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& &.- / & &"- # & &u- v & &#f- g# & &/N- O/ &  & &c#&U``- V & &H- I & &5- 6 & &l- m & &Q- R & &-@- . &  & &Kq&}3``- 4} & & )- * & &-  & &-  & &-  & &-  &  & & &Q- R & &>|- }? & &,m- n- & &R- S & &=- > & &@@-  &  & &|&R>`0- ?R & &U3P- 4U & &W%`- &W & &\p- \ & &d- d & &q@- q &  & & F&/3`- 40 & &)- * & &-  & &-  & &-  & &@@-  &  & &ZO}r&|n``- o| & &`- a & &T- U & &D- E & &2- 3 & &@-  &  & &x)&usF- Gtu & &z%&|^8- 9_| & &~M%- &N~ & &5 -  6 & &-  & &-  &  &  &  &  &  & &TNPP & -- '--'&--&TNPP &--'&--&TNPP &T 4  @@``@T 4@T 4 @`@lL <ArialNew RomanbbL <Times New RomanbbL <Monotype SortsbbL <Symbole SortsbbL <Courier 10cpibb$PqD* Db(b(b((((bddhhhRddhhhddhhhddhhhddhhh}Db(bbbUFKdhhhURuKdhhhUF?dhhhddhhhddhhh}Db(b b    bddhhhRddhhhddhhhddhhhddhhh}   (b 8b`t:D$42dd35//0011u8 @ p%p@ H./04x$xa(  (bp@ 8b (  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  b (bap 8bD  8.04x$xOԟ{a  (b@ p 8 b  z b (bp@ P 8b9b$ģb 0\W 4J5Chemistry Tutorialh/X/ /, /(X0H040dd(111  b (b p 8bs9bC$ģb 0Y<T#5REDOX REACTIONS by Dr John G Wrighth/X///0040TFKd40TFKd40TFKd(11#1 ' (bpP 8|bK(bbbObject1ClipMS_ClipArt_Gallery.5-Microsoft Clip Gallery@  $NAME7DEFINEDINNAVIGATORFalse6 HOTSPOTTYPE PlayOl &FMicrosoft ClipArt GalleryMS_ClipArt_GalleryMS_ClipArt_Gallery9qScience & MedicineC:\MSOFFICE\clipart\pcsfiles\SCIENCE.PCSj@jn=j=@jj|jA=j=@j@j=H j{jH`Dj8@0 N8@0 jdjxj@@PS.Science & Medicine(C:\MSOFFICE\clipart\pcsfiles\SCIENCE.PCSCompObjxOle10NativeObject2&FA;I "A;I "Ole  &FMicrosoft ClipArt GalleryMS_ClipArt_GalleryMS_ClipArt_Gallery9qScience & MedicineC:\MSOFFICE\clipart\pcsfiles\SCIENCE.PCSj@jn=j=@jj|jA=j=@j@j=PH j{j*PH`Dj8@ N8@ jdjxj@@PS.Science & MedicineVerb)BRANCHTO1 $* O (b@j O 8b9b$b 0y 37+5Press PgDn or left click for the next slide</,/+/X0H0+40dd(11+1 4 (bo  8b9bt$b 0 5The Wright Stuff</,//X0H040dd(111@ p%p@ H./04x$xa(  (bp@ 8b (  b (bap 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011   8.04x$xOԟ{a  (b@ p 8b P    b (bp`  8bU9b%$ģb 0| dY5Oxidation and Reduction Oxidation Numbers (this set of 18 slides) Disproportionation Balancing Half Equations Electrochemical Cells Problems The individual topics can be viewed as separate slide shows. If a slide show is missing from the web page, it means it is being updated and improved. (Or perhaps I just havent finished writing it yet!)</,/Y/0040UFKd*40UFKd40UFKd40UFKd40UFKd40UFKd(11Y1 K  b (b  8b9bs$ģb 0 5Redox Reactions</,//(X0H040dd(111  (bp`u ` 8bH9b$b 0|< D5 The tutorials are divided into five main sections, each covering a separate topic concerning redox reactions, plus a problems section. h/X///X0H040dd(111f @ p%p@ H./04x$xa(  (bp@ 8b (  b (bap 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011   8.04x$xOԟ{a  (b@ p 8'}b  S  b (b  8b9b{$ģb 0 #5Oxidation and Reduction</,//(X0H040dd(111   b (bp  8b9b$ģb 0 t/5It is assumed that you have read the first tutorial on Oxidation and Reduction and are familiar with the meaning of these terms, especially the electron-based definitions. Students on introductory level courses may only need to read the previous section on Oxidation and Reduction, but those on advanced courses, such as A-level and above, will need to study both tutorials, with the emphasis being placed on the modern electron-based definition of oxidation and reduction. </,//0040UFKd.40UFKd(111 $* . @ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011   8.04x$xOԟ{a  (b@ p 8'}b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111 a  b (bp  8b9b$ģb 0 t1%5 This section is about calculating the oxidation state or oxidation number of an element in a compound. It uses a set of rules which require little or no previous knowledge of the chemical being examined. (The oxidation number is similar to the valency of the element but has a + or - sign.)