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Before you start it would be helpful to… Recall the layout of the periodic table Be able to balance simple equations REDOX.

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Presentation on theme: "Before you start it would be helpful to… Recall the layout of the periodic table Be able to balance simple equations REDOX."— Presentation transcript:

1 Before you start it would be helpful to… Recall the layout of the periodic table Be able to balance simple equations REDOX

2 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

3 OXIDATION GAIN OF OXYGEN 2Mg + O 2 ——> 2MgO magnesium has been oxidised as it has gained oxygen REMOVAL (LOSS) OF HYDROGEN C 2 H 5 OH ——> CH 3 CHO + H 2 ethanol has been oxidised as it has ‘lost’ hydrogen OXIDATION & REDUCTION - Definitions

4 REDUCTION GAIN OF HYDROGEN C 2 H 4 + H 2 ——> C 2 H 6 ethene has been reduced as it has gained hydrogen REMOVAL (LOSS) OF OXYGEN CuO + H 2 ——> Cu + H 2 O copper(II) oxide has been reduced as it has ‘lost’ oxygen However as chemistry became more sophisticated, it was realised that another definition was required

5 ... OXIDATION Removal (loss) of electrons species will get less negative or more positive REDUCTION Gain of electrons species will become more negative or less positive REDOX When reduction and oxidation take place OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H OXIDATION & REDUCTION - Definitions

6 ... OXIDATION Removal (loss) of electrons species will get less negative or more positive REDUCTION Gain of electrons species will become more negative or less positive REDOX When reduction and oxidation take place OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H OXIDATION & REDUCTION - Definitions OIL - Oxidation Is the Loss of electrons RIG - Reduction Is the Gain of electrons

7 Used to...tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atomsNa in Na= 0neutral already... no need to add any electrons cationsNa in Na + = +1need to add 1 electron to make Na + neutral anionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral OXIDATION NUMBER

8 Q. What are the oxidation states of the elements in the following? a) C b) Fe 3+ c) Fe 2+ d) O 2- e) He f) Al 3+ Q. What are the oxidation states of the elements in the following? a) C b) Fe 3+ c) Fe 2+ d) O 2- e) He f) Al 3+ Used to...tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atomsNa in Na= 0neutral already... no need to add any electrons cationsNa in Na + = +1need to add 1 electron to make Na + neutral anionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral

9 OXIDATION NUMBER Q. What are the oxidation states of the elements in the following? a) C (0)b) Fe 3+ (+3)c) Fe 2+ (+2) d) O 2- (-2) e) He (0) f) Al 3+ (+3) Q. What are the oxidation states of the elements in the following? a) C (0)b) Fe 3+ (+3)c) Fe 2+ (+2) d) O 2- (-2) e) He (0) f) Al 3+ (+3) Used to...tell if oxidation or reduction has taken place work out what has been oxidised and/or reduced ATOMS AND SIMPLE IONS The number of electrons which must be added or removed to become neutral atomsNa in Na= 0neutral already... no need to add any electrons cationsNa in Na + = +1need to add 1 electron to make Na + neutral anionsCl in Cl¯ = -1 need to take 1 electron away to make Cl¯ neutral

10 OXIDATION NUMBER MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x -2 = Zero

11 because CO 2 is a neutral molecule, the sum of the oxidation states must be zero for this, one element must have a positive OS and the other must be negative OXIDATION NUMBER Explanation MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x -2 = Zero

12 HOW DO YOU DETERMINE WHICH IS THE POSITIVE ONE? the more electronegative species will have the negative value 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 NUMBER MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x +2 = Zero

13 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 OXIDATION NUMBER MOLECULES The SUM of the oxidation states adds up to ZERO ELEMENTSH in H 2 = 0both are the same and must add up to Zero COMPOUNDSC in CO 2 = +4 O in CO 2 = -21 x +4 and 2 x +2 = Zero

14 OXIDATION NUMBER in SO 4 2- the oxidation state of S = +6there is ONE S O = -2there are FOUR O’s +6 + 4(-2) = -2 so the ion has a 2- charge COMPLEX IONS The SUM of the oxidation states adds up to THE CHARGE e.g.NO 3 - sum of the oxidation states = - 1 SO 4 2- sum of the oxidation states = - 2 NH 4 + sum of the oxidation states = +1 Examples

15 OXIDATION NUMBER What is the oxidation state (OS) of Mn in MnO 4 ¯ ? 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 MnO 4 ¯is +7 +7 + 4(-2) = - 1 COMPLEX IONS The SUM of the oxidation states adds up to THE CHARGE e.g.NO 3 - sum of the oxidation states = - 1 SO 4 2- sum of the oxidation states = - 2 NH 4 + sum of the oxidation states = +1 Examples

16 HYDROGEN +1 except 0atom (H) and molecule (H 2 ) -1hydride ion, H¯ in sodium hydride NaH OXYGEN -2 except 0atom (O) and molecule (O 2 ) -1in hydrogen peroxide, H 2 O 2 +2in F 2 O FLUORINE -1 except 0atom (F) and molecule (F 2 ) OXIDATION NUMBER CALCULATING OXIDATION NUMBER - 1 Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other values

