colourless (-2) yellow ppt (0) sulphur hydrogen peroxideH 2 O 2(aq) → O 2( g) + 2H + + 2e - colourless (-1) colourless (0) oxygen gas iodide ion2I - (aq) → I 2(s) + 2e - colourless (-1) grey (0) iodine bromide ion2Br - (aq) → Br 2(aq) + 2e - colourless (-1) orange (0) bromine Sulfur dioxideSO 2(g) + 2H 2 O → SO 4 2- (aq) + 4H + + 2e - colourless (+4) colourless (+6) sulphate ion Reductants – L3 carbonC (s) + H 2 O → CO (g) + 2H + + 2e - carbon black (0) colourless (+2) monoxide

13.  The Oxidation Number (ON) gives the ‘degree’ of oxidation or reduction of an element.  They are assigned to a INDIVIDUAL atom using the following rules. Elements = 0 e.g. Fe H 2 Hydrogen atom (not as element) = +1 e.g. HCl H 2 SO 4 Except Hydrides = -1 e.g. LiH Oxygen atom (not as element) = -2 e.g. MnO 4 - CO 2 Except peroxides = -1 H 2 O 2 ON 00 +1 -2 ON Oxidation Numbers

14. The Oxidation Number (ON) gives the ‘degree’ of oxidation or reduction of an element. They are assigned to a INDIVIDUAL atom using the following rules. Elements Each atom = 0 e.g. Fe H 2 Hydrogen atom (not as element) = +1 e.g. HCl H 2 SO 4 Except Hydrides = -1 e.g. LiH Oxygen atom (not as element) = -2 e.g. MnO 4 - CO 2 Except peroxides = -1 H 2 O 2 ON Oxidation Numbers

15. Monatomic ions = charge e.g Fe Cl Polyatomic ions Sum of =charge e.g. MnO 4 Because Total charge = -1 And Oxygen = -2 +7 + (4x -2) = -1 Molecules Sum of on atoms =charge e.g. CO 2 Because Total charge = 0 And Oxygen = -2 +4 +(2x-2) = 0 ON +2 -2+7 +4-2 2+ - - MnOcO Oxidation Numbers

16. Monatomic ions = charge e.g Fe 3+ Cl - Polyatomic ions Sum of =charge e.g. MnO 4 - Because Total charge = -1 And Oxygen = -2 +7 + (4x -2) = -1 Molecules Sum of =zero e.g. CO 2 =0 Because Total charge = 0 And Oxygen = -2 +4 +(2x-2) = 0 ON -2+7 +4-2 MnOCO Oxidation Numbers

17. Oxidation is a loss of electrons and causes an increase in Oxidation of Fe Fe Fe + e- Fe has increased (+2 to +3) caused by a loss of electrons e- 2+ ON 3+2+ ON Reduction is a gain of electrons and causes an decrease in Reduction of MnO 4 MnO 4 + 5e- Mn Mn has decreased (+7 to +2) caused by a gain in electrons e- - ON 2+- ON +3 +2 +7 OXIDATION and REDUCTION always occur together. The electrons lost by one atom are gained by another atom. This is called a REDOX reaction. Join the two half equations together. Oxidation Numbers

18. Using ON to identify Redox Reactions What has been oxidised and what has been reduced? STEP ONE – write the ON for each atom STEP TWO – Identify the atom that has had its ON increased. It is Oxidised C has increased ON (0 to +4) so C is Oxidised. STEP THREE – Identify the atom that has decreased ON. It is reduced. Fe has decreased ON (+3 to 0) so Fe is Reduced. +3-20 0+4-2

19. 1. Write half equation by identifying reactant and product 2. Balance atoms that are not O or H 3. Balance O by adding H 2 O and H by adding H 4. Balance charge by adding electrons Fe Atoms already balanced There are no O or H atoms to balance Fe Fe + e 2+ = 3+ (-1) 1 electron to balance charge Balance the half equation for the oxidation of Fe to Fe 3+2+ + 3+2+3+2+ Fe Fe + e 2+3+ Balancing half equations

