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2012 topic 09 oxidation and reduction reactions
 

2012 topic 09 oxidation and reduction reactions

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  • Dear David

    Nice to see ythat ou are back using my work again! In the past six months I have had a couple of comments from teachers around the world telling me that someone is using my work. I know that I have mentioned this before but I am now insisting that you change MUCH to MOST on the second page and give me the credit eg content courtesy of Jonathan Hopton ON ALL SLIDES SHOWS WHERE YOU USE MY WORK.



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    2012 topic 09 oxidation and reduction reactions 2012 topic 09 oxidation and reduction reactions Presentation Transcript

    • IB Chemistry Power Points Topic 09 Oxidation and Reductionwww.pedagogics.ca
    • Much taken from REDOXA guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS
    • OXIDATION & REDUCTION – Simplified Definitions OXIDATION is a GAIN OF OXYGEN 2Mg + O2 ——> 2MgO magnesium has been oxidised as it has gained oxygen is the REMOVAL (LOSS) OF HYDROGEN C2H5OH ——> CH3CHO + H2 ethanol has been oxidised as it has ‘lost’ hydrogen
    • OXIDATION & REDUCTION – Simplified Definitions REDUCTION is a GAIN OF HYDROGEN C 2 H4 + H 2 ——> C2H6 ethene has been reduced as it has gained hydrogen is the REMOVAL (LOSS) OF OXYGEN CuO + H2 ——> Cu + H2Ocopper(II) oxide has been reduced as it has ‘lost’ oxygen However as chemistry became more sophisticated, it was realised that another definition was required
    • OXIDATION & REDUCTION – Better Definitions OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H... OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTION Gain of electrons ‘RIG’ species will become more negative or less positive REDOX When reduction and oxidation take place
    • OIL - Oxidation Is the Loss of electronsRIG - Reduction Is the Gain of electrons
    • OXIDATION NUMBERS (STATES) Used to... tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced construct half equations and balance redox equations FOR ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutralAtoms Na = 0 neutral already ... no need to add any electronsCations Na+ = +1 need to add 1 electron to make Na+ neutralAnions Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral Q. 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 STATES MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTS H in H2 = 0 both are the same and must add up to Zero COMPOUNDS C in CO2 = +4 O in CO2 = -2 1 x +4 and 2 x -2 = Zero Explanation • because CO2 is a neutral molecule, the sum of the oxidation states must be ? • for this, one element must have a positive OS and the other must be ?HOW DO YOU DETERMINE THE VALUE OF AN ELEMENT’S OXIDATION STATE?• from its position in the periodic table and/or• the other element(s) present in the formula (oxygen is almost always -2 etc)HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE?• the more electronegative species will have the negative value
    • OXIDATION STATES COMPLEX IONS The 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 = +1 Example SO42-in SO42- the oxidation state of S = +6 there is ONE S O = -2 there are FOUR O’s +6 + 4(-2) = -2 so the ion has a 2- charge
    • OXIDATION STATES COMPLEX IONS The 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 = +1 Example What is the oxidation number 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 STATES CALCULATING OXIDATION STATE - 1 Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other valuesHYDROGEN +1 except 0 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 F2OHALOGENS -1 except 0 atom (X) and molecule (X2) Q. Give the oxidation state of the element other than O, H or F in... SO2 NH3 NO2 NH4+ IF7 Cl2O7 NO3¯ 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 = -2 2 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 = -2 7 x -2 = -14 overall neutral Cl = +7NO3¯ O = -2 3 x -2 = - 6 overall -1 N = +5NO2¯ O = -2 2 x -2 = - 4 overall -1 N = +3SO32- O = -2 3 x -2 = - 6 overall -2 S = +4S2O32- O = -2 3 x -2 = - 6 overall -2 S = +2MnO42- O = -2 4 x -2 = - 8 overall -2 Mn = +6
    • OXIDATION STATES CALCULATING OXIDATION STATE - 2 The 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 +4 Mn 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 +7
    • OXIDATION STATES CALCULATING OXIDATION STATE - 2 The 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 +4 Mn 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 +7 Q. 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 +1
    • OXIDATION STATES CALCULATING OXIDATION STATE - 2 The 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 +5
    • OXIDATION STATES THE 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 statedichromate(VI) for Cr2O72- Cr is in the +6 oxidation statephosphorus(V) chloride for PCl5 P is in the +5 oxidation statephosphorus(III) chloride for PCl3 P is in the +3 oxidation stateQ. Name the following... PbO2 SnCl2 SbCl3 TiCl4 BrF5
