1. Balancing Chemical Equations A chemical reaction is a process by which one set of chemicals is transformed into a new set of chemicals. A chemical equation uses standard chemical symbols to describe the changes occurring during a reaction General form:reactants  products Example: Water Skeletal equationH 2 + O 2  H 2 O Hydrogen and oxygen combine to form water. But the equation is not balanced. There are 2 oxygen atoms on the reactant side of the equation, but only one on the product side. If we place a 2 in front of the water on the product side, we will balance the oxygens. H 2 + O 2  2H 2 O

2. We now need to balance the hydrogens  Place a 2 before the hydrogens on the reactant side of the equation 2H 2 + O 2  2H 2 O Balanced Note: Diatomic molecules Equation for formation of water included hydrogens and oxygens with subscript 2. Other elements which occur in this way are Nitrogen, Fluorine, Chlorine, Bromine and Iodine. These elements occur naturally as diatomic (meaning 2 atoms) molecules The stoichiometric coefficients multiplying the chemical formulas tell you the relative numbers of moles of each substance that reacts or is produced in a chemical reaction. Therefore, we can conclude from the balanced equation for water that 2 moles of hydrogen and one mol of oxygen combine to form water.

3. A chemical equation may also tell you what physical state the reactants and products are in. The following state symbols are used: s (solid) l (liquid) g (gas) aq (aqueous solution) These are found as subscripts after each reactant and product. Example Balance the following equation: Na (s) + H 2 O (l)  NaOH (aq) + H 2(g) Step 1: How many atoms of each element are present on the reactant and product side? ReactantsProducts NaH O 121 HO 1 31

4. Let us balance the hydrogen atoms first Na + 2H 2 O  2NaOH + H 2 Na HO HO 14224 2 Next we balance the sodium atoms 2Na + 2H 2 O  2NaOH + H 2 Na HO HO 242 242 Step 2: Add state symbols 2Na (s) + 2H 2 O (l)  2NaOH (aq) + H 2(g)

5. Questions Balance the following equations: (a) P 2 O 5(s) + H 2 O (l)  H 3 PO 4(aq) Answer: P 2 O 5(s) + 3H 2 O (l)  2H 3 PO 4(aq) (b) NO 2(g) + H 2(g)  NH 3(g) + H 2 O (g) Answer: NO 2(g) + 3 ½ H 2(g)  NH 3(g) + 2H 2 O (g) (c) SO 2(g) + O 2(g)  SO 3(g) Answer: 2SO 2(g) + O 2(g)  2SO 3(g)

6. Balancing redox reactions Redox (meaning reduction-oxidation) reactions involves transfer of electrons Involves two processes oxidation is loss reduction is gain Can split a redox reaction into two half reactions: The oxidation reaction which involves a loss of electrons The reduction reaction which involves a gain of electrons To keep track of electrons in redox reactions, use oxidation numbers Oxidation number: the charge an atom has or appears to have when electrons are distributed according to certain rules OIL RIG Whenever oxidation occurs, reduction must also occur

7. Rules for assigning oxidation numbers (O.N.) 1.A free element (or uncombined element) is assigned O.N. of zero 2.The sum of O.N.s of atoms in a neutral compound must equal zero 3.The charge of a monoatomic ion is same as its oxidation number, e.g. Li + has O.N. of +1Fe +3 has O.N. of +3 4.In their compounds, group 1 elements have an O.N. of +1, group 2 elements have an O.N. of +2. 5.The sum of O.N. of atoms in polyatomic ion equals the charge on the ion 6.The O.N. of halogens is usually -1. Exception: when they are bonded to a more electonegative element 7.The O.N. of oxygen in most compounds is -2. Exceptions: H 2 O 2 and peroxide ion, O 2 -2 (O.N. of -1), and the compound OF 2 (O.N. +2) 8.The O.N. of hydrogen is +1, Exception: when bonded to a less electronegative element, i.e. a metal, where it’s O.N. is -1, e.g. LiH, NaH