</,/%/X0H0%40TFKd(11%1-@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bp  8b:9b $ģb 0 25The oxidation number of an element is the charge which the atom would have if the element was acting as an ion in the species being studied. This can seem a bit strange at times, especially when you know that the compound you are examining is covalent. But it enables us to easily work out whether an element is oxidised or reduced in a reaction. We just compare the oxidation number before and after the reaction and work out whether it has gained or lost electrons. There are a series of simple rules we use to calculate the oxidation number of an element.h/X///0040UFKd40UFKd(1121 $*  : (bS] 8b9bz$b 0yg"5First, the definition.</,//X0H040dd(111$ @ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111 W  b (bp  8b9b$ģb 0 t'5The rules are given in order of importance. If you reach a rule which gives you data to work from and a later rule appears to contradict the earlier result, just ignore it, you use the earlier result. I.e. later rules do not overrule earlier decisions that you have already made. </,//X0H040UFKd(111 @ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b W M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bp  8b9b$ģb 0 tv5The oxidation number of an uncombined element is zero. I.e. the element in the element itself is zero. The algebraic sum of the oxidation numbers of all the atoms in a compound equals zero. The algebraic sum of the oxidation numbers of all the atoms in an ion equals the charge on the ion. Algebraic sum means take account of the sign of the charge. Learn these rules.h/X/$/R/00 0i40UFKdW40UFKdg40UFKdO40UFKd(11v1 $* t@ p%p@ H./04x$xa(  (bp@ 8b (  b (bap 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8+b   M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bp`  8b9b$ģb 0| _;5Fluorine is always ALWAYS -1 in its compounds. Group I metals are always ALWAYS +1 in compounds. Group II metals are always +2 in their compounds. Oxygen is almost always -2 in compounds, except in peroxides (H2O2, ROOR, etc.) where it is -1. (Or when overruled by one of the above rules.) Halides are often -1, but other numbers are often possible. Hydrogen can be +1 or -1. (If it is the first element given in a formula, it is usually +1, if given second or third, it may be -1. If in doubt, the most electronegative element is +ve and the other element will be -ve.)//////f/00/40UFKd240UFKd240UFKd40UFKd<40UFKd40UFKd(11;1 $*   (bS-  8bD9b$b 0I 5But where do you start? We find that some atoms always have the same oxidation number in their compounds or ions. As stated earlier, the first rules overrule the later rules.</,//X0H040dd(111.@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bpp P 8b9b$ģb 0 45Calculate the oxidation number of Mn in KMnO4. K and O are in our table of known oxidation numbers, and KMnO4 is a compound. K is +1 (Group I metal) and oxygen is almost always -2. K + Mn + (4 x O) = 0 (thats a zero) So +1 + Mn + (4 x -2) = 0 i.e. +1 + Mn - 8 = 0 i.e. Mn - 7 = 0 so Mn = +7 (We say +7 and not 7, because it is a charge.) Easy? Its just simple arithmatic, not chemistry.//,//@//'/00/40UFKdO40TFKd940TFKd'40TFKd40TFKd40TFKd40TFKd>40TFKd340TFKd(111 $* \ (b 8b9b$b 0bD85Lets look at these rules in action in a simple problem.</,/8/X0H0840dd(1181@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b y  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bp  8b9b$ģb 0 tAI5Calculate the oxidation number of chlorine in the ion ClO3- Oxygen comes before chlorine in our known table, so its rule takes preference, and it has a charge of -2. Cl + (3 x O) = -1 ( the -1 is the charge on this ion) Cl + (3 x -2) = -1 Cl - 6 = -1 Cl = +5 (Again, its +5 and not 5 as its the charge on the ion.)//9////00<40UFKdl40TFKd940TFKd40TFKd40TFKdF40TFKd(11I1 $* W (bS 8b9b$b 0S?35Heres another example, this time involving an ion.