    • OXIDATION STATES THE 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 statedichromate(VI) for Cr2O72- Cr is in the +6 oxidation statephosphorus(V) chloride for PCl5 P is in the +5 oxidation statephosphorus(III) chloride for PCl3 P is in the +3 oxidation stateQ. Name the following... PbO2 lead(IV) oxide SnCl2 tin(II) chloride SbCl3 antimony(III) chloride TiCl4 titanium(IV) chloride BrF5 bromine(V) fluoride
    • REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and HREDOX When reduction and oxidation take place +7OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive O +6 R X +5 EREDUCTION Gain of electrons ‘RIG’ I +4 D species will become more negative or less positive D +3 U A +2 C T +1 T I 0 I O -1 O N -2 N -3 -4
    • REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and HREDOX When reduction and oxidation take place +7OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive O +6 R X +5 EREDUCTION Gain of electrons ‘RIG’ I +4 D species will become more negative or less positive D +3 U A +2 C T +1 T I 0 IREDUCTION in O.N. Species has been REDUCED -1 O O e.g. Cl is reduced to Cl¯ (0 to -1) N -2 N -3INCREASE in O.N. Species has been OXIDISED -4 e.g. Na is oxidised to Na+ (0 to +1)
    • REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species 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- —> SO2
    • REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species 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 +4
    • OXIDATION STATES - Review CALCULATING OXIDATION STATE – MOST IMPORTANTQ. What is the oxidation state of each element in the following compounds/ions ? CH4 PCl3 NCl3 CS2 ICl5 BrF3 PCl4+ H3PO4 NH4Cl H2SO4 MgCO3 SOCl2
    • OXIDATION STATES CALCULATING OXIDATION STATE - 2Q. What is the oxidation state of each element in the following compounds/ions ? CH4 C=-4 H = +1 PCl3 P = +3 Cl = -1 NCl3 N = +3 Cl = -1 CS2 C = +4 S = -2 ICl5 I = +5 Cl = -1 BrF3 Br = +3 F = -1 PCl4+ P = +5 Cl = -1 H3PO4 P = +5 H = +1 O = -2 NH4Cl N = -3 H = +1 Cl = -1 H2SO4 S = +6 H = +1 O = -2 MgCO3 Mg = +2 C = +4 O = -2 SOCl2 S = +4 Cl = -1 O = -2
    • END OF PART ONE
    • WRITING & BALANCING REDOX HALF EQUATIONS
    • WRITING & BALANCING REDOX HALF EQUATIONS Example 1 Iron(II) being oxidised to iron(III)Step 1 Fe2+ ——> Fe3+Step 2 +2 +3Step 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)
    • In the case of reactions occurring in acidic solutions, more effort is required towrite the half reactions.If the charges on the species (ions and electrons) on either side of theequation do not balance then add sufficient H+ ions to one of the sides tobalance the chargesIf equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution No need to balance Step 1 MnO4¯ ———> Mn2+ Mn; equal numbers
    • In the case of reactions occurring in acidic solutions, more effort is required towrite the half reactions.If the charges on the species (ions and electrons) on either side of theequation do not balance then add sufficient H+ ions to one of the sides tobalance the chargesIf equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Overall charge on MnO4¯ is -1; sum of the ON’s of all atoms must add up to -1 Oxygen is in its usual oxidation state of -2; four oxygen atoms add up to -8 To make the overall charge -1, Mn must be in oxidation state +7 ... [+7 + (4x -2) = -1]
    • In the case of reactions occurring in acidic solutions, more effort is required towrite the half reactions.If the charges on the species (ions and electrons) on either side of theequation do not balance then add sufficient H+ ions to one of the sides tobalance the chargesIf equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Step 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 equation
    • In the case of reactions occurring in acidic solutions, more effort is required towrite the half reactions.If the charges on the species (ions and electrons) on either side of theequation do not balance then add sufficient H+ ions to one of the sides tobalance the chargesIf equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Step 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 equation
    • In the case of reactions occurring in acidic solutions, more effort is required towrite the half reactions.If the charges on the species (ions and electrons) on either side of theequation do not balance then add sufficient H+ ions to one of the sides tobalance the chargesIf equation still doesn’t balance, add sufficient water molecules to one side Example 2 MnO4¯ being reduced to Mn2+ in acidic solution Step 1 MnO4¯ ———> Mn2+ Step 2 +7 +2 Step 3 MnO4¯ + 5e¯ ———> Mn2+ Step 4 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ Step 5 MnO4¯ + 5e¯ + 8H+ ———> Mn2+ + 4H2O now balanced Everything balances apart from oxygen and hydrogen O LHS = 4 RHS = 0 H LHS = 8 RHS = 0 You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced
    • BALANCING REDOX HALF EQUATIONS Watch out for cases when the species is present in different amounts on either side of the equation ... IT MUST BE BALANCED FIRST Example 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 Cr @ +6 2 Cr @ +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 balanced
