PF1009 2024 12 Redox Reactions PDF

Pharmaceutical Chemistry Redox Reactions Dr. J.J. Keating 1  Oxidation Numbers The effective charge of an atom The oxidation number of an elemental substance is zero The oxidation number of a monoatomic ion = charge numb...

Pharmaceutical Chemistry Redox Reactions Dr. J.J. Keating 1  Oxidation Numbers The effective charge of an atom The oxidation number of an elemental substance is zero The oxidation number of a monoatomic ion = charge number of that ion. Rules for assigning oxidation numbers The sum of the oxidation numbers of all atoms in the species is equal to its total charge. Atoms in their elemental form = 0 For elements of Group I, +1 For elements of Group II, +2 For elements of Group III (except B), +3 for M3+; +1 for M+ For elements of Group IV (except C, Si), +4 for M4+; +2 for M2+ For H, +1 in combination with non-metals, –1 in combination with metals For F, –1 in all its compounds For O, –2 unless combined with F, –1 in peroxides (O22–), –½ in superoxides (O2–), –⅓ in ozonides (O3–). 2  Redox Reactions Oxidation is electron loss. Reduction is electron gain. 2Mg + O2 → 2MgO A redox reaction is a reaction in which oxidation and reduction take place. Oxidising agent – removes electrons and becomes reduced in a reaction. An element in the oxidising agent undergoes a decrease in oxidation number. Reducing agent – supplies electrons and becomes oxidised in a reaction. An element in the reducing agent undergoes an increase in oxidation number. 2Mg + O2 → 2MgO (exothermic) 3  Redox Reactions in Acidic Solution O O HO OH MnO4– + H2C2O4 → Mn2+ + CO2 KMnO4 oxalic acid (H2C2O4) potassium permanganate ethane-1,2-dioic acid 1. Identify the oxidised and reduced species from the changes in oxidation numbers. 2. Write two skeletal equations for the half-reactions. 3. Balance by inspection all elements in the half-reactions except O, H and the charge. 4. Balance O atoms in each half-reaction by adding H2O. 5. Balance H atoms in each half-reaction by adding H+. 6. Balance the electric charges in each half-reaction by adding electrons. 7. Prepare the two half-reactions for summation by making the number of electrons in the same in both. To do this, you may have to initially independently multiply one or both reactions across the reaction by a factor. 8. Combine the two half-reactions by adding them together (treat them as adding two simultaneous equations) and simplify (combine common species across the balanced equation). 9. Check the equation is balanced by the numbers of each atoms and overall charge on each side of the equation. 2MnO4– + 6H+ + 5H2C2O4 → 2Mn2+ + 10CO2 + 8H2O 4  Redox Reactions in Basic Solution KMnO4 potassium permanganate MnO4– + Br– → MnO2 + BrO3– BrO3 – bromate As per balancing redox reactions in acidic aqueous solution, but add the following steps between Step 5 and 6 on the previous slide: After step 5, add the same number of OH– ions to both sides of each half-reaction as there are H+ in each half-reaction. Combine H+ and OH– in each half-reaction to form H2O. If, at this stage, H2O is present on both sides of each half-reaction, bring H 2O to one side of each half-reaction. Continue with Step 6 on the previous slide. 2MnO4– + H2O + Br– → 2MnO2 + 2OH– + BrO3– 5  Redox Reactions in Basic Solution NO3– Zn(OH4)2– NH3 Zn + NO3– → Zn(OH)42– + NH3 nitrate zinc tetrahydroxide ammonia The unbalanced redox reaction shown above takes place in aqueous base, and involves the reaction of zinc metal and nitrate ions, forming zinc tetrahydroxide and ammonia. The balanced reaction is: 4Zn + 7OH– + NO3– + 6H2O → 4Zn(OH)42– + NH3 Check the balanced reaction is correct: 4 x Zn → 4 x Zn 1xN → 1xN 16 x O → 16 x O 19 x H → 19 x H 8 e– → 8 e– 6  Element Activity Series List of elements arranged such that a metal can reduce the cations formed by any of the metals below it in the list. Zn can reduce Cu2+, but not vice versa. Zn + Cu2+ → Zn2+ + Cu 7