</,/3/X0H0340dd(1131#@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111 V  b (bp  8bz9bJ$ģb 0 t5Calculate the oxidation number of phosphorus in POCl2F Oxygen is almost always -2, chlorine is usually -1 and fluorine is always ALWAYS -1. As P is the unknown, we can reason that although Cl can have other values, then -1, the commonest value, is the one to use (otherwise the problem is unsolvable). P + O + (2 x Cl) + F = 0 P - 2 - 2 -1 = 0 P - 5 = 0 P = +5 But phosphorus can have other values for its oxidation number, depending on the compound under investigation.//4///00740UFKd40TFKd40TFKd40TFKd 40TFKd40TFKdo40TFKd(111 $*@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b i  M  b (b  8b9bu$ģb 0 5Oxidation numbers</,//(X0H040dd(111   b (bp  8b 9b$ģb 0 L5What is the oxidation number for P in NaH2PO3? Na = +1, H = +1 usually, and O = -2 almost always. Na + 2H + P + 3O = 0 + 1 + 2 + P - 6 = 0 P - 3 = 0, so P = +3 What is charge on the Cr in K2Cr2O7 ? K and O are in the table of fixed values. 2K + 2Cr + 7O = 0 + 2 + 2Cr - 14 = 0 2Cr - 12 = 0, so 2Cr = +12, and Cr = +6//)///////////00/40UFKd640TFKd40TFKd40TFKd40TFKd&40UFKd+40TFKd40TFKd40TFKd,40TFKd(11L1 $*@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111    b (bp` P 8b79b$ģb 0| 435To avoid any confusion when an element can have several oxidation numbers, the oxidation number is usually mentioned in the compounds name, written as Roman numerals in brackets after the element to which it refers. In names like elementate(X), the number refers to element and not the associated oxygens. So if we look at the examples weve just done, we get the following names:- KMnO4 potassium manganate(VII) NaClO3 sodium chlorate(V) POCl2F phosphorus(V) oxydichlorofluoride NaH2PO3 sodium dihydrogenphosphate(III) K2Cr2O7 potassium dichromate(VI)//// ////(////#///////00840UFKdL40UFKd 40TFKd40TFKd*40TFKd)40TFKd!40TFKd(1131 $*@ p%p@ H./04x$xa(  (bp@ 8b (  b (bap 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation numbers</,//(X0H040dd(111 D  b (bp  8bh9b8$ģb 0 t5Check for yourself that the numbers in the following compounds names are the same as the oxidation number of the element with which they are associated LiNO3 lithium nitrate(V), PtCl4 platinum(IV) chloride NaNO2 sodium nitrate(III), NaBrO4 sodium bromate(VII) POCl3 phosphorus(V) oxychloride CrCl3 chromium(III) chloride, TiO titanium(II) oxide TiO2 titanium(IV) oxide Did you get them all right?////"//!//"////$//=//7/0040UFKdD40UFKd>40TFKd%40TFKd?40TFKd 40TFKd40TFKd(111 $*@ p%p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bp  8b9b$ģb 0 dT85Consider the equation below. 2FeCl2 + Cl2 2FeCl3 The oxidation number of the iron changes thus: Fe2+ Fe3+ ( + e- of course) sometimes written as Fe(II) Fe(III) + e- i.e. Fe2+ has been oxidised. This example is easy to spot as an example of oxidation, but sometimes its a bit harder.//#////////4////// //*//////n/00:40UFKd40UFKd(1181 $* h (b]-  8b9b$b 0I PD5An important use of oxidation numbers is in the recognition of oxidation and reduction. They let us quickly see when an element has gained or lost electrons, and hence been oxidised or reduced. If the oxidation number becomes more positive, oxidation has occurred, becoming more negative means that reduction has occurred.</,/D/X0H0D40dd(11D1& @ p% p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD   b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bp  8b9b$ģb 0 Um5 I- IO3- Is this oxidation or reduction? Calculating the oxidation numbers reveals all. In I-, the iodine is -1, while in IO3- it is +5. (Dont take my word for it, calculate the oxidation numbers yourself.) So the iodide ion has lost electrons, and been oxidised to +5. I- IO3- + 6e- which could also be written as I(I) I(V) + 6e- /x////////X///////////////(////00040UFKd140TFKd 40UFKd(11m1 $*  (bS- 9 8b9b$b 0I xl5Consider a reaction where the following change occurs (the other reagents are not important at this stage).