REDOX A guide for A level students

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REDOXA guide for A level students2008 SPECIFICATIONSKNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHINGREDOXINTRODUCTIONThis Powerpoint show is one of several produced to help…
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REDOXA guide for A level students2008 SPECIFICATIONSKNOCKHARDY PUBLISHINGKNOCKHARDY PUBLISHINGREDOXINTRODUCTIONThis Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards.Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available.Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at...www.knockhardy.org.uk/sci.htmNavigation is achieved by...either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboardREDOX
  • CONTENTS
  • Definitions of oxidation and reduction
  • Calculating oxidation state
  • Use of H, O and F in calculating oxidation state
  • Naming compounds
  • Redox reactions
  • Balancing ionic half equations
  • Combining half equations to form a redox equation
  • Revision check list
  • REDOX
  • Before you start it would be helpful to…
  • Recall the layout of the periodic table
  • Be able to balance simple equations
  • OXIDATION & REDUCTION - DefinitionsOXIDATIONGAIN OF OXYGEN2Mg + O2——> 2MgOmagnesium has been oxidised as it has gained oxygenREMOVAL (LOSS) OF HYDROGENC2H5OH ——> CH3CHO + H2ethanol has been oxidised as it has ‘lost’ hydrogenOXIDATION & REDUCTION - DefinitionsREDUCTIONGAIN OF HYDROGENC2H4 + H2 ——> C2H6ethene has been reduced as it has gained hydrogenREMOVAL (LOSS) OF OXYGENCuO + H2 ——> Cu + H2Ocopper(II) oxide has been reduced as it has ‘lost’ oxygenHowever as chemistry became more sophisticated, it was realised that another definition was requiredOXIDATION & REDUCTION - DefinitionsOXIDATION AND REDUCTION IN TERMS OF ELECTRONSOxidation and reduction are not only defined as changes in O and H...OXIDATIONRemoval (loss) of electrons ‘OIL’species will get less negative or more positiveREDUCTIONGain of electrons ‘RIG’species will become more negative or less positiveREDOXWhen reduction and oxidation take placeOXIDATION & REDUCTION - DefinitionsOXIDATION AND REDUCTION IN TERMS OF ELECTRONSOxidation and reduction are not only defined as changes in O and H...OXIDATIONRemoval (loss) of electrons ‘OIL’species will get less negative or more positiveREDUCTIONGain of electrons ‘RIG’species will become more negative or less positiveREDOXWhen reduction and oxidation take placeOIL -Oxidation Is the Loss of electronsRIG -Reduction Is the Gain of electronsOXIDATION STATESUsed to... tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equationsATOMS AND SIMPLE IONSThe number of electrons which must be added or removed to become neutralatomsNa in Na = 0 neutral already ... no need to add any electronscationsNa in Na+ = +1 need to add 1 electron to make Na+ neutralanionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutralOXIDATION STATESUsed to... tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equationsATOMS AND SIMPLE IONSThe number of electrons which must be added or removed to become neutralatomsNa in Na = 0 neutral already ... no need to add any electronscationsNa in Na+ = +1 need to add 1 electron to make Na+ neutralanionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutralQ. What are the oxidation states of the elements in the following? a) C b) Fe3+ c) Fe2+ d) O2- e) He f) Al3+OXIDATION STATESUsed to... tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equationsATOMS AND SIMPLE IONSThe number of electrons which must be added or removed to become neutralatomsNa in Na = 0 neutral already ... no need to add any electronscationsNa in Na+ = +1 need to add 1 electron to make Na+ neutralanionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutralQ. What are the oxidation states of the elements in the following? a) C (0) b) Fe3+(+3) c) Fe2+ (+2) d) O2-(-2) e) He (0) f) Al3+ (+3)OXIDATION STATESMOLECULESThe SUM of the oxidation states adds up to ZEROELEMENTSH in H2 = 0 both are the same and must add up to ZeroCOMPOUNDSC in CO2 = +4 O in CO2 = -2 1 x +4 and 2 x -2 = ZeroOXIDATION STATESMOLECULESThe SUM of the oxidation states adds up to ZEROELEMENTSH in H2 = 0 both are the same and must add up to ZeroCOMPOUNDSC in CO2 = +4 O in CO2 = -2 1 x +4 and 2 x -2 = ZeroExplanation
  • because CO2 is a neutralmolecule, the sumoftheoxidationstatesmust be zero
  • for this, oneelementmusthave a positiveOS and the othermustbenegative
  • OXIDATION STATESMOLECULESThe SUM of the oxidation states adds up to ZEROELEMENTS H in H2 = 0 both are the same and must add up to ZeroCOMPOUNDS C in CO2 = +4 O in CO2 = -2 1 x +4 and 2 x +2 = Zero
  • HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?
