1. 1. OXIDATION REDUCTION (a) Addition of oxygen Removal of oxygen (b) Removal of hydrogen Addition of hydrogen (c) Loss of electron Gain of electron (d) Increase in O.S.Decrease in O.S. IVREDOX REACTIONS A. Definition O I L R I G Oxidation Is Loss, Reduction Is Gain

2. 2. A redox reaction consists of 2 half-reactions: reduction and oxidation occurring simultaneously. e.g. Cu 2+ + Zn ———  Cu + Zn 2+ Cu 2+ gets reduced. Zn gets oxidised. Oxidising agent Reducing agent

3. B.Rules for assigning oxidation numbers (or oxidation states) 1. The oxidation number of an uncombined element is zero. e.g. Cl 2 O.S. of Cl = 0 MgO.S. of Mg = 0 2. The oxidation number of an element in its simple ion is the charge of the ion. e.g. Fe 2+ O.S. of Fe= +2 Cl - O.S. of Cl= -1 O 2- O.S. of O= -2 Note the sign (+,-) is b4 the no.

4. 3. The sum of the oxidation number of all the atoms in a complex ion or compound is equal to the charge. e.g. OH - ion:sum of O.S= (-2) + (+1) = -1 (charge on the ion.) FeCl 2 :sum of O.S= (+2) + 2  (-1) = 0 (charge of neutral cpd=0)

5. 4. In covalent compounds, the oxidation number of an element is dependent on the number of bonds and its electronegativity. e.g. Electronegativity of elements power of element to attract electrons towards itself increases across the Periodic Table. (  no. of protons  greater tendency to attract e - ) decreases down the group (  atomic size  lower tendency to attract e - ) CO 2 molecule O.S. of C = +4 C formed 4 covalent bonds & is less electronegative than O. C =OO =

6. (a) oxygen is –2, exceptions in F 2 O and peroxides, eg. H 2 O 2, BaO 2, NaO 2, etc. (b) hydrogen is +1, exceptions in metal hydrides, eg. NaH, CaH 2, etc. From the above rules, it can be deduced readily the usual oxidation number of Eg 5.1 Give the oxidation number of the underlined element. SO 3 2- SO 3 2-, CrO 4 2-, P 4 O 10, S 2 O 3 2-, S 4 O 6 2-, N 2 O 5CrO 4S 4 O 6 2-

7. SO 3 2- ion: Let the ox. no. of S in SO 3 2- be x. x + 3  (-2) = -2 x = +4

8. CrO 4 2- : Let the ox. no. of Cr in CrO 4 2- be x. x + 4  (-2) = -2 x = +6

9. S 4 O 6 2- ion: Let the ox. no. of S in S 4 O 6 2- be x. 4  x + 6  (-2) = -2 4x = +10 x = +2.5

10. Steps for balancing redox equations in acid medium: (1) Identify the oxidation half-equation and reduction half-equation. e.g.MnO 4 - + Fe 2+  Mn 2+ + Fe 3+ acidic medium C.Balancing Redox Equations Fe 2+ (aq) ——  Fe 3+ (aq) MnO 4 - (aq) ——  Mn 2+ (aq) [R] [O] +7 +2 +3

11. MnO 4 - (aq)  Mn 2+ (aq) ( a) balancing the element that undergoes changes in oxidation number. (2) Balance the half-equation by (3) Adding the 2 half-equations in a way to eliminate the electrons. {e - gained = e - lost} (b) adding H 2 O(l) if there is insufficient oxygen, + 4H 2 O(l) (c) adding H + (aq) if there is insufficient hydrogen, + 8H + (aq) (d) adding e - to balance the charge. + 5e - + e - MnO 4 - (aq) + 8H + (aq) +5Fe 2+ (aq)  Mn 2+ (aq) +4H 2 O(l) + 5Fe 3+ (aq) 5 5 Fe 2+ (aq)  Fe 3+ (aq)

