1. Oxidation & Reduction

2. Put the sign in front of the number (ion charge other way) Put the sign in front of the number (ion charge other way) Oxygen always -2 (except in H 2 O 2 : -1) Oxygen always -2 (except in H 2 O 2 : -1) Hydrogen always +1 (except hydride: -1) Hydrogen always +1 (except hydride: -1) Sum of ON in compounds = 0 Sum of ON in compounds = 0 Elements = 0 Elements = 0 Monoatomic ion ON = charge Monoatomic ion ON = charge Sum of polyatomic ON = charge Sum of polyatomic ON = charge if you get stuck, assume grp1 = +1, grp2 = +2 if you get stuck, assume grp1 = +1, grp2 = +2 With complex formulae, use polyatomic ion together With complex formulae, use polyatomic ion together Oxidation numbers

3. ON Practice 1.ClO - 2.CrO 4 2- 3.CO 2 4.SO 3 2- 5.H 2 S 6.NH 3 7.N 2 8.NO 9.HNO 2 10.HNO 3 11.HClO 12.HClO 2 13.HClO 3 14.HClO 4 15.Cl 2 O 16.Fe(OH) 2 17.KBrO 3 18.NaOCl 19.H 2 O 2 20.MnO 4 - -2+1 -2+6 -2+4 -2+4 +1-2 +1-3 0 -2+2 -2+1+3 -2+1+5 -2+1+1 -2+1+3 -2+1+5 -2+1+7 -2 +1 -2+1+2 -2+1+5 -2+1+1 +1 -2+7

4. The easiest way to do this is work out all ON then compare atoms before and after the arrow. Identifying oxidants and reductants If the ON increases, it is being oxidised and is then therefore a reductant If the ON decreases (is reduced), it is being reduced and is then therefore an oxidant If the ON does not change, it is a spectator ion If the atom is within a compound, give that compound as the species that is being reduced/oxidised

5. Appearance & state of reductants & oxidants Reduced form Oxidised Form FormulaAppearanceFormulaAppearance Cu Brown solid Cu 2+ Blue ion SO 2 SO 4 2- Mn 2+ MnO 4 - Purple ion H2O2H2O2H2O2H2O2 O2O2O2O2 H2OH2OH2OH2O H2O2H2O2H2O2H2O2 Cr 3+ Green ion Cr 2 O 7 2- Orange ion Fe 2+ Green ion Fe 3+ Orange ion Cl - Cl 2 Pale green gas I-I-I-I- I 2 (aq) Brown solution H2H2H2H2 H+H+H+H+ Zn Grey solid Zn 2+ Br - Br 2 (aq) Orange solution

6. It doesn’t matter how simple or complex the half equations are you can follow the same steps: Write balanced redox equations Balance atoms that aren’t O or H Balance atoms that aren’t O or H Balance oxygens by adding waters Balance oxygens by adding waters Balance hydrogens by adding H+ Balance hydrogens by adding H+ Add electrons to the more positive side Add electrons to the more positive side Remember, these are ionic equations so you must identify the ions involved in the reaction from the description To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms

7. Write balanced redox equations Balance atoms that aren’t O or H Balance atoms that aren’t O or H Balance oxygens by adding waters Balance oxygens by adding waters Balance hydrogens by adding H+ Balance hydrogens by adding H+ Add electrons to the more positive side Add electrons to the more positive side This method is used when the reaction takes place in acid or neutral conditions To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms

8. Write balanced redox equations Balance atoms that aren’t O or H Balance atoms that aren’t O or H Balance oxygens by adding waters Balance oxygens by adding waters Balance hydrogens by adding H+ Balance hydrogens by adding H+ Cancel out H+ by adding OH- to both sides Cancel out H+ by adding OH- to both sides Cancel out extra waters etc. Cancel out extra waters etc. Add electrons to the more positive side Add electrons to the more positive side Balancing in alkaline conditions alters the method. Look for the clues in the question To combine the 2 half equations, multiply them up so the number of electrons are the same. Cancel like terms

9. Titration Recap n = cV n = moles (mol) c = concentration (molL -1 ) V = volume (L) m = cV m = mass (g) c = concentration (gL -1 ) V = volume (L) n = m/M m = mass (g) M = molar mass (gmol -1 ) n = moles (mol) To convert gL -1 to molL -1 = gL -1 /M = molL -1 To convert molL -1 to gL -1 = molL -1 X M = gL -1