20. 1. Write half equation by identifying reactant and product 2. Balance atoms that are not O or H 3. Balance O by adding H 2 O and H by adding H 4. Balance charge by adding electrons MnO 4 MnAtoms already balanced MnO 4 + 8H Mn + 4H 2 O Balance O by adding 4H 2 O and H by adding 8 H+ MnO 4 + 8H Total charge +7 Mn + 4H 2 O Total charge +2 Add 5 electrons (e-) Balance the half equation for the reduction of MnO 4 to Mn 2+ - + - + - + - MnO 4 + 8H + 5e Mn + 4H 2 O 2+ - + - Balancing half equations

21. Rules e.g. Cr 2 O 7 2- →Cr 3+ 1. Assign oxidation numbers and identify element oxidised or reduced. (+6)(-2)(+3) Cr 2 O 7 2- →Cr 3+ 2. Balance atom no. for element oxidised or reduced (other than oxygen and hydrogen) Cr 2 O 7 2- →2Cr 3+ 3. Balance the Oxygen using H 2 O Cr 2 O 7 2- →2Cr 3+ + 7H 2 O 4. Use H + (acidic conditions) to balance the hydrogen 14H + + Cr 2 O 7 2- + 6e - →2Cr 3+ + 7H 2 O 5. Use OH - (in alkaline conditions) to cancel any H + [same amount on both sides] 6. Balance charge by adding electrons (LHS on oxidants RHS on reductants) 14H + + Cr 2 O 7 2- + 6e - →2Cr 3+ + 7H 2 O 7. Check balance of elements and charges. Balancing half equations

22. Electrochemistry Electrochemistry is the chemistry of reactions involving the transfer of electrons, redox reactions. In year 12 the focus was on electrolytic cells or electrolysis. This involves a non-spontaneous reaction in which an external source of electricity provides electrons with the energy required to bring about a redox reaction. In year 13 the focus is on electrochemical cells in which spontaneous redox reactions use the energy released from a chemical reaction to generate electric current. These are called Galvanic cells or batteries.

23. Galvanic Cells and Salt Bridges Under normal conditions a redox reaction occurs when an oxidising agent is in contact with a reducing agent. If the two half reactions are physically separated, the transfer of electrons is forced to take place through an external metal wire. As the reaction progresses a flow of electrons occurs. This only happens if there is a full circuit so that there is no net build-up of charge. To complete this circuit the separate solutions are connected using a salt bridge which allows ions to flow and transfer charge. Typically the salt bridge is a glass tube filled with a gel prepared using a strong electrolyte such as KNO 3(aq) (which contains ions that do not react with the electrodes or species in the solutions. The anions (NO 3 - ) and cations (K + ) can move through the salt bridge so that charge does not build up in either cell as the redox reaction proceeds.

24. Galvanic cells The oxidation and reduction reactions that occur at the electrodes are called half-cell reactions. Zn electrode (anode, oxidation) Zn(s)  Zn 2+ (aq) + 2e  Cu electrode (cathode, reduction) Cu 2+ (aq) + 2e   Cu(s )

25. Electromotive Force The reduced and oxidised substances in each cell form a redox couple. The 2 couples in this cell (the Daniel cell) are Zn 2+ |Zn and Cu 2+ |Cu. By convention, when writing redox couples, the oxidised form is always written first. The fact that electrons flow from one electrode to the other indicates that there is a voltage difference between the two electrodes. This voltage difference is called the electromotive force or emf of the cell and can be measured by connecting a voltmeter between the two electrodes. The emf is therefore measured in volts and is referred to as the cell voltage or cell potential. A high cell potential shows that the cell reaction has a high tendency to generate a current of electrons. Obviously the size of this voltage depends on the particular solutions and electrodes used, but it also depends on the concentration of ions and the temperature at which the cell operates. ZnSO 4(aq) CuSO 4(aq) Anode (Zn) Cathode (Cu) Salt Bridge

