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1. OXIDATION REDUCTION (a) Addition of oxygen Removal of oxygen

Increase in O.S. Decrease in O.S. O xidation I s L oss, R eduction I s G ain. O I L R I G. IV REDOX REACTIONS. A. Definition. 1. OXIDATION REDUCTION (a) Addition of oxygen Removal of oxygen (b) Removal of hydrogen Addition of hydrogen

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1. OXIDATION REDUCTION (a) Addition of oxygen Removal of oxygen

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  1. Increase in O.S. Decrease in O.S. Oxidation Is Loss, Reduction Is Gain O I L R I G IV REDOX REACTIONS A. Definition 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)

  2. Zn gets oxidised. Cu2+ gets reduced. 2. A redox reactionconsists of 2 half-reactions: reduction and oxidation occurring simultaneously. e.g. Cu2+ + Zn ——— Cu + Zn2+ Reducing agent Oxidising agent

  3. 2. The oxidation number of an element in its simple ion is the charge of the ion. e.g. Fe2+ O.S. of Fe = +2 Cl- O.S. of Cl = -1 O2- O.S. of O = -2 B. Rules for assigning oxidation numbers (or oxidation states) 1. The oxidation number of an uncombined element is zero. e.g. Cl2 O.S. of Cl = 0 Mg O.S. of Mg = 0 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.) FeCl2: 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. CO2 molecule O.S. of C = +4 C formed 4 covalent bonds & is less electronegative than O. O = C = O • 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-)

  6. From the above rules, it can be deduced readily the usual oxidation number of (a) oxygen is –2, exceptions in F2O and peroxides, eg. H2O2, BaO2, NaO2, etc. (b) hydrogen is +1, exceptions in metal hydrides, eg. NaH, CaH2, etc. Eg 5.1 Give the oxidation number of the underlined element. SO32-, CrO42-, P4O10, S2O32-, S4O62- , N2O5

  7. SO32- ion: Let the ox. no. of S in SO32- be x. x + 3  (-2) = -2 x = +4

  8. CrO42-: Let the ox. no. of Cr in CrO42- be x. x + 4  (-2) = -2 x = +6

  9. S4O62- ion: Let the ox. no. of S in S4O62- be x. 4 x + 6  (-2) = -2 4x = +10 x = +2.5

  10. MnO4-(aq) —— Mn2+(aq) Fe2+(aq) —— Fe3+(aq) C. Balancing Redox Equations Steps for balancing redox equations in acid medium: e.g. MnO4- + Fe2+ Mn2+ + Fe3+ acidic medium (1) Identify the oxidation half-equation and reduction half-equation. [R] +2 +7 [O] +3 +2

  11. 5 (2) Balance the half-equation by (a) balancing the element that undergoes changes in oxidation number. (b) adding H2O(l) if there is insufficient oxygen, (c) adding H+(aq) if there is insufficient hydrogen, (d) adding e- to balance the charge. MnO4-(aq)  Mn2+(aq) + 4H2O(l) + 8H+(aq) + 5e- Fe2+(aq)  Fe3+(aq) + e- (3) Adding the 2 half-equations in a way to eliminate the electrons. {e- gained = e- lost} MnO4-(aq) + 8H+(aq) +5Fe2+(aq)  Mn2+(aq) +4H2O(l) + 5Fe3+(aq)

  12. Some important reactants and products in constructing half-equations in acid medium are given below. (a) MnO4-(aq)  Mn2+(aq) (b) Cr2O72-(aq)  Cr3+(aq) (c) Fe2+(aq)  Fe3+(aq) (d) Cl2(aq)  Cl-(aq) (e) I2(aq)  I-(aq) (f) H2O2(aq)  O2(aq) ; H2O2 acts as reducing agent. (g) H2O2(aq)  H2O(l) ; H2O2 acts as oxidising agent. (h) IO3-(aq)  I2(aq) (i) S2O32-(aq)  S4O62-(aq) ; iodine-thiosulphate reaction (j) C2O42-(aq)  CO2(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) MnO4- + Cl-  Mn2+ + Cl2 (b) MnO4- + H2O2  Mn2+ + O2 (c) H2O2 + I-  H2O + I2 (d) Cr2O72- + NO2-  Cr3+ + NO3- (e) PbO2(s) + Cl-(aq)  Pb2+(aq) + Cl2(g)

  14. D. Redox Titrations 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 FeSO4 . (NH4)2SO4 .xH2O were dissolved & made up to 250 cm3 of acidified aq. solution. 25.0 cm3 of this solution required 22.50 cm3 of 0.0150 mol dm-3 KMnO4 for complete reaction. Calculate x in the formula FeSO4 . (NH4)2SO4 .xH2O.

  16. E.g. (b) A standard solution is made by dissolving 1.145 g of K2Cr2O7 and making up to 250 cm3. A 25.0 cm3 portion is added to an excess of KI and dilute sulphuric acid. The iodine liberated is titrated with sodium thiosulphate solution. 34.70 cm3 of thiosulphate solution is needed. Calculate the conc. of the thiosulphate ion in mol dm-3.

  17. E.g. (c) 10.0 cm3 of a given solution of hydrogen peroxide was diluted to 250 cm3 with distilled water. 25.0 cm3 of this diluted hydrogen peroxide solution then required 16.70 cm3 of 0.0250 mol dm-3 KMnO4 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 dm3 at s.t.p., given off when by 1 dm3 of the solution is decomposed according to the equation 2H2O2(aq)  2H2O(l) + O2(g)]

  20. E.g. (d) 25.0 cm3 of a solution containing ethanedioic acid and sodium ethanedioate required 15.55 cm3 of 0.100 mol dm-3 NaOH solution for neutralisation. Another 25.0 cm3 of the solution which contained the ethanedioic acid and sodium ethanedioate required 34.55 cm3 of 0.0205 mol dm-3 KMnO4 solution for oxidation in acidic conditions at about 60oC. Calculate the mass of each (anhydrous) constituent per dm3 of the solution containing ethanedioic acid and sodium ethanedioate. [RAM values : H=1, C=12, O=16, Na=23]

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