    • BALANCING REDOX HALF EQUATIONSQ. Balance the following half equations... Na —> Na+ Fe2+ —> Fe3+ I2 —> I¯ C2O42- —> CO2 H2O2 —> O2 H2O2 —> H2O NO3- —> NO NO3- —> NO2 SO42- —> SO2 REMINDER 1 Work out the formula of the species before and after the change; balance if required 2 Work out the 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 all 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 the equation still doesn’t balance, add sufficient water molecules to one side
    • BALANCING REDOX HALF EQUATIONSQ. Balance the following half equations... Na —> Na+ + e- Fe2+ —> Fe3+ + e- I2 + 2e- —> 2I¯ C2O42- —> 2CO2 + 2e- H2O2 —> O2 + 2H+ + 2e- H2O2 + 2H+ + 2e- —> 2H2O NO3- + 4H+ + 3e- —> NO + 2H2O NO3- + 2H+ + e- —> NO2 + H2O SO42- + 4H+ + 2e- —> SO2 + 2H2O
    • COMBINING HALF EQUATIONSA combination of two ionic half equations, one involving oxidation and the otherreduction, produces a REDOX equation. The equations are balanced as follows...Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)
    • COMBINING HALF EQUATIONSA combination of two ionic half equations, one involving oxidation and the otherreduction, produces a REDOX equation. The equations are balanced as follows...Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O Reduction
    • COMBINING HALF EQUATIONSA combination of two ionic half equations, one involving oxidation and the otherreduction, produces a REDOX equation. The equations are balanced as follows...Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O ReductionStep 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1
    • COMBINING HALF EQUATIONSA combination of two ionic half equations, one involving oxidation and the otherreduction, produces a REDOX equation. The equations are balanced as follows...Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O ReductionStep 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯
    • OXIDIZING AGENT AND REDUCING AGENTIn the reaction between manganate(VII) and iron(II)MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯The manganate (VII) ion is reduced (gains electrons) and is calledthe oxidizing agent.The iron(II) ion is oxidized (loses electrons) and is called thereducing agent.
    • COMBINING HALF EQUATIONSA combination of two ionic half equations, one involving oxidation and the otherreduction, produces a REDOX equation. The equations are balanced as follows...Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)Step 1 Fe2+ ——> Fe3+ + e¯ Oxidation MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O ReductionStep 2 5Fe2+ ——> 5Fe3+ + 5e¯ multiplied by 5 MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O multiplied by 1Step 3 MnO4¯ + 5e¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+ + 5e¯Step 4 MnO4¯ + 8H+ + 5Fe2+ ——> Mn2+ + 4H2O + 5Fe3+
    • COMBINING HALF EQUATIONSA combination of two ionic half equations, one involving oxidation and the otherreduction, produces a REDOX equation. The equations are balanced as follows...Step 1 Write out the two half equationsStep 2 Multiply the equations so that the number of electrons in each is the sameStep 3 Add the two equations and cancel out the electrons on either sideStep 4 If necessary, cancel any other species which appear on both sides Q. Construct balanced redox equations for the reactions between... Mg (OX) and H+ (RED) Fe2+ (OX) and Cr2O72- (RED) MnO4¯ (RED) and H2O2 (OX) MnO4¯ (RED) and C2O42- (OX) S2O32- (OX) and I2 (RED) Cr2O72- (RED) and I- (OX)
    • BALANCING Mg ——> Mg2+ + 2e¯ (x1) H+ + e¯ ——> ½ H2 (x2)REDOX Mg + 2H+ ——> Mg2+ + H2EQUATIONS Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1) Fe2+ ——> Fe3+ + e¯ (x6) Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2OANSWERS MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2) H2O2 ——> O2 + 2H+ + 2e¯ (x5) 2MnO4¯ + 5H2O2 + 6H+ ——> 2Mn2+ + 5O2 + 8H2O MnO4¯ + 5e¯ + 8H+ ——> Mn2+ + 4H2O (x2) C2O42- ——> 2CO2 + 2e¯ (x5) 2MnO4¯ + 5C2O42- + 16H+ ——> 2Mn2+ + 10CO2 + 8H2O 2S2O32- ——> S4O62- + 2e¯ (x1) ½ I2 + e¯ ——> I¯ (x2) 2S2O32- + I2 ——> S4O62- + 2I¯ Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1) I¯ ——> ½ I2 + e¯ (x6) Cr2O72- + 14H+ + 6I ¯ ——> 2Cr3+ + 3I2 + 7H2O
    • END OF PART TWO
    • Reactivity Series
    • Displacement reactionsA displacement reaction is one where a MORE REACTIVE metal will DISPLACE a LESS REACTIVE metal from a compound. Magnesium Copper sulphate Mg Cu SO4 The magnesium DISPLACES the copper from copper sulphate Mg SO4 Cu Magnesium sulphate Copper
    • We have looked at several reactions: K Fe + Cu(NO3)2 Cu + Fe(NO3)2 Na Li + H2O LiOH + H2 Li No, Ni is Ca Yes, Li is below NaSuch experiments reveal trends. The activity series ranks the Mgrelative reactivity of metals. above Zn AlIt allows us to predict if certain chemicals will undergo Zn Yes, Al issingle displacement reactions when mixed: metals near the Fetop are most reactive and will displace metals near the above Cubottom. Ni SnQ: Which of these reactions occur? Yes, Fe is Pb Fe + CuSO4 Cu + Fe2(SO4)3 Cu above H Ni + NaCl NR (no reaction) Cu Li + ZnCO3 Zn + Li2CO3 Hg Al + CuCl2 Cu + AlCl3 Ag Au
    • Trends in Oxidation and ReductionActive metals: Lose electrons easily Are easily oxidized Are strong reducing agentsActive nonmetals: Gain electrons easily Are easily reduced Are strong oxidizing agents