8. Examples Assign oxidation numbers to all the elements in the following: (a) Li 2 O Lithium: Group 1 element  Oxidation number +1 Oxygen: Oxidation number -2 (b) HNO 3 Hydrogen: +1Oxygen: -2 HNO 3 is neutral  adding the O.N.s of the atoms together must equal zero (+1) + N + 3(-2) = 0  N = +5 (c) Cr 2 O 7 2- Sum of O.N.s equals charge on ion  here the sum equals -2 Oxygen: Oxidation number -2  2Cr + 7(-2) = -2 2Cr -14 = -2 Cr = +6

9. Questions Assign oxidation numbers to each of the following: (a) ZnCl 2 Answer: Zn: +2Cl: -1 (b) SO 3 Answer: O: -2S: +6 (c) MnO 4 - Answer: O: -2Mn: +7

10. Defining Oxidising and Reducing agents Remember that for an oxidation to occur, a reduction must also occur OIL RIG Oxidation can be defined (in terms of electon transfer) as the loss of electrons from a species Reduction can be defined as the gain of electrons by a species An oxidising agent (or electron acceptor) must itself be reduced A reducing agent (or electron donor) must itself be oxidised The action of both oxidising and reducing agents can be represented in the form of a half-equation

11. Identifying Oxidising and Reducing Agents Example Formation of calcium oxide (CaO) 2Ca (s) + O 2(g)  2CaO (s) Ca and O 2 (reactants): neutral atoms Calcium oxide (product): ionic compound Ca 2+ O 2- ions Can write the reaction as two half-reactions: 2Ca  2Ca 2+ O 2  2O 2- Here, calcium is oxidised; i.e. it loses electrons to oxygen It reduces the oxygen  it acts as a reducing agent Oxygen is reduced; i.e. it gains electrons from calcium It oxidises the calcium  it acts as an oxidising agent + 4e - OIL RIG

12. Example PbO (s) + CO (g)  Pb (s) + CO 2(g) Identify the oxidising and reducing agents in the following reaction: +2-2+2-20+4-2 Pb increases O.N. from +2 to 0 PbO is reduced; acts as oxidising agent C decreases O.N. from +2 to +4 CO is oxidised; acts as reducing agent Step 1: Assign oxidation numbers: PbO (s) + CO (g)  Pb (s) + CO 2(g) Step 2: Write half equations: Pb +2 + 2e - Pb 0 C +2 C +4 + 2e - Step 3: Identify oxidising and reducing agents:

13. Questions Identify the oxidising and reducing agents and write the half reactions in the following: (c) 4Fe +3O 2  2Fe 2 O 3 Answer: Fe  Fe 3+ + 3e - O 2 + 4e -  2O 2- O 2 : oxidising agent Fe: reducing agent (b) Si + 2F 2  SiF 4 Answer: Si  Si 4+ + 4e - F 2 + 2e -  2F - F 2 : oxidising agent Si : reducing agent (a) Cl 2 + 2NaBr  2NaCl + Br 2 Answer: 2Br --  Br 2 + 2e - Cl 2 + 2e -  2Cl - Cl 2 : oxidising agent Br - : reducing agent

14. Balancing redox equations Example Balance the equation showing the oxidation of Fe 2+ ions to Fe 3+ ions by dichromate ions (Cr 2 O 7 2- ) in an acidic medium Fe 2+ + Cr 2 O 7 2-  Fe 3+ + Cr 3+ Step 1: Identify oxidising and reducing agents and write half reactions Fe 2+ : +2 Cr 2 O 7 2- : 2Cr + 7(-2) = -2 2Cr = 12 Cr = +6 Fe 3+ : +3 Cr 3+ : +3 Fe 2+  Fe 3+ Iron loses one electron, i.e. it is oxidised. It donates one electron to dichromate, i.e. it is the reducing agent. Cr 2 6+  Cr 3+ Chromium gains three electrons, i.e. it is reduced. It gains its electrons from iron, i.e. it is the oxidising agent. Oxidation reaction: Fe 2+  Fe 3+ + e - Reduction reaction: Cr 2 6+ + 3e -  Cr 3+ Half equation:

15. Step 2: Balance each kind of atom other that H and O Fe 2+  Fe 3+ + e - Cr 2 O 7 2- + 3e -  2Cr 3+ Step 3: Balance oxygen atoms by using H 2 O Fe 2+  Fe 3+ + e - Cr 2 O 7 2- + 3e -  2Cr 3+ + 7H 2 O Step 4: Balance H atoms by using H + ions Fe 2+  Fe 3+ + e - 14H + + Cr 2 O 7 2- + 3e -  2Cr 3+ + 7H 2 O Step 5: Use electrons as needed to obtain a charge that is balanced Fe 2+  Fe 3+ + e - 14H + + Cr 2 O 7 2- +3e -  2Cr 3+ + 7H 2 O +9+6 Add three electrons to the reactant side to balance the charges Fe 2+  Fe 3+ + e - 14H + + Cr 2 O 7 2- + 6e -  2Cr 3+ + 7H 2 O Fe 2+ + Cr 2 O 7 2-  Fe 3+ + Cr 3+ +14-2-3+60+3+2

16. Step 6: Ensure the number of electrons gained equals the number of electrons lost and add the 2 half reactions together Fe 2+  Fe 3+ + e - 14H + + Cr 2 O 7 2- + 6e -  2Cr 3+ + 7H 2 O ×6 ×1 6Fe 2+  6Fe 3+ + 6e - 14H + + Cr 2 O 7 2- + 6e -  2Cr 3+ + 7H 2 O 14H + + Cr 2 O 7 2- + 6Fe 2+ + 6e -  2Cr 3+ + 7H 2 O + 6Fe 3+ + 6e - 14H + + Cr 2 O 7 2- + 6Fe 2+  2Cr 3+ +7H 2 O + 6Fe 3+ Balanced Check to ensure that all atoms and charges are balanced.

17. Step 1: Identify the oxidising and reducing agents, then write the half equations Step 2: Balance each atom, except H and O Step 3: Balance the O atoms (using H 2 O) Step 4: Balance the H atoms (using H + ) Step 5: Check the charges are balanced for each half equation Step 6: Ensure the numbers of electrons are equal (i.e. there are the same numbers lost and gained) Step 7: Add the two equations together Finally: check that all the charges and the numbers of atoms balance! Question Balance the equation for the redox reaction of MnO 4 - with Fe 2+ to produce Mn 2+ and Fe 3+ in an acidic medium. Procedure:

18. Question Balance the equation for the redox reaction of MnO 4 - with Fe 2+ to produce Mn 2+ and Fe 3+ in an acidic medium. Answer Step 1: Identify oxidising and reducing agents and write half reactions MnO 4 -  Mn 2+ Mn + 4(-2) = -1  Mn = +7 MnO 4 - gains 5 electrons. It acts as the oxidising agent as it is reduced. Mn : +2 MnO 4 - + 5e -  Mn 2+ Reduction reaction Fe 2+  Fe 3+ Fe loses an electron. It acts as the reducing agent as it is oxidised. Fe 2+  Fe 3+ + e - Oxidation reaction

19. Step 2: Balance each kind of atom other than H and O. Balanced in this case Step 3: Balance O atoms by adding H 2 O MnO 4 - + 5e -  Mn 2+ + 4H 2 O Fe 2+  Fe 3+ + e - Step 4: Balance H atoms by using H + ions 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O Fe 2+  Fe 3+ + e - Step 5: Use electrons as needed to obtain a charge that is balanced Fe 2+  Fe 3+ + e - 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O +2 Already balanced!

20. Step 6: Ensure the number of electrons gained equals the number of electrons lost and add the two half reactions together Fe 2+  Fe 3+ + e - 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O × 5 × 1 5Fe 2+  5Fe 3+ + 5e - 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O 8H + + MnO 4 - + 5Fe 2+ + 5e -  Mn 2+ + 5Fe 3+ + 4H 2 O + 5e - 8H + + MnO 4 - + 5Fe 2+  Mn 2+ + 5Fe 3+ + 4H 2 O Check to ensure that all atoms and charges are balanced. Balanced

