The document defines redox reactions as reactions involving the transfer of electrons between reactants. It defines oxidation as the loss of electrons and reduction as the gain of electrons. It provides examples of writing half reactions and ionic equations for redox reactions involving magnesium and hydrochloric acid.

1. Redox Reactions in terms of electrons transfer

2. Oxidation = loss of electrons A substance that loses electrons is said to be oxidised. Reduction = gain of electrons. A substance that gains electrons is said to be reduced. Remember LEO ( L oss of E lectrons is O xidation) GER ( G ain of E lectrons is R eduction) OR OIL ( O xidation I s L oss of electrons) RIG ( R eduction I s G ain of electrons)

3. Example (In terms of electrons) 1. Mg (s) + 2HCl (aq) -> MgCl 2 (aq) + H 2 (g) Ionic eqn: Recall the steps in writing ionic eqn First, split all compounds which are in (aq) into their ions, do NOT split all those in (s), (l) or (g) as they do not contain free ions. Mg (s) + 2H + (aq) + 2Cl - -> Mg 2+ (aq) + 2Cl - (aq) + H 2 (g) Cancel ions which appear on both sides Mg (s) + 2H + (aq) + 2Cl - -> Mg 2+ (aq) + 2Cl - (aq) + H 2 (g) Final eqn: Mg (s) + 2H + (aq) -> Mg 2+ (aq) + H 2 (g)

4. Example (In terms of electrons) 1. Mg (s) + 2H + (aq) -> Mg 2+ (aq) + H 2 (g) What happens to Mg in the reaction? From the ionic eqn, we can see that Mg has become Mg 2+ , this conversion must have involved the transfer of electrons, and can be represented by means of a half-eqn . What are half equations? Half equations are used to identify and represent redox reactions. In the writing of half-eqn, The atoms and charges must be balanced. Electrons are used to balance the charges on the two sides of the equation. They can be combined to give an overall ionic eqn for the reaction.

5. Steps in writing half eqn: First from the half eqn, we know that Mg has become Mg 2+ So we write, Mg (s) -> Mg 2+ (aq) Check the charges on both sides of eqn Mg (s) -> Mg 2+ (aq) Notice that Mg being a atom has a charge of 0 while Mg 2+ has a charge of +2. To balance the eqn, charges on both sides must be balanced, and this is done by adding electrons to the eqn. Recall electrons carry negative charge, so to balance the eqn, 2 electrons must be added to the right hand side of the eqn. Mg (s) -> Mg 2+ (aq) + 2e - Final half-eqn Example (In terms of electrons) 0 +2 0 +2 -2

6. The electrons can be moved to the left hand side of eqn and the eqn becomes Mg (s) - 2e - -> Mg 2+ (aq) Example (In terms of electrons) It can be seen that each magnesium atom loses 2 electrons to form a magnesium ion, hence magnesium is oxidised.

7. Which substance is reduced? Write the half eqn which involves the conversion of H + to H 2 Steps: Balance the no. of H on both sides. 2 H + (aq) -> H 2 (g) Check the charges on both sides of eqn 2H + (aq) -> H 2 (g) To balance the charges on both sides of eqn, electrons must be added to the left hand side of the eqn. 2H + (aq) + 2e - -> H 2 (g) The half-eqn is now complete. From the half eqn, we see that hydrogen ions have gained electrons to form hydrogen gas, hence hydrochloric acid is being reduced. Example (In terms of electrons) +2 0 +2 0 -2

8. Example (In terms of electrons) Note: Addition of the 2 half-eqns give the overall ionic eqn for the reaction Mg (s) -> Mg (aq) + 2e - Eqn 1 2H + (aq) + 2e - -> H 2 (g) Eqn 2 Adding Eqn 1 and 2, we get Mg (s) + 2H + (aq) + 2e - -> Mg (aq) + 2e - + H 2 (g) Notice 2 electrons appear on both sides of eqn and should be cancelled to yield the final ionic eqn: Mg (s) + 2H + (aq) -> Mg (aq) + H 2 (g)

9. Practice 1 (Pg 3) Chlorine reacts with iron (II) ions according to the two half-eqn below: Cl 2 + 2e - -> Cl 2 Eqn 1 Fe 2+ -> Fe 3+ + e - Eqn 2 Write the overall ionic eqn for the reaction between chlorine and iron (III) ions. Notice eqn 1 involves 2e - while eqn 2 involves 1 e - . Hence there is a need to multiply eqn 2 by 2 throughout so that it has the same no. of electron as eqn 1. 2 Fe 2+ -> 2 Fe 3+ + 2 e - Now add the two eqns together, we get Cl 2 + 2e - + 2Fe 2+ -> Cl 2 + 2Fe 3+ + 2e - Cancel the 2e - on both sides, we will get the final eqn to be: Cl 2 + 2Fe 2+ -> Cl 2 + 2Fe 3+

10. Practice 1 (Pg 3) Under acidic medium, manganate (VII) ion, MnO 4 - , can be converted to manganese (II) ions. Write a half-eqn for the conversion. Based on the information given, write the following eqn: MnO 4 - -> Mn 2+ Notice that the left hand side (LHS) of eqn has 4 oxygen atoms while right hand side (RHS) has none. Balance the oxygen atoms by putting 4H 2 O molecules on the right hand side of the eqn. MnO 4 - -> Mn 2+ + 4H 2 O Notice that now RHS has 8 H atoms while LHS has none. Add 8H + (Q says under acidic medium, so it is OK to add H + ) to the LHS of eqn. MnO 4 - + 8H + -> Mn 2+ + 4H 2 O Now check the charge on both sides of eqn. LHS: -1+8 =+7 while RHS: +2 To balance the charges, add 5e - to the LHS of eqn. We get: MnO 4 - + 8H + + 5e - -> Mn 2+ + 4H 2 O

11. Practice 1 (Pg 3) For the following reaction, identify the substance that is being oxidised and reduced. Explain your answer in terms of gain/loss of electrons and write half equations to illustrate the reduction and oxidation process. Zn (s) + CuSO 4 (aq) -> ZnSO 4 (aq) + Cu (s) Half eqn for oxidation: Zn (s) -> Zn 2+ (aq) + 2e - Zinc is oxidised because each atom of zinc loses 2 electrons to form zinc ions . Half eqn for reduction: Cu 2+ (aq) + 2e - -> Cu (s) Copper (II) sulfate is reduced because each copper (II) ion gains 2 electrons to form a copper atom .