26. Cell Diagrams Galvanic cells can be represented using cell diagrams. This is a type of short hand notation that follows a standard IUPAC convention. For the copper/zinc cell the standard cell diagram is Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) The vertical lines represent phase boundaries and || represents the salt bridge. The cathode (reduction reaction) is always shown on the right hand side and the anode (oxidation) on the left in a standard cell diagram. The electrons thus move from left to right in the standard cell diagram, representing a spontaneous redox reaction. The electrodes are always written in at the beginning and end of a cell diagram. This occurs both if the metal is involved in the redox reaction (as in the Daniel cell above where the electrodes are the Cu and Zn), and also if an inert electrode is used. In each half cell the reactant appears first, followed by the product.

27. Cell Diagrams An inert electrode must be used in cells in which both species in a redox couple are in aqueous solution (MnO 4 - and Mn 2+ ). The inert electrodes are commonly either platinum, Pt(s) or graphite, C(s) electrodes. Since the two species in the redox couple are in solution, they are separated by a comma rather than a vertical line. egCu(s) | Cu 2+ (aq) || MnO 4  (aq), Mn 2+ (aq) | Pt(s) The cell diagram shows two half cells linked. Each half cell consists of the oxidant, the reductant and the electrode (which may be the oxidant or reductant). The two half cells above are Cu(s)|Cu 2+ (aq) and MnO 4  (aq), Mn 2+ (aq)|Pt(s).

28. Standard Electrode Potentials Under standard conditions (when the pressure of hydrogen gas is 1 atm, and the concentration of acid is 1 mol L -1 ) the potential for the reduction reaction is assigned a value of zero. 2H + (aq) + 2e  → H 2 (g)E o = 0.00 V The superscript o denotes standard state conditions. When the hydrogen electrode acts as a cathode, H + ions are reduced, whereas when it acts as an anode, H 2 gas is oxidised.

29. Standard Electrode Potential The overall cell voltage is the sum of the electric potential at each electrode. If one of the electrode potentials is known, and the overall cell voltage is measured, then the potential of the other electrode can be calculated by subtraction. Clearly it is best if all electrode potentials are measured relative to a particular electrode. In this way, a scale of relative values can be established. The standard hydrogen electrode (SHE) is used as the standard reference electrode, and it has arbitrarily been given a value of 0.00 V.

30. Standard Electrode Reduction Potential For any redox couple, the standard electrode (reduction) potential is the voltage obtained under standard conditions when that half-cell is connected to the standard hydrogen electrode. For example, the electrode potential of a Zn 2+ |Zn electrode can be measured by connecting it to a hydrogen electrode. Experimentally, the more positive terminal is always where reduction is occurring in a spontaneous reaction. In example (a) reduction occurs in the hydrogen electrode (positive electrode) while oxidation occurs in the Zn 2+ |Zn compartment (negative electrode). The cell diagram for this electrochemical cell is Zn(s) | Zn 2+ (aq) || H + (aq), H 2 (g) | Pt(s)

31. Standard Reduction Potentials Using the standard reduction potentials for many half reactions have been measured under standard conditions (at 25 o C). Standard reduction potentials are provided in examinations. The table can be used to decide the relative strength of species as oxidants or reductants. The species on the left in the couple with the most positive reduction potential, will be the strongest oxidising agent or oxidant. E.g it is F 2 (g) (NOT F 2 / F  ). This means F 2 has the greatest tendency to gain electrons. As the electrode potential decreases, the strength as an oxidant decreases. Conversely the strongest reducing agent or reductant would have the least positive (or most negative) e.g. Li(s). This means Li has the greatest tendency to lose electrons.

32. Using reduction potentials to determine E o cell In any electrochemical cell, the standard cell potential (voltage), E 0 cell, is the difference between the reduction potentials of the two redox couples involved. The couple with the more positive reduction potential will be the reduction half-cell (cathode). This means that the E o cell for any combination of electrodes can be predicted using the relationship E o cell = E o (reduction half-cell) - E o (oxidation half-cell) ORE o cell = E o (cathode) - E o (anode) OR E o cell = E o (RHE) - E o (LHE) (where RHE is the right hand electrode and LHE is the left hand electrode in the standard cell diagram).

