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Edexcel AS Chemistry: (2.2) Redox - Reduction & Oxidation

Defining reduction and oxidation wrt electron transfer, oxidation numbers, half-equations, completing full ionic equatns
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J Amuah-Fuster

on 8 July 2016

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Transcript of Edexcel AS Chemistry: (2.2) Redox - Reduction & Oxidation

Ni (s) + Pb2+ (aq)  Pb (s) + Ni2+ (aq) REDOX Reduction & Oxidation REDOX Oxidation - losing electrons Reduction - gaining electrons What is Redox? Half-Equations What are they, and how are they written? Some useful ionic-half equations (common oxidising agents) Questions Balancing equations using oxidation numbers Answers a. demonstrate an understanding of:
i. oxidation number — the rules for assigning oxidation numbers
ii. oxidation and reduction as electron transfer
iii. oxidation and reduction in terms of oxidation number changes
iv. how oxidation number is a useful concept in terms of the classification of reactions as redox and as disproportionation
b. write ionic half-equations and use them to construct full ionic equations. Lesson Objectives Questions Some useful ionic-half equations (common reducing agents) Some useful ionic-half equations (common reducing agents) Some useful ionic-half equations (common oxidising agents) Questions Questions Questions More questions Answers Rules for assigning oxidation numbers Chlorine oxidises iron(II) ions to iron(III).

Write the two ionic half-equations and hence derive the overall equation. An example The electrons are on the left-hand side for the reduction equation and on the right-hand side for the oxidation equation.

If the reaction takes place in acidified conditions then H+ ions are added to the left-hand side and water molecules to the right-hand side.

Balance the equations for the atoms.

Balance the electrons so that the half-equation balances for charge. Rules for writing half equations 1. Fe2+ (aq)  Fe3+ (aq) + e-
2. Br2 (aq) + 2e-  2Br- (aq)
(the equation can be halved: ½Br2 (aq) + e-  Br- (aq))
3. Cr2O72- (aq) + 14H+ (aq) + 6e-  2Cr3+ (aq) + 7H2O (l) Answers An oxidising agent is a species (atom, molecule or ion) that oxidises another species by removing one or more electrons. When an oxidising agent reacts, it is itself reduced.
A reducing agent reduces another species by giving it one or more electrons. When a reducing agents reacts it is itself oxidised. Oxidising and reducing agents Rusting – a redox reaction A redox reaction is one in which reduction and oxidation takes place.

Oxidation occurs when a species loses one or more electrons.

Reduction occurs when a species gains one or more electrons. Electron transfer in redox reactions Oxidation and reduction And another Another example Oxidation numbers allow redox equations to be balanced in terms of the charges on the species and the atoms present.
If the oxidation number of a species decreases then it is said to have been reduced.
If the oxidation number of a species increases then it is said to have been oxidised. Oxidation and reduction in terms of oxidation number Show your working – you will be required to do this very exercise in the exam. Questions Working out oxidation numbers The example of copper displacing silver ions in solution was an example of how half-equations can be written from a redox equation. Deducing half-equations from redox equations Sn2+ (aq)  Sn4+ (aq) + 2e-
SO32- (aq) + H2O (l)  SO42- (aq) + 2H+ (aq) + 2e-
PbO2 (s) + 4H+ (aq) + 2e-  Pb2+ (aq) + 2H2O (l)
MnO2 (s) + 4H+ (aq) + 2e- (aq)  Mn2+ (aq) + 2H2O (l) Answers - 1 The half equations must be balanced to show the same numbers of electrons being lost and being gained. Writing half equations (3) Half equations are written to show the gain and loss of electrons. The next step is to balance the half equations: Writing half equations (2) The electron configurations of Fe and O will tell you how many electrons will be gained lost or gained. Writing half equations (1) The oxidation number of an element in a compound or ion is the charge that the element would have if the compound were fully ionic. Redox reactions can also be described in terms of their oxidation numbers – it allows us to follow shifts in electron density.
In simple ionic compounds the oxidation number of the species involved are the charges on the ions, e.g., in MgCl2, the oxidation number of Mg is 2+ and each chloride ion has an oxidation number of -1. Oxidation numbers [in the last question, ignore the complex ion, OH-, when showing the oxidised and reduced species] Mg (s) + FeSO4 (aq)  Fe (s) + MgSO4 (aq)


