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)

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)