33. Using reduction potentials to determine E o cell For example, consider the cell Zn(s) | Zn 2+ (aq) || Ag + (aq) | Ag(s) Reduction reaction is Ag + (aq) + e   Ag(s) E o (Ag + /Ag) = +0.80V Oxidation reaction is Zn(s)  Zn 2+ (aq) + 2e  E o (Zn 2+ /Zn) = -0.76V E o cell = E o (Ag + /Ag) - E o (Zn 2+ /Zn) = 0.80 - (-0.76) V = +1.56V 1. Any metal that is more reactive (lower E o value) will reduce the cation of a less reactive metal because the E o cell value for the reaction will be positive. 2. A positive standard cell potential suggests that the reaction occurs spontaneously from left to right, as shown in a standard cell diagram. In reality the reaction may not appear to proceed as it may be very slow due to a high activation energy. Another reason a reaction may not proceed is if the surface of the metal is coated with an oxide. Aluminium oxide on the surface of aluminium can protect the aluminium from undergoing a spontaneous oxidation reaction

34. Predicting whether a Reaction will occur It is possible to use E o values to predict whether a reaction will occur. This simply involves identifying which species must be reduced and which species must be oxidised if the reaction is to proceed spontaneously. The appropriate reduction potentials are then substituted into the equation. E o cell = E o (cathode) - E o (anode) where E o (cathode) is the reduction potential for the half cell where reduction occurs and E o (anode) is the reduction potential for the half cell where oxidation occurs. If the E o cell calculated is positive, then the reaction will occur spontaneously. Conversely, a negative cell potential means the reaction will not proceed.

35. Predicting whether a Reaction will occur Corrosion – an everyday application of a redox reaction Corrosion is the term usually applied to the deterioration of metals by an electrochemical process. One example is the rusting of iron in the presence of water and oxygen. Although the reactions involved in rusting are quite complex, the main steps are as follows. Step 1 - Oxidation occurs at a region of the iron’s surface, the anode. Fe(s)  Fe2+(aq) + 2e  Eo red = -0.44 V Step 2 - Electrons travel to some other region of the metal’s surface where a variety of cathode reactions can occur. Step 3 - In acidic medium, atmospheric oxygen is quickly reduced to H2O. The acidity of the solution can, in part, be due to the presence of dissolved acidic gases such as CO2 and SO2.

36. Example Can a solution of acidified potassium permanganate oxidise the Fe 2+ present in a solution of iron (II) nitrate? (Note in questions such as this you will have to recognise that ions such as sodium and potassium are spectator ions.) The unbalanced equation for the reaction would be: MnO 4  + Fe 2+  Mn 2+ + Fe 3+ Reduction reaction is MnO 4   Mn 2+ E o (MnO 4  /Mn 2+ ) = +1.51 V Oxidation reaction is Fe 2+  Fe 3+ E o (Fe 3+ /Fe 2+ ) = +0.77 V E o cell = E o (MnO 4  /Mn 2+ ) - E o (Fe 3+ /Fe 2+ ) = +1.51 - +0.77 = 0.74 V Since the cell potential is a positive value (>0.00 V) the reaction should proceed. If it had been negative then the reaction would not proceed. Predicting whether a Reaction will occur

37. Discussion format of E° Cells 1.Movement of electrons from anode (oxidation > LEO) lower E° value to cathode (reduction > GER) higher E° value 2. Half equations at each electrode anode X - → X + e - cathode X + e - → X - 3. Movement of salt bridge solutions Anions → Anode Cations → Cathode 4. Overall summary Oxidation at the anode → salt bridge anions balance cations produced at the electrodes Reduction at the cathode → salt bridge cations balance cations removed at electrode

38. E° Cells Summary CCR AAO nionnion nodenode xidationxidation ationation athodeathode eductioneduction LEO GER