10. Steps in titration calculations 1.Write balanced equation 2.Calculate the amount of known substance 3.Use equation to work out unknown substance 4.Calculate concentration/mass of unknown Calculate conc. Of HCl if 20.0 mL reacts with 21.7 mL of 0.0492 molL -1 sodium carbonate. Summarise what you have. Convert mL to L Na 2 CO 3 + 2HCl  2NaCl + CO 2 + H 2 O n(Na 2 CO 3 ) = 0.0492 X 21.7X10 -3 n = 1.068 X 10 -3 mol Write down 1 more s.f. then you need but leave all numbers in calculator Use numbers from equation: unknown/known then X this by the n of known n(HCl) = 2 n(Na 2 CO 3 ) 1 n(HCl) = 2 X n(Na 2 CO 3 ) = 2.136 X 10 -3 mol C(HCl) = n/V = 2.136X10 -3 / 20.0X10 -3 = 0.107 molL -1

11. Balanced equation: Known:Unknown: Ratio: Volume:Volume: Conc:Conc: Amount:Amount:

12. Titration expectations AchievedMeritExcellence At least 2 titres fall within 0.6mL. Average titre within 0.6mL of expected At least 3 titres fall within 0.4mL. Average titre within 0.4mL of expected At least 3 titres fall within 0.2mL. Average titre within 0.2mL of expected Only titres within 0.6mL range used. Minor error in calculation allowed. Only titres within 0.4mL range used. Composition correctly determined. Only titres within 0.2mL range used. Composition correct with units and s.f

13. Titration calculations 25.0 mL samples of HCl solution were titrated against standard sodium carbonate solution of concentration 0.9955 molL-1. Titres of 29.3 mL, 27.6 mL, 27.8 mL and 27.8 mL were obtained. Calculate the concentration of the HCl solution Balanced equation: Na 2 CO 3 + 2HCl  2NaCl + CO 2 + H 2 O Known: Na 2 CO 3 Unknown:HCl Ratio: 1: 2 Volume: 27.73 X10 -3 L Volume: 25.0 X10 -3 L Conc: 0.9955 molL -1 Conc: 2.21 molL -1 Amount: 2.761 X10 -2 mol Amount: 5.521 X10 -2 mol n(HCl) = 2 n(Na 2 CO 3 ) 1 n(HCl) = 2 X n(Na 2 CO 3 )

14. Titration calculations Jaime titrated 10.0 mL aliquots of sulfuric acid against 0.01122 molL-1 sodium hydroxide solution. She got titres of: 19.82 mL, 19.62 mL, 18.65 mL, 19.68 mL, 19.65 mL. Calculate the concentration of sulfuric acid. Balanced equation: H 2 SO 4 + 2NaOH  Na 2 SO 4 + 2H 2 O Known:NaOHUnknown: H 2 SO 4 Ratio: 2: 1 Volume: 19.65 X10 -3 L Volume: 10.0 X10 -3 L Conc: 0.0112 molL -1 Conc: 0.0110 molL -1 Amount: 2.201 X10 -4 mol Amount: 1.100 X10 -4 mol n(H 2 SO 4 ) = 1 n(NaOH) 2 n(H 2 SO 4 ) = 0.5 X n(NaOH)

15. More complex titration problems Often used to calculate purity of a substance or % composition A 5.026g sample of an ore of iron was dissolved in 50mL of dilute sulfuric acid. The iron was converted to Fe 2+ (aq). The resulting solution was titrated against 0.064 02molL -1 KMnO 4 solution and required 30.68mL to oxidise all the iron. Calculate the mass of iron in the ore hence the percentage of iron in the ore. You need to pick out the values you can work with and figure out the ionic equation C(MnO 4 - )= 0.06402molL -1 V(MnO 4 - )=30.68 X 10 -3 L n = cv n(MnO 4 - )= 1.96413 X 10 -3 mol Keep numbers in calculator. Write 6 figures in standard form. Give final answer to 3 s.f. Write your 2 half equations to work out the full ionic equation Fe 2+  Fe 3+ + e - MnO 4 - + 8H + + 5e -  Mn 2+ +4H 2 O MnO 4 - + 8H + + 5Fe 2+  Mn 2+ +4H 2 O + 5Fe 3+ KU U/K X n(MnO 4 - ) 5 X n(MnO 4 - ) ------------1

