AS Chemistry Oxidation & Reduction PDF

AS CHEMISTRY 3.1.7 OXIDATION & REDUCTION COMMON OXIDATION STATES POSITIVE NEGATIVE GROUP 1 = +I F = -I GROUP 2 = +II...

AS CHEMISTRY 3.1.7 OXIDATION & REDUCTION COMMON OXIDATION STATES POSITIVE NEGATIVE GROUP 1 = +I F = -I GROUP 2 = +II O = -II H = +I Cl = -I Ag = +I Br = -I Zn = +II I = -I Pb = +II or +IV N = -III Al = + III S = -II (Transition metals are variable) P = -III Fe = +II or +III Cu = +II (sometimes +I) C = +II or +IV Most common oxidation states, but VARIABLE depending on what they are bonded to. They may even be POSITIVE when bonded to a more electronegative element. i.e. F or O RULES FOR DEDUCING OXIDATION STATES How To Deduce Oxidation States RULES 1. All elements have an oxidation state of 0 2. The overall oxidation state of a compound = 0 3. The overall oxidation state of an ion = the charge AQA www.chemistrycoach.co.uk © scidekick ltd 2024 AS CHEMISTRY 3.1.7 OXIDATION & REDUCTION OXIDATION & REDUCTION Many reactions that you come across are REDOX reactions. This means that electrons are transferred during the reaction and, therefore, the oxidation states of some elements change. Remember OIL RIG In any reaction, compare the oxidation states of the elements in the products compared to the reactants. Oxidation An element has been OXIDISED if its oxidation state has Is become more positive / less negative. Loss of electrons An element has been REDUCED if its oxidation state has become more negative / less positive. You can then say if those elements have gained or lost Reduction electrons. Is The change in oxidation state related directly to the number of electrons that have been transferred. Gain of electrons e.g. if the oxidation state has increased by 2, it has lost 2 electrons If the oxidation state has decreased by 3, it has gained 3 electrons OXIDISING & REDUCING AGENTS In exam questions, you may be asked to identify the “oxidising agent” or “reducing agent”. How To Identify Oxidising Oxidising Agents & Reducing Agents REMOVE electrons from another element and gains them itself. Hence, the oxidising agent in any reaction is the one that has been reduced. Reducing Agents DONATES electrons to another element and loses them itself. Hence, the reducing agent in any reaction is the one that has been oxidised. AQA www.chemistrycoach.co.uk © scidekick ltd 2024 AS CHEMISTRY 3.1.7 OXIDATION & REDUCTION HALF-EQUATIONS Half-equations are way of showing the movement of electrons in an equation for REDOX reaction. Take the following REDOX reaction: Ca + CuSO4 → CaSO4 + Cu Oxidation States: 0 +II +II 0 Ca is oxidised from 0 to +II. This means it has lost 2 electrons. Half-Equation: Ca → Ca2+ + 2e- Cu is reduced from +II to 0. This means it has gained 2 electrons. Half-Equation: Cu2+ + 2e- → Cu So, a half-equation literally shows half of an equation, showing which species gain and lose electron, and how many. How To Write Half-Equations AQA www.chemistrycoach.co.uk © scidekick ltd 2024 AS CHEMISTRY 3.1.7 OXIDATION & REDUCTION REDOX IN ACIDIC CONDITIONS Sometimes, a REDOX reaction requires acidic conditions (H+(aq)) in order to take place. This is because the oxidising agent contains oxygen. The H+ ions are needed in order to combine with this oxygen to produce H2O as a product. One example is potassium manganate (VII), KMnO4, which is a common oxidising agent. Here, we ignore the K as it doesn’t get involved in the reaction. So we show it in equation as MnO4- 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O +II +VII +III +II 5Fe are oxidised from +II to +III. Each Fe2+ loses one electron, so this means they have lost 5 electrons in total. 5Fe2+ → 5Fe3+ + 5e- Mn is reduced from +VII to +II. This means it has gained 5 electrons. However, it needs H+ ions in order to produce H2O with the oxygen in the MnO4- ion. MnO4- + 5e- + 8H+ → Mn2+ + 4H2O AQA www.chemistrycoach.co.uk © scidekick ltd 2024 AS CHEMISTRY 3.1.7 OXIDATION & REDUCTION COMBINING HALF-EQUATIONS In the exam, you could be asked to take two half equations and combine them to write the overall equation for the reaction. e.g. K reacts with Cu2+ ions in a REDOX reaction to produce K+ and Cu. K → K+ + e- Cu2+ + 2e- → Cu 1. Balance the electrons in the two equations. In this case, we must double everything in the potassium half equation. We now have… 2K → 2K+ + 2e- Cu2+ + 2e- → Cu 2. Now the electrons are balanced, we literally combine the two equations. Both sets of reactants on one side an both sets of products on the other… 2K + Cu2+ + 2e- → 2K+ + 2e- + Cu 3. The electrons now cancel, so we can remove them from the equation to give the overall equation for the reaction… 2K + Cu2+ → 2K+ + Cu How To Combine Half-Equations AQA www.chemistrycoach.co.uk © scidekick ltd 2024