12. Some important reactants and products in constructing half-equations in acid medium are given below. (a) MnO 4 - (aq)  Mn 2+ (aq) (b) Cr 2 O 7 2- (aq)  Cr 3+ (aq) (c) Fe 2+ (aq)  Fe 3+ (aq) (d) Cl 2 (aq)  Cl - (aq) (e) I 2 (aq)  I - (aq) (f) H 2 O 2 (aq)  O 2 (aq) ; H 2 O 2 acts as reducing agent. (g) H 2 O 2 (aq)  H 2 O(l) ; H 2 O 2 acts as oxidising agent. (h) IO 3 - (aq)  I 2 (aq) (i) S 2 O 3 2- (aq)  S 4 O 6 2- (aq) ; iodine-thiosulphate reaction (j) C 2 O 4 2- (aq)  CO 2 (g) ; oxidation of ethanedioate ion E.g. Construct a balanced half-equation for each of the above reactions.

13. E.g. Balance the following equations using the half- equation method (include state symbols). (a) MnO 4 - + Cl -  Mn 2+ + Cl 2 (b) MnO 4 - + H 2 O 2  Mn 2+ + O 2 (c) H 2 O 2 + I -  H 2 O + I 2 (d) Cr 2 O 7 2- + NO 2 -  Cr 3+ + NO 3 - (e) PbO 2 (s) + Cl - (aq)  Pb 2+ (aq) + Cl 2 (g)

14. D.Redox Titrations Oxidising agent Reducing agent Indicator (a) KMnO 4 / H + Fe 2+, C 2 O 4 2-, H 2 O 2, FeC 2 O 4 None required (end-pt: (b) K 2 CrO 7 / H + Fe 2+ N-phenylanthranilic acid (end-pt: (c) I 2 (liberated from rxn bet. I - and OA) S 2 O 3 2- Starch (end-pt: or None (end-pt: pink green to reddish-violet dark blue to colourless pale yellow to colourless

15. E.g. (a) 8.49 g of iron(II) ammonium sulphate crystals FeSO 4. (NH 4 ) 2 SO 4. xH 2 O were dissolved & made up to 250 cm 3 of acidified aq. solution. 25.0 cm 3 of this solution required 22.50 cm 3 of 0.0150 mol dm -3 KMnO 4 for complete reaction. Calculate x in the formula FeSO 4. (NH 4 ) 2 SO 4. xH 2 O.

16. E.g. (b) A standard solution is made by dissolving 1.145 g of K 2 Cr 2 O 7 and making up to 250 cm 3. A 25.0 cm 3 portion is added to an excess of KI and dilute sulphuric acid. The iodine liberated is titrated with sodium thiosulphate solution. 34.70 cm 3 of thiosulphate solution is needed. Calculate the conc. of the thiosulphate ion in mol dm -3.

17. E.g. (c) 10.0 cm 3 of a given solution of hydrogen peroxide was diluted to 250 cm 3 with distilled water. 25.0 cm 3 of this diluted hydrogen peroxide solution then required 16.70 cm 3 of 0.0250 mol dm -3 KMnO 4 for titration in acidic conditions. Calculate the (i) conc. of the diluted hydrogen peroxide solution in mol dm -3.

18. (ii) conc. of the original hydrogen peroxide solution in mol dm -3.

19. (iii) volume strength of the original hydrogen peroxide solution. [Volume strength is the volume of oxygen, in dm 3 at s.t.p., given off when by 1 dm 3 of the solution is decomposed according to the equation 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g)]

20. E.g. (d) 25.0 cm 3 of a solution containing ethanedioic acid and sodium ethanedioate required 15.55 cm 3 of 0.100 mol dm -3 NaOH solution for neutralisation. Another 25.0 cm 3 of the solution which contained the ethanedioic acid and sodium ethanedioate required 34.55 cm 3 of 0.0205 mol dm -3 KMnO 4 solution for oxidation in acidic conditions at about 60 o C. Calculate the mass of each (anhydrous) constituent per dm 3 of the solution containing ethanedioic acid and sodium ethanedioate. [RAM values : H=1, C=12, O=16, Na=23]