  • the more electronegative species will have the negativevalue
  • electronegativity increases across a period and decreases down a group
  • O is further to the right than C in the periodic table so it has the negative value
  • OXIDATION STATESMOLECULESThe SUM of the oxidation states adds up to ZEROELEMENTS H in H2 = 0 both are the same and must add up to ZeroCOMPOUNDS C in CO2 = +4 O in CO2 = -2 1 x +4 and 2 x +2 = Zero
  • HOW DO YOU DETERMINE THE VALUE OF
  • AN ELEMENT’S OXIDATION STATE?
  • from its positionintheperiodictable and/or
  • the other element(s) present in the formula
  • OXIDATION STATESCOMPLEX IONSThe SUM of the oxidation states adds up to THE CHARGE e.g. NO3-sum of the oxidation states = - 1 SO42- sum of the oxidation states = - 2 NH4+ sum of the oxidation states = +1Examplesin SO42- the oxidation state of S = +6 there is ONE SO = -2 there are FOUR O’s+6 + 4(-2) = -2 so the ion has a 2- chargeOXIDATION STATESCOMPLEX IONSThe SUM of the oxidation states adds up to THE CHARGE e.g. NO3-sum of the oxidation states = - 1 SO42- sum of the oxidation states = - 2 NH4+ sum of the oxidation states = +1Examples
  • What is the oxidation state (OS) of Mn in MnO4¯ ?
  • the oxidation state of oxygen in most compounds is - 2
  • there are 4 O’s so the sum of its oxidation states - 8
  • overall charge on the ion is - 1
  • therefore the sum of all the oxidation states must add up to - 1
  • the oxidation states of Mn four O’s must therefore equal - 1
  • therefore the oxidation state of Mn in MnO4¯is +7
  • +7 + 4(-2) = - 1
  • OXIDATION STATESCALCULATING OXIDATION STATE - 1Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other valuesHYDROGEN+1 except0 atom (H) and molecule (H2)-1 hydride ion, H¯ in sodium hydride NaHOXYGEN -2 except 0 atom (O) and molecule (O2)-1 in hydrogen peroxide, H2O2+2 in F2OFLUORINE -1 except 0 atom (F) and molecule (F2)OXIDATION STATESCALCULATING OXIDATION STATE - 1Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other valuesHYDROGEN+1 except0 atom (H) and molecule (H2)-1 hydride ion, H¯ in sodium hydride NaHOXYGEN -2 except 0 atom (O) and molecule (O2)-1 in hydrogen peroxide, H2O2+2 in F2OFLUORINE -1 except 0 atom (F) and molecule (F2)Q.Give the oxidation state of the element other than O, H or F in...SO2NH3NO2NH4+IF7Cl2O7NO3¯ NO2¯ SO32-S2O32-S4O62-MnO42- What is odd about the value of the oxidation state of S in S4O62- ?OXIDATION STATESA. The oxidation states of the elements other than O, H or F areSO2 O = -2 2 x -2 = - 4 overall neutral S = +4NH3 H = +1 3 x +1 = +3 overall neutral N = - 3NO2 O = -22 x -2 = - 4 overall neutral N = +4NH4+ H = +1 4 x +1 = +4 overall +1 N = - 3IF7 F = -1 7 x -1 = - 7 overall neutral I = +7Cl2O7 O = -27 x -2 = -14 overall neutral Cl = +7 (14/2)NO3¯ O = -23 x -2 = - 6 overall -1 N = +5NO2¯ O = -22 x -2 = - 4 overall -1 N = +3SO32- O = -23 x -2 = - 6 overall -2 S = +4S2O32- O = -23 x -2 = - 6 overall -2 S = +2 (4/2)S4O62- O = -26 x -2 = -12 overall -2 S = +2½ ! (10/4)MnO42- O = -24 x -2 = - 8 overall -2 Mn = +6 What is odd about the value of the oxidation state of S in S4O62- ? An oxidation state must be a whole number (+2½ is the average value)OXIDATION STATESCALCULATING OXIDATION STATE - 2The position of an element in the periodic table can act as a guideMETALS• have positive values in compounds • value is usually that of the Group Number Al is+3 • where there are several possibilities the values go no higher than the Group No. Sn can be+2 or +4Mn can be +2,+4,+6,+7NON-METALS • mostly negative based on their usual ion Cl usually-1 • can have values up to their Group No. Cl +1 +3 +5 or +7OXIDATION STATESCALCULATING OXIDATION STATE - 2The position of an element in the periodic table can act as a guideMETALS• have positive values in compounds • value is usually that of the Group Number Al is+3 • where there are several possibilities the values go no higher than the Group No. Sn can be+2 or +4Mn can be +2,+4,+6,+7NON-METALS • mostly negative based on their usual ion Cl usually-1 • can have values up to their Group No. Cl +1 +3 +5or+7Q. What is the theoretical maximum oxidation state of the following elements?Na P Ba Pb S Mn Cr What will be the usual and the maximum oxidation state in compounds of?Li Br Sr O B N+1OXIDATION STATESCALCULATING OXIDATION STATE - 2The position of an element in the periodic table can act as a guideA.What is the theoretical maximum oxidation state of the following elements?Na P Ba Pb S Mn Cr+1 +5 +2 +4 +6 +7 +6 What will be the usual and the maximum oxidation state in compounds of?Li Br Sr O B NUSUAL+1 -1 +2 -2 +3 -3 or +5MAXIMUM+1 +7 +2 +6 +3 +5OXIDATION STATESCALCULATING OXIDATION STATE - 2Q. What is the oxidation state of each element in the following compounds/ions ? CH4 PCl3 NCl3 CS2 ICl5 BrF3 PCl4+ H3PO4 NH4Cl H2SO4 MgCO3 SOCl2OXIDATION STATESCALCULATING OXIDATION STATE - 2Q. What is the oxidation state of each element in the following compounds/ions ? CH4C = - 4 H = +1 PCl3P = +3 Cl = -1 NCl3N = +3 Cl = -1 CS2C = +4 S = -2 ICl5I = +5 Cl = -1 BrF3Br = +3 F = -1 PCl4+P = +4 Cl = -1 H3PO4P = +5 H = +1 O = -2 NH4Cl N = -3 H = +1 Cl = -1 H2SO4S = +6 H = +1 O = -2 MgCO3Mg = +2 C = +4 O = -2 SOCl2S = +4 Cl = -1 O = -2OXIDATION STATESTHE ROLE OF OXIDATION STATE IN NAMING SPECIESTo avoid ambiguity, the oxidation state is often included in the name of a speciesmanganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2sulphur(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 Q. Name the following... PbO2 SnCl2 SbCl3 TiCl4 BrF5OXIDATION STATESTHE ROLE OF OXIDATION STATE IN NAMING SPECIESTo avoid ambiguity, the oxidation state is often included in the name of a speciesmanganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2sulphur(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 Q. Name the following... PbO2lead(IV) oxide SnCl2tin(II) chlorideSbCl3antimony(III) chlorideTiCl4titanium(IV) chlorideBrF5bromine(V) fluorideREDOX REACTIONSOXIDATION AND REDUCTION IN TERMS OF ELECTRONSOxidation and reduction are not only defined as changes in O and HREDOXWhen reduction and oxidation take placeOXIDATIONRemoval (loss) of electrons ‘OIL’species will get less negative or more positiveREDUCTIONGain of electrons ‘RIG’species will become more negative or less positiveREDOX REACTIONSOXIDATION AND REDUCTION IN TERMS OF ELECTRONSOxidation and reduction are not only defined as changes in O and HREDOXWhen reduction and oxidation take placeOXIDATIONRemoval (loss) of electrons ‘OIL’species will get less negative or more positiveREDUCTIONGain of electrons ‘RIG’species will become more negative or less positiveREDUCTION in O.S. Species has been REDUCED e.g. Cl is reduced to Cl¯ (0 to -1)INCREASE in O.S.Species has been OXIDISED e.g. Na is oxidised to Na+ (0 to +1)REDOX REACTIONSOXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. INCREASE in O.S.Species has been REDUCEDSpecies has been OXIDISEDQ. State if the changes involve oxidation (O) or reduction (R) or neither (N)Fe2+ —> Fe3+ I2 —> I¯ F2 —> F2O C2O42- —> CO2 H2O2 —> O2 H2O2 —> H2O Cr2O72- —> Cr3+ Cr2O72- —> CrO42- SO42- —> SO2REDOX REACTIONSOXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. INCREASE in O.S.Species has been REDUCEDSpecies has been