Chapter 7 - Oxidation, reduction and redox reactions

Naomi Moylan-Torke
Note by Naomi Moylan-Torke, updated more than 1 year ago More Less
Naomi Moylan-Torke
Created by Naomi Moylan-Torke almost 4 years ago
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AS - Level AS Chemistry (Physical Chemistry) Note on Chapter 7 - Oxidation, reduction and redox reactions, created by Naomi Moylan-Torke on 02/07/2016.

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7.1 - Oxidation and reduction

7.1 - Oxidation and reductionRedox reactions:-short for reduction-oxidation-if one reagent is reduced, it itself is an oxidising agentAlso called electron transfer reactions.-if a reagent has been oxidised, it has lost elections-if a reagent has been reduced, it has gain electrons (originally removal of oxygen)copper and oxygen -> copper oxide Cu -> Cu(2+) + 2e- because Cu now has a positive charge, meaning it has lose electrons, it has been OXIDISEDremember: the higher the negative number, the MORE it has been REDUCEDOxidationIsLossReductionIsGainoxidising and reducing agents:~reducing agents are electron donors (give them away)~oxidising agents are electron acceptors (get given them)

7.2 - Oxidation states Oxidation states are used to see what has been oxidised and what has been reduced in a redox reaction. ~in an ionic compound, they tell us how many e- have been lost/gained~in a molecule, they tell us about the distribution between elements of different electronegativity-every element in its uncombined state = 0-positive number = lost electrons (oxidation)-negative number = gained electrons (reduction)The more positive the number, the more it has been oxidised and the more negative the number, the more it has been reduced.Rules: 1) uncombined = 02) some elements always have the same number, for others, sometimes it changes3) sum of all o. numbers in a compound = 0 (they aren't charged)4) sum of all o. states in a complex ion = the charge of the ion (NH4+ = +1)5) the most electronegative element in a compound always has a - oxidation stateelement o. state e.g. H +1 (in metal hydrides is -1) HCl Group 1 +1 NaCl Group 2 +2 CaCl2 Al +3 AlCl3 O -2 (in peroxides and OF2 = +2) Na2O F -1 NaF Cl -1 (except w/ F and O working out oxidation states:1. remember rules (e.g. H=+1)2. using basic maths work out the values for the other elementse.g. 1:phosphorus pentachloride, PCL51. there are 5 Cl, each with a charge of -12. if Cl5=-5, then P has to equal +5e.g. 2: red copper oxide, Cu2O1. oxygen=-22. Cu2 = +2(3. divide Cu2 by 2 to get the value for EACH Cu element)+2/2= +1, Cu=+1nb: if an element's oxidation state has not changed, it is a spectator ion.

7.3 - Redox equationswhen an element is reduced, it GAINS electrons and its o. state goes downmake sure you balance equations for not only ATOMS but CHARGE as well1. write oxidation states2. balance atoms and charge3. make sure it's balanced correctly

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