Ni (s) + Pb(NO3)2 (aq)  Pb (s) + Ni(NO3)2 (aq)


CaBr2 (aq) + Cl2 (g)  CaCl2 (aq) + Br2 (aq)


2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g) Convert these displacement reactions into their ionic form and use this to show where the oxidation and reduction are taking place. Then write the half equations for each reaction. Question 2 Once you have written the two half equations, the number of electrons in both equations need to be balanced.
The atoms may need to be balanced again in one or both equations.
Then add the two equations together, omitting the electrons. In the exam you will be asked to write half equations, and using these, the full redox equations. Half-equations to redox equations Write a half-equation for the oxidation of Fe2+ ions to Fe3+ ions.
Write a half-equation for the reduction of bromine.
Write a half-equation for the reduction of dichromate ions, Cr2O72-, to Cr3+ ions in acid solution. Write half equations for the following reactions: Writing half equations Check that equations are balanced for charge and atoms. Remember H2O may need to be added to equations to balance the atoms present. Sn2+ (aq)  Sn4+ (aq)
SO32- (aq)  SO42- (aq) + H+ (aq)
PbO2 (s) + H+ (aq)  Pb2+ (aq)
MnO2 (s) + H+ (aq)  Mn2+ (aq) Write balanced half-equations for the following: Questions - 1 The silver ions are the oxidising agent because they get reduced, and the copper atoms are the reducing agents because they get oxidised. …don’t forget to balance it the half equations so that the number of electron in both equations are the same. Writing the half equations Answer lead(IV) oxide, tin(II) chloride, antimony(III) chloride, titanium(IV) chloride, bromine(V) fluoride Q. Name these: PbO2 , SnCl2 , SbCl3 , TiCl4 , BrF5 manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2
sulphur(VI) oxide for SO3 S is in the +6 oxidation state
dichromate(VI) for Cr2O72- Cr is in the +6 oxidation state
phosphorus(V) chloride for PCl5 P is in the +5 oxidation state
phosphorus(III) chloride for PCl3 P is in the +3 oxidation state Many of the transition elements have more than one oxidation state. Stock notation is used to identify the oxidation number of the element that precedes it. Stock notation At GCSE the processes of oxidation and reduction were used to describe the reactions involved in the extraction of metals. You even used half equations to describe the reactions that take place at the electrode during the electrolysis of aluminium. Oxidation is not just about the addition of oxygen –
the definition has extended to the transfer of electrons. GCSE knowledge Disproportionation reactions are an examiner’s favourite. You will be required to work out oxidation numbers and identify an equation shows disproportionation. +1 -1 0 Cl2 + 2NaOH  NaOCl + NaCl + H2O In a disproportionation reaction an element must have at least three oxidation states – the initial one, one higher and one lower. Disproportionation is a redox reaction in which an element in a single species is both oxidised and reduced at the same time. Disproportionation reactions There are examples for you to try in your own time in the spiral homework sheet that goes with this topic.
