Summary

These notes cover redox reactions, including definitions, examples, and how to identify oxidizing and reducing agents. It details reactions like the oxidation of magnesium and reduction of copper (II) oxide, along with calculating oxidation numbers.

Full Transcript

PAKISTAN INTERNATIONAL SCHOOL, JEDDAH (ES) Unit: 11 REDOX REACTIONS MNEMONIC – INTENDED TO ASSIST THE MEMORY OIL ---------------------- Oxidation is loss (of electron and hydrogen) RIG ---------------------- Reduction is gain (of electron and hydr...

PAKISTAN INTERNATIONAL SCHOOL, JEDDAH (ES) Unit: 11 REDOX REACTIONS MNEMONIC – INTENDED TO ASSIST THE MEMORY OIL ---------------------- Oxidation is loss (of electron and hydrogen) RIG ---------------------- Reduction is gain (of electron and hydrogen) Example: Oxidation of Magnesium: When Magnesium burns in oxygen, Magnesium oxide is formed. As Magnesium has gained oxygen so we say that Magnesium has been oxidized. Example: Reduction of Copper (II) Oxide: We pass hydrogen/natural gas over heated Copper (II) oxide. The black copper (II) oxide changes to pink copper. As oxygen is removed from the copper (II) oxide so we say that copper (II) oxide has been reduced by hydrogen. The hydrogen is the reducing agent because it has removed the oxygen from copper (II) oxide, it has become oxidized. The copper (II)oxide is the oxidizing agent because it gives its oxygen to hydrogen to form water, it has been reduced. Oxidation state: a number assigned to an element in chemical combination Number represents the number of electrons lost or gained by an atom of that element in the compound. Rules for assigning oxidation state: The oxidation state of an uncombined element / molecule is 0 e.g, Mg , O2. In an ionic compound, the oxidation state of the element is the same as the charge on its ions. MgO = Mg+2 O-2 =0 The oxidation states in a compound add up to zero. H2O = 2H+1O-2 = 0 Increase in oxidation state is known as oxidation. Example: Fe+2 → Fe+3 Decrease in oxidation state is known as reduction. Example: Cl2 → 2Cl-1 Half Ionic equations: In order to predict which substance has oxidised or reduced we have to look at the oxidation states of the atoms involved and it is done by writing; i. Half equations for oxidation and reduction. ii. Over all ionic equation Example: Cl2 + 2KBr → 2KCl + Br2 Identify the oxidising and reducing agent. 0 decreases to -1 Cl2 + 2KBr → 2KCl + Br2 -1 increases to 0 Cl2 + 2e- → 2Cl-1 (reduction) 2Br-1 → Br2 + 2e- (oxidation) Over all ionic equation Cl2 + 2Br-1 → 2Cl-1 + Br2

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