</,/l/X0H0l40dd(11l1&!@ p%!p@ H./04x$xa(  (bp@ 8b (  b (b`p 8bD!  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011  8.04x$xOԟ{a  (b@ p 8b  M  b (b  8b9bu$ģb 0 5Oxidation Numbers</,//(X0H040dd(111   b (bpp P 8b 9b $ģb 0 4M !53MnO2 + KClO3 + 6KOH 3K2MnO4 + KCl + 3H2O Mn = +4 Mn = +6 Cl = +5 Cl = -1 i.e Mn has lost electrons and been oxidised, while Cl has gained electrons and been reduced. 4FeCO3 + O2 2Fe2O3 + 4CO2 Fe = +2 Fe = +3 O = 0 O = -2 C = +4 C = +4 i.e. Fe has lost electrons and been oxidised, while the oxygen has gained electrons and been reduced. The carbon hasnt changed. Check these oxidation numbers for yourself. ////// ////// ////!//q//////////////!//%///0040UFKdO40UFKd(11!1 $*  (bS# / 8b9b$b 0? t5Here are a few more examples showing the use of oxidation numbers to discover whether oxidation or reduction occurs.</,/t/X0H0t40dd(11t1[ H@ p%Hp@ H./04x$xa(  (bp@ 8b (  b (bap 8bDH  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011   8.04x$xOԟ{a  (b@ p 8'}b  C  b (b  8b9bk$ģb 0 5The End</,//(X0H040dd(111   b (bp  8b9b$ģb 0 th5I hope you have enjoyed this tutorial. It is the second in a series of tutorials on oxidation and reduction. The tutorials become progressively more advance, but unless it says so on the frist couple of pages, are all intended for A-level students. I also hope you have increased your understanding of chemistry, learned something useful about chemistry and that it will increase your marks in examinations. Bye for now, Dr John G Wright, The Wright Stuff www.sky-web.net h/X//&/0040UFKd40UFKd(111 I@ p%Ip@ H./04x$xa(  (bp@ 8b (  b (b`p 8bDI  b (bp0  8bt9bD$Db 0 t 42dd 3 5//0011   8.04x$x###ԟ{a  (b@ p 8b  =  b (b  8b9be$ģb 0  5 </,//(X0H040dd(111 =  b (bp  8b9be$ģb 0 t 5 </,//X0H040TFKd(111  @ p!$7̙33$7$7ff3333f$7333MMM$7f$7f$73 8.04x$xOԟ{a  (b@ p 8'}b  ^ b (bp  8b9b$ģb 0 t.R5Click to edit Master text styles Second Level Third Level Fourth Level Fifth Level//!/ / / / /x0h0!40UFKd 40UuKd 40UF@d 40dd 40dd(11R1 \ b (b  8b9b$ģb 0 , 5Click to edit Master title style</,/ /(X0H0 40dd(11 1 ` <  (blh 8O2b  d  (bP@ 8b  (bpP 8b  (bP` 8b  (bP 8b  (b P 8b  (bP 8b  (b@P 8b  (bP0 8b  (b`P 8b  (bPP 8b  (bP 8b  (bPp 8b  (bP 8b  (b0P 8b  (bP  8b  (bPP 8b  (bP@ 8b  (bpP 8b  (bP` 8b  (bP 8b  (b P 8b  (bP 8b  (bPV 8b  (bP 8b  (bPv 8b  (bP 8b  (b6P 8b  (bP& 8b  (bVP 8b % (bs9 8b9be$ţb 0V7: 5*</,//X0H040dd(111 ' (b@p 8lbK(bܤbbObject2ClipMS_ClipArt_Gallery.5-Microsoft Clip Gallery@  $NAME7DEFINEDINNAVIGATORFalse6 HOTSPOTTYPE PlayOleVerb)CompObj xOle10NativePersistentStorage Directory8">Current User#e(C:\MSOFFICE\clipart\pcsfiles\SCIENCE.PCS"PowerPoint DocumentDr John G WrightMicrosoft (R) PowerPoint (R) Windows _ ՜.+,0\ hp  On-screen ShowPBRANCHTO1 $*'p@  0H./04x$xa(  (bp@ 8b k _ b (bp0  8b9b$ģb 0 t /S5Click to edit Master notes styles Second Level Third Level Fourth Level Fifth Level//"/  /  /  /  / x0h0"40dd 40dd 40dd 40dd 40dd(11S1  b (b`p 8bDdp@ H./04x$xa(  (bp@ 8b  04BdBd b b$de, Header$9SummaryInformation( RDocumentSummaryInformation8%n Times New RomanMonotype SortsArialSymbolCourier 10cpi azures.pptMicrosoft Clip GalleryChemistry TutorialRedox ReactionsOxidation and ReductionOxidation NumbersOxidation NumbersOxidation NumbersOxidation NumbersOxidation NumbersOxidation NumbersOxidation NumbersOxidation NumbersOxidation numbersOxidation NumbersOxidation numbersOxidation N      !"#$%&'()* Oh+'0tR0 LX |   Chemistry TutorialhemDr John G Wrightlr Jr J1c:\msoffice\powerpnt\template\sldshow\azures.pptiDr John G Wrightnt10JMicrosoft PowerPoint 4.0lat@f$n@@`o @@iDI "J-GPoM  P('& &&#TNPPh0D & TNPP &&TNPP     'A x(xʦ """)))UUUMMMBBB999|PP3f3333f333ff3fffff3f3f̙f3333f3333333333f3333333f3f33ff3f3f3f3333f3333333f3̙33333f333ff3ffffff3f33f3ff3f3f3ffff3fffffffff3fffffff3f̙ffff3ff333f3ff33fff33f3ff̙3f3f3333f333ff3fffff̙̙3̙f̙̙̙3f̙3f3f3333f333ff3fffff3f3f̙3ffffffffff!___www