21. Balance the following equation, which takes place in an acidic medium: Question MnO 4 - + SO 2  Mn 2+ + SO 4 2- Step 1: Identify oxidising and reducing agents and write half reactions MnO 4 - : Mn +4(-2) = -1 Mn - 8 = -1 Mn = +7 Mn 2+ : +2 MnO 4 - gains 5 electrons: it acts as the oxidising agent as it is reduced MnO 4 - + 5e -  Mn 2+ Reduction reaction SO 2 : S + 2(-2) = 0 S - 4 = 0 S = +4 SO 4 2- : S + 4(-2) = -2 S - 8 = -2 S = +6 SO 2 loses 2 electrons: it acts as the reducing agent as it is oxidised SO 2  SO 4 2- + 2e - Oxidation reaction

22. Step 2: Balance each kind of atom other than H and O Balanced in this case Step 3: Balance O atoms by using H 2 O MnO 4 - + 5e -  Mn 2+ + 4H 2 O 2H 2 O + SO 2  SO 4 2- + 2e - Step 4: Balance H atoms by using H + ions 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O 2H 2 O + SO 2  SO 4 2- + 2e - + 4H + Step 5: Use electrons as needed to obtain a charge that is balanced 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O +2 Already balanced! MnO 4 - + SO 2  Mn 2+ + SO 4 2- MnO 4 - + 5e -  Mn 2+ SO 2  SO 4 2- + 2e - -5+20+8

23. 2H 2 O + SO 2  SO 4 2- + 2e - + 4H + Already balanced 00 Step 6: Ensure number of electrons gained equals number of electrons lost and add the two half reactions together 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O × 2 2H 2 O + SO 2  SO 4 2- + 2e - + 4H + × 5 16H + + 2MnO 4 - + 10e -  2Mn 2+ + 8H 2 O 10H 2 O + 5SO 2  5SO 4 2- + 10e - + 20H + 16H + + 2MnO 4 - + 10H 2 O + 5SO 2  2Mn 2+ + 8H 2 O + 5SO 4 2- + 20H + 2MnO 4 - + 2H 2 O + 5SO 2  2Mn 2+ + 5SO 4 2- + 4H + 0 0-2 +4 Balanced Check to ensure that all atoms and charges are balanced.

24. Questions Write balanced equations to represent the following reactions in an acidic solution: (a) I - + SO 4 2-  I 2 + S Answer: 6I - + 8H + + SO 4 2-  3I 2 + S + 4H 2 O (b) MnO 4 - + H 2 C 2 O 4  Mn 2+ + CO 2 Answer: 2MnO 4 - + 5H 2 C 2 O 4 + 6H +  2Mn 2+ + 10CO 2 + 8H 2 O (c) ClO 3 - + Cl -  Cl 2 + ClO 2 Answer: 2Cl - + 2ClO 3 - + 4H +  Cl 2 + 2ClO 2 + 2H 2 O

25. In basic solution Example Write a balanced equation to represent the oxidation of iodide ion (I - ) by permanganate ion (MnO 4 - ) in basic solution to yield molecular iodine (I 2 ) and manganese (IV) oxide (MnO 2 ) Step 1: Identify oxidising and reducing agents and write half equations I -  I 2 O.N. of I - is -1 O.N of I 2 is 0 I - loses an electron: it acts as the reducing agent as it is oxidised I -  I 2 + e - Oxidation reaction MnO 4 -  MnO 2 MnO 4 - : Mn + 4(-2) = -1 Mn = +7 MnO 2 : Mn + 2(-2) = 0 Mn = +4 MnO 4 - gains 3 electrons: it acts as the oxidising agent as it is reduced MnO 4 - + 3e -  MnO 2 Reduction reaction Skeletal equation: I - + MnO 4 - I 2 + MnO 2

26. Step 2: Balance each kind of atom other than H and O 2I -  I 2 + e - MnO 4 - + 3e -  MnO 2 Step 3: Balance the O atoms by using H 2 O 2I -  I 2 + e - MnO 4 - + 3e -  MnO 2 + 2H 2 O Step 4: Balance H atoms by using H + ions 2I -  I 2 + e - 4H + + MnO 4 - + 3e -  MnO 2 + 2H 2 O