16. Mole ratios 25.0 mL of diluted hydrogen peroxide solution reacts with 31.1 mL of 0.0184 mol L –1 MnO 4 – solution. 2MnO 4 - + 5H 2 O 2 + 6H +  2Mn 2+ + 5O 2 + 8H 2 O 20.0 mL of 0.114 mol L –1 oxalic acid reacted with 23.7 mL of MnO 4 – solution. 5C 2 O 4 2- + 2MnO 4 - + 16H +  10CO 2 + 2Mn 2+ + 8H 2 O

17. Titration problems with 2 equations A student prepared a 0.01636molL -1 solution of KBrO 3, then took 25.0mL samples of this solution, added 1g of KI crystals and 15mL dilute sulfuric acid, and titrated the liberated iodine against sodium thiosulfate solution. An average of 27.32mL of thiosulfate were required. What is the concentration of the thiosulfate? BrO 3 - + 6I - + 6H +  Br - + 3H 2 O + 3I 2 I 2 + 2S 2 O 3 2-  2I - + S 4 O 6 2- You need to figure out the known and unknown in each equation C(BrO 3 - )= 0.01636molL -1 V(BrO 3 - )=25 X 10 -3 L n = cv n(BrO 3 - )= 4.09000 X 10 -4 mol K U Means iodine is made in the first reaction that is then used in the second reaction U K U X U K K n(I 2 ) n(I 2 ) n(BrO 3 - ) n(S 2 O 3 2- ) n(S 2 O 3 2- ) n(I 2 ) n(I 2 ) 3 X 2 1 1 n(S 2 O 3 2- ) = 6 X n(BrO 3 - )

18. Mole ratios 20.0 mL of a 0.0175 mol L –1 KBrO 3 solution is reacted with KI and the iodine liberated reacted with 28.9 mL of thiosulfate solution. BrO 3 - + 6I - + 6H +  Br - + 3I 2 + 3H 2 O I 2 + 2S 2 O 3 2-  2I - + S 4 O 6 2- 10.0 mL of a Cu 2+ solution was reacted with KI and the iodine liberated titrated against 0.025 mol L –1 thiosulfate solution. 26.4 mL of thiosulfate were required. 2Cu 2+ + 4I -  2CuI + I 2 I 2 + 2S 2 O 3 2-  2I - + S 4 O 6 2-

19. Electrochemistry Copper nitrate Zinc

20. Electrochemistry

21. Electrochemistry Zinc metal disappears, blue colour fades and copper is deposited Zn (s) + Cu(NO 3 ) 2(aq) ➞ Zn(NO 3 ) 2(aq) + Cu (s) Zn (s) ➞ Zn 2+ (aq) + 2e - Zn (s) ➞ Zn 2+ (aq) + 2e - Cu 2+ (aq) + 2e - ➞ Cu (s) Cu 2+ (aq) + 2e - ➞ Cu (s) These 2 half-equations can occur in separate beakers as long as there’s a path for the electrons to travel (wire) and the ions to travel (salt bridge) These 2 half-equations can occur in separate beakers as long as there’s a path for the electrons to travel (wire) and the ions to travel (salt bridge)

22. Electrochemistry Zinc nitrate (Zn 2+ ) Zinc (Zn) Copper (Cu) Copper nitrate (Cu 2+ ) NO 3 -  NO 3 - K+ K+ K+ K+  Salt bridge allows ions to move to keep the system electrically neutral V e- e - flow from Zn leaving Zn 2+ in beaker Zn  Zn 2+ + 2e - Oxidation at anode e - flow to copper beaker to combine with Cu 2+ to form Cu Cu 2+ + 2e -  Cu Reduction at cathode Please note that electrons move from –ve to +ve. Anode is –ve in electrochemical cells!! Remember; RED CAT If voltage is +ve, e- move L to R. If voltage is –ve, this is reversed