OXIDISEDQ. State if the changes involve oxidation (O) or reduction (R) or neither (N)Fe2+ —> Fe3+ O +2 to +3 I2 —> I¯ R 0 to -1 F2 —> F2O R 0 to -1 C2O42- —> CO2 O +3 to +4 H2O2 —> O2 O -1 to 0 H2O2 —> H2O R -1 to -2 Cr2O72- —> Cr3+ R +6 to +3 Cr2O72- —> CrO42- N +6 to +6 SO42- —> SO2 R +6 to +4BALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideBALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 1 Iron(II) being oxidised to iron(III) Step 1 Fe2+ ——> Fe3+ Step 2 +2+3 Step 3 Fe2+ ——> Fe3+ + e¯ now balanced An electron (charge -1) is added to the RHS of the equation... this balances the oxidation state change i.e. (+2) ——> (+3) + (-1)As everything balances, there is no need to proceed to Steps 4 and 5BALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 2 MnO4¯ being reduced to Mn2+ in acidic solutionBALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 2 MnO4¯ being reduced to Mn2+ in acidic solutionStep 1 MnO4¯ ———> Mn2+No need to balance Mn; equal numbersBALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 2 MnO4¯ being reduced to Mn2+ in acidic solutionStep 1 MnO4¯ ———> Mn2+Step 2 +7 +2Overall charge on MnO4¯ is -1; sum of the OS’s of all atoms must add up to -1Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]BALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 2 MnO4¯ being reduced to Mn2+ in acidic solutionStep 1 MnO4¯ ———> Mn2+Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+The oxidation states on either side are different; +7 —> +2 (REDUCTION)To balance; add 5 negative charges to the LHS [+7 + (5 x -1) = +2]You must ADD 5 ELECTRONS to the LHS of the equationBALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 2 MnO4¯ being reduced to Mn2+ in acidic solutionStep 1 MnO4¯ ———> Mn2+Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+Total charges on either side are not equal; LHS = 1- and 5- = 6- RHS = 2+Balance them by adding 8 positive charges to the LHS [ 6- + (8 x 1+) = 2+ ]You must ADD 8 PROTONS (H+ ions) to the LHS of the equationBALANCING REDOX HALF EQUATIONS1 Work out formulae of the species before and after the change; balance if required 2 Work out oxidation state of the element before and after the change 3 Add electrons to one side of the equation so that the oxidation states balance 4 If the charges on the species (ions and electrons) on either side of the equation do not balance then add sufficient H+ ions to one of the sides to balance the charges 5 If equation still doesn’t balance, add sufficient water molecules to one sideExample 2 MnO4¯ being reduced to Mn2+ in acidic solutionStep 1 MnO4¯ ———> Mn2+Step 2 +7 +2Step 3 MnO4¯ + 5e¯ ———> Mn2+Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O now balancedEverything balances apart from oxygen and hydrogen O LHS = 4 RHS = 0 H LHS = 8 RHS = 0You must ADD 4 WATER MOLECULES to the RHS; the equation is now balancedBALANCING REDOX HALF EQUATIONSWatch out for cases when the species is present in different amounts oneither side of the equation ... IT MUST BE BALANCED FIRSTExample 3 Cr2O72- being reduced to Cr3+ in acidic solutionStep 1 Cr2O72- ———> Cr3+ there are two Cr’s on LHS Cr2O72- ———> 2Cr3+ both sides now have 2Step 2 2 @ +6 2 @ +3 both Cr’s are reducedStep 3 Cr2O72- + 6e¯ ——> 2Cr3+ each Cr needs 3 electronsStep 4 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+Step 5 Cr2O72- + 6e¯ + 14H+ ——> 2Cr3+ + 7H2O now balancedBALANCING REDOX HALF EQUATIONSQ. Balance the following half equations...Na —> Na+ Fe2+ —> Fe3+ I2 —> I¯ C2O42- —> CO2 H2O2 —> O2 H2O2 —> H2O NO3- —> NONO3- —> NO2 SO42- —
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