27. Step 5: Since the reaction occurs in a basic medium, for each H + ion we add an equal number of OH - ions to both sides of the equation. Where H + and OH - ions appear on the same side of the equation, they may be combined to form H 2 O. 4H + + 4OH - + MnO 4 - + 3e -  MnO 2 + 2H 2 O + 4OH - 4H 2 O + MnO 4 - + 3e -  MnO 2 + 2H 2 O + 4OH - 2H 2 O + MnO 4 - + 3e -  MnO 2 + 4OH - Step 6: Use electrons as needed to obtain a charge that is balanced 2I -  I 2 + e - -2 Add one electron to the product side to balance charges 2I -  I 2 + 2e - 2H 2 O + MnO 4 - + 3e -  MnO 2 + 4OH - -4 Already balanced! -20 -30-40

28. Step 7: Ensure the number of electrons gained equals the number of electrons lost and add the two half reactions together 2I -  I 2 + 2e - × 3 2H 2 O + MnO 4 - + 3e -  MnO 2 + 4OH - × 2 6I -  3I 2 + 6e - 4H 2 O + 2MnO 4 - + 6e -  2MnO 2 + 8OH - 2MnO 4 - + 6I - + 4H 2 O  2MnO 2 + 3I 2 + 8OH - Balanced Check to ensure that all atoms and charges are balanced.

29. Question Balance the following redox equation which occurs in basic solution Mn 2+ + H 2 O 2  MnO 2 + H 2 O Step 1: Identify the oxidising and reducing agents and write half reactions Mn 2+  MnO 2 Mn 2+ : +2MnO 2 : Mn +2(-2) = 0 Mn = +4 Mn 2+ loses two electrons: it acts as the reducing agent as it is oxidised Mn 2+  MnO 2 + 2e - Oxidation reaction H 2 O 2  H 2 O H 2 O 2 : 2(+1) + 2O = 0 O = -1 H 2 O: 2(+1) + O = 0 O = -2 H 2 O 2 gains one electron: it acts as the oxidising agent as it is reduced H 2 O 2 + e -  H 2 OReduction reaction

30. Step 2: Balance each kind of atom other than H and O. Already balanced in this case Step 3: Balance the O atoms by using H 2 O 2H 2 O + Mn 2+  MnO 2 + 2e - H 2 O 2 + e -  2H 2 O Step 4: Balance the H atoms by using H + 2H 2 O + Mn 2+  MnO 2 + 2e - + 4H + 2H + + H 2 O 2 + e -  2H 2 O Step 5: For each H + ion, add equal no. of OH - to both sides of equation 2H 2 O + Mn 2+ + 4OH -  MnO 2 + 2e - + 4H + + 4OH - 2H 2 O + Mn 2+ + 4OH -  MnO 2 + 2e - + 4H 2 O Mn 2+ + 4OH -  MnO 2 + 2e - + 2H 2 O Mn 2+  MnO 2 + 2e - H 2 O 2 + e -  H 2 O

31. 2H + + 2OH - + H 2 O 2 + e -  2H 2 O + 2OH - 2H 2 O + H 2 O 2 + e -  2H 2 O + 2OH - H 2 O 2 + e -  2OH - Step 6: Use electrons as needed to obtain a charge that is balanced Mn 2+ + 4OH -  MnO 2 + 2e - + 2H 2 O -2 H 2 O 2 + e -  2OH - -2 Add one electron to the reactant side to balance charges H 2 O 2 + 2e -  2OH - Step 7: Ensure no. of electrons gained equals no. of electrons lost and add the half reactions together Mn 2+ + 4OH -  MnO 2 + 2e - + 2H 2 O H 2 O 2 + 2e -  2OH - Mn 2+ + H 2 O 2 + 2OH -  MnO 2 + 2H 2 O +2-40-20 0

32. Questions Write balanced equations to represent the following reactions in a basic solution (a) Fe(OH) 2 + MnO 4 -  MnO 2 + Fe(OH) 3 Answer: 3Fe(OH) 2 + MnO 4 - + 2H 2 O  MnO 2 + 3Fe(OH) 3 + OH - (b) Bi(OH) 3 + SnO 2 2-  SnO 3 2- + Bi Answer: 2Bi(OH) 3 + 3SnO 2 2-  2Bi + 3H 2 O + 3SnO 3 2-