23. Electrochemistry The voltmeter connecting the 2 half-cells is measuring the electromotive force (emf or E) The voltmeter connecting the 2 half-cells is measuring the electromotive force (emf or E) We can use emf figures to compare the strength of reductants and oxidants We can use emf figures to compare the strength of reductants and oxidants To compare you must use the standard half-cell with an emf of 0 (H + /H 2 ) and standard conditions: 25  C (298K), 1.0molL -1, 1.0atm (101.3 kPa) To compare you must use the standard half-cell with an emf of 0 (H + /H 2 ) and standard conditions: 25  C (298K), 1.0molL -1, 1.0atm (101.3 kPa) In these comparisons, the hydrogen half-cell is on the right-hand side and connected to the positive terminal of the voltmeter to that cell In these comparisons, the hydrogen half-cell is on the right-hand side and connected to the positive terminal of the voltmeter to that cell

24. Cell diagrams Zinc nitrate (Zn 2+ ) NO 3 -  NO 3 - K+ K+ K+ K+  Zinc (Zn) Copper (Cu) Copper nitrate (Cu 2+ ) Ve- Zn (s) /Zn 2+ (aq) //Cu 2+ (aq) /Cu (s) // represents the salt bridge / represents change of phase Write left hand electrode first as oxidation Write right hand electrode last as reduction

25. Zn(s)/ Zn 2+ (aq)// Ag + (aq)/ Ag(s) Pb(s)/ Pb 2+ (aq)// Fe 3+ (aq), Fe 2+ (aq)/ C(s)

26. Cu(s)/ Cu 2+ (aq)// MnO 4 – (aq), Mn 2+ (aq)/ C(s) Pt(s)/ Cl - (aq)/ Cl 2 (g)// BrO 3 - (aq),Br 2 / C(s)

27. Calculating E  (cell) E  (cell) = E  (red) – E  (ox) If E  (cell) is positive: cell on left experiences oxidation and cell on right experiences reduction. Reaction is spontaneous. If E  (cell) is positive: cell on left experiences oxidation and cell on right experiences reduction. Reaction is spontaneous. If E  (cell) is negative: cell on left experiences reduction and cell on right experiences oxidation. Reaction is non- spontaneous. If E  (cell) is negative: cell on left experiences reduction and cell on right experiences oxidation. Reaction is non- spontaneous.

28. Calculating E  (cell) Calculate the E  (cell) for the following cell, then write the overall cell equation. Zn(s)/Zn 2 +(aq)//Cu 2 +(aq)/Cu(s) E  (Zn 2+ /Zn) = -0.76VE  (Cu 2+ /Cu) = +0.34V E  (cell)is positive so occurs as written in the cell diagram: Zn  Zn 2+ + 2e - Cu 2+ + 2e -  Cu Zn + Cu 2+  Zn 2+ + Cu E  (cell) = E  (red) – E  (ox) = +0.34V –(-0.76V) = +0.34V –(-0.76V) = +1.10V = +1.10V - a – becomes a +! Reduction always on right There are many different methods but this one always works!

29. Which reaction is reduction? E  (cell) = E  (red) – E  (ox) = +1.33V –(-0.20V) = +1.33V –(-0.20V) = +1.53V = +1.53V positive: spontaneous. The reaction will occur Calculate the E  (cell) for the following cell. C(s)/C 2 O 4 2- (aq)/CO 2 (g)//Cr 2 O 7 2- (aq),Cr 3+ (aq)/C(s) E  (CO 2 /C 2 O 4 2- ) = -0.20VE  (Cr 2 O 7 2- (aq),Cr 3+ (aq)/C(s) = +1.33V

30. Will sulfur precipitate when H 2 S gas is bubbled through NiSO 4 solution? E  (S/H 2 S) = +0.17VE  (Ni 2+ /Ni) = -0.23V E  (cell) = E  (red) – E  (ox) = -0.23V –(+0.17V) = -0.23V –(+0.17V) = -0.40V = -0.40V Negative: Non-spontaneous. The reaction will not occur; sulfur will not precipitate H 2 S → S Ni 2+ → Ni H 2 S/S//Ni 2+ /Ni

31. Strongest & weakest Standard reduction potentials (E  Cell) are written with reductants on the right Standard reduction potentials (E  Cell) are written with reductants on the right Strongest reductant is the species on the right with the most negative E  Strongest reductant is the species on the right with the most negative E  Strongest oxidant is the species on the left with the most positive E  Strongest oxidant is the species on the left with the most positive E  Weakest oxidant = strongest reductant = most –ve Weakest oxidant = strongest reductant = most –ve Weakest reductant = strongest oxidant = most +ve Weakest reductant = strongest oxidant = most +ve