It is unlikely you will get a question as complicated as this at AS. You will more likely get a question in which the equation is formed after working out the balanced half-equation. 2MnO4-(aq) + 16H+(aq) + 5Sn2+(aq)  2Mn2+(aq) + 5Sn4+(aq) + 8H2O(l) 5. Now balance the H+ and H2O. 4. Multiply Sn by 5 and Mn by 2. 2MnO4- (aq) + H+ (aq) + 5Sn2+ (aq)  2Mn2+ (aq) + 5Sn4+ (aq) + H2O (l) continued Ag+ (aq) + Cu(s)  Cu2+(aq) + Ag(s) It is better to use the ionic equation for this reaction to illustrate the species that is oxidised and the species that is reduced. 2AgNO3(aq) + Cu(s)  Cu(NO3)2(aq) + 2Ag(s) Copper will displace silver ions in solution to form copper ions and silver metal. Another example 2Ag+ (aq) + Cu(s)  Cu2+(aq) + 2Ag(s) Ag+ (aq) + Cu(s)  Cu2+(aq) + Ag(s) One final point….the ionic equation we used was not balanced in terms of electron transfer.
Therefore, when it comes to redox equations they must be balanced in terms of the atoms present and the electrons transferred. Balanced half equations [R]; H+ in water [O] 2H+ (aq) + 2e-  H2 (g) 2Na (s)  2Na+ (aq) + 2e- 2Na (s) + 2H2O (l)  2Na+ (aq) + 2OH- (aq) + H2 (g) 2Na (s) + 2H2O (l)  2NaOH (aq) + H2 (g) Answer (d) [R] [O] Pb2+ (aq) + 2e-  Pb (s) Ni (s)  Ni2+ (aq) + 2e- Ni (s) + Pb(NO3)2 (aq)  Pb (s) + Ni(NO3)2 (aq) Answer (b) [R] [O] Cl2 (g) + 2e-  2Cl- (aq) 2Br- (aq)  Br2 (aq) + 2e- 2Br- (aq) + Cl2 (g)  2Cl- (aq) + Br2 (aq) CaBr2 (aq) + Cl2 (g)  CaCl2 (aq) + Br2 (aq) Answer (c) [R] [O] Fe2+ (aq) + 2e-  Fe (s) Mg (s)  Mg2+ (aq) + 2e- Mg (s) + Fe2+ (aq) Fe (s) + Mg2+ (aq) Mg (s) + FeSO4 (aq)  Fe (s) + MgSO4 (aq) Answer (a) +4 +2 +2 +7 MnO4- (aq) + H+ (aq) + Sn2+ (aq)  Mn2+ (aq) + Sn4+ (aq) + H2O (l) 1. Identify the species that have been reduced and oxidised. MnO4- (aq) + H+ (aq) + Sn2+ (aq)  Mn2+ (aq) + Sn4+ (aq) + H2O (l) Use the oxidation number method to balance this redox equation. An example 3. The change in oxidation number must be the same for each species. Multiply the change to achieve this and then do the same to the appropriate species in the equation. Sn increases by +2; Mn decreases by -5 2. Calculate the change in oxidation number for the reduced and oxidised species. +4 +2 +2 +7 MnO4- (aq) + H+ (aq) + Sn2+ (aq)  Mn2+ (aq) + Sn4+ (aq) + H2O (l) continued ? Deduce the oxidation number for the remaining species:
+12/2 = +6 Total oxidation state for 2 x Cr = +12 (?) + (-14) = -2 Use the rules to work out the sum of oxidation numbers for each species in the formula. 7 x (-2) = -14 The ion has an overall charge of -2. Hence, the sum of the oxidation numbers must equal -2. Cr O Oxidation number of elements in ions Cl2 (aq) + 2Fe2+ (aq)  2Fe3+ (aq) + 2Cl- (aq) Cl2 (aq) + 2e- + 2Fe2+ (aq)  2Fe3+ (aq) + 2Cl- (aq) + 2e- The two equations are added together omitting the electrons Cl2 (aq) + 2e-  2Cl- (aq) 2Fe2+ (aq)  2Fe3+ (aq) + 2e- Equation 1 will need to be multiplied by 2 to balance the number of electrons with equation 2 Cl2 (aq) + 2e-  2Cl- (aq) Fe2+ (aq)  Fe3+ (aq) + e- Answer ? Deduce the oxidation number for the remaining species. Oxidation number of S: = +6 Add up them up Total so far: (+2) + (-8) = -6 Use the rules to work out the sum of oxidation numbers for each species in the formula. -8 +2 -2
-2
-2
-2 +1
+1 The molecule has an overall charge of zero. Hence, the sum of the oxidation numbers must equal zero. H SO Oxidation number of elements in molecules Element Oxidation number Li, Na, K +1 Mg, Ca, Sr, Ba +2 Al +3 N (except in some oxides) -1, +3 or +5 O (except in peroxides (-1) and superoxides (-1) or when combined with F. -2 S -2, +2, +4 or +6 F -1 Cl, Br, I -1, +1, +3, +5 or +7 Common oxidation numbers of some elements in their compounds At AS level these ideas will be extended to different types of reactions and develop further your confidence in the use of equations. unbalanced balanced Ni (s) + Pb2+ (aq)  Pb (s) + Ni2+ (aq) 2 4 2- 2 7 Therefore, Oxidation state for Cr = +6
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