17 HYDROGEN +1 except 0atom (H) and molecule (H 2 ) -1hydride ion, H¯ in sodium hydride NaH OXYGEN -2 except 0atom (O) and molecule (O 2 ) -1in hydrogen peroxide, H 2 O 2 +2in F 2 O FLUORINE -1 except 0atom (F) and molecule (F 2 ) OXIDATION STATES Q. Give the oxidation state of the element other than O, H or F in... SO 2 NH 3 NO 2 NH 4 + IF 7 Cl 2 O 7 NO 3 ¯NO 2 ¯SO 3 2- S 2 O 3 2- S 4 O 6 2- MnO 4 2- What is odd about the value of the oxidation state of S in S 4 O 6 2- ? Q. Give the oxidation state of the element other than O, H or F in... SO 2 NH 3 NO 2 NH 4 + IF 7 Cl 2 O 7 NO 3 ¯NO 2 ¯SO 3 2- S 2 O 3 2- S 4 O 6 2- MnO 4 2- What is odd about the value of the oxidation state of S in S 4 O 6 2- ? CALCULATING OXIDATION NUMBER - 1 Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other values

18 OXIDATION NUMBER A. The oxidation states of the elements other than O, H or F are SO 2 O = -22 x -2 = - 4overall neutralS = +4 NH 3 H = +13 x +1 = +3overall neutralN = - 3 NO 2 O = -2 2 x -2 = - 4 overall neutralN = +4 NH 4 + H = +14 x +1 = +4overall +1N = - 3 IF 7 F = -17 x -1 = - 7overall neutralI = +7 Cl 2 O 7 O = -2 7 x -2 = -14 overall neutralCl = +7 (14/2) NO 3 ¯ O = -2 3 x -2 = - 6 overall -1N = +5 NO 2 ¯ O = -2 2 x -2 = - 4 overall -1N = +3 SO 3 2- O = -2 3 x -2 = - 6 overall -2S = +4 S 2 O 3 2- O = -2 3 x -2 = - 6 overall -2S = +2 (4/2) S 4 O 6 2- O = -2 6 x -2 = -12 overall -2S = +2½ ! (10/4) MnO 4 2- O = -2 4 x -2 = - 8overall -2Mn = +6 What is odd about the value of the oxidation state of S in S 4 O 6 2- ? An oxidation state must be a whole number (+2½ is the average value)

19 METALS have positive values in compounds value is usually that of the Group Number Al is +3 NON-METALS mostly negative based on their usual ionCl usually -1 OXIDATION NUMBER CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide

20 OXIDATION STATES CALCULATING OXIDATION STATE - 2 Q. What is the oxidation state of each element in the following compounds/ions ? CH 4 PCl 3 NCl 3 CS 2 ICl 5 BrF 3 PCl 4 + H 3 PO 4 NH 4 Cl H 2 SO 4 MgCO 3 SOCl 2

21 OXIDATION STATES CALCULATING OXIDATION STATE - 2 Q. What is the oxidation state of each element in the following compounds/ions ? CH 4 C = - 4H = +1 PCl 3 P = +3Cl = -1 NCl 3 N = +3Cl = -1 CS 2 C = +4S = -2 ICl 5 I = +5Cl = -1 BrF 3 Br = +3F = -1 PCl 4 + P = +4Cl = -1 H 3 PO 4 P = +5H = +1O = -2 NH 4 ClN = -3H = +1Cl = -1 H 2 SO 4 S = +6H = +1O = -2 MgCO 3 Mg = +2H = +4O = -2 SOCl 2 S = +4Cl = -1O = -2

22 manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO 2 sulphur(VI) oxide for SO 3 S is in the +6 oxidation state dichromate(VI) for Cr 2 O 7 2- Cr is in the +6 oxidation state phosphorus(V) chloride for PCl 5 P is in the +5 oxidation state phosphorus(III) chloride for PCl 3 P is in the +3 oxidation state OXIDATION STATES THE ROLE OF OXIDATION STATE IN NAMING SPECIES To avoid ambiguity, the oxidation state is often included in the name of a species Q. Name the following...PbO 2 SnCl 2 SbCl 3 TiCl 4 BrF 5

23 OXIDATION STATES Q. Name the following...PbO 2 lead(IV) oxide SnCl 2 tin(II) chloride SbCl 3 antimony(III) chloride TiCl 4 titanium(IV) chloride BrF 5 bromine(V) fluoride manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO 2 sulphur(VI) oxide for SO 3 S is in the +6 oxidation state dichromate(VI) for Cr 2 O 7 2- Cr is in the +6 oxidation state phosphorus(V) chloride for PCl 5 P is in the +5 oxidation state phosphorus(III) chloride for PCl 3 P is in the +3 oxidation state THE ROLE OF OXIDATION STATE IN NAMING SPECIES To avoid ambiguity, the oxidation state is often included in the name of a species

24 REDOXWhen reduction and oxidation take place OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTIONGain of electrons ‘RIG’ species will become more negative or less positive REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H

25 REDOXWhen reduction and oxidation take place OXIDATION Removal (loss) of electrons ‘OIL’ species will get less negative or more positive REDUCTIONGain of electrons ‘RIG’ species will become more negative or less positive REDUCTION 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 REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H

26 REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species has been OXIDISED REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Q. State if the changes involve oxidation (O) or reduction (R) or neither (N) Fe 2+ —>Fe 3+ I 2 —>I¯ F 2 —> F 2 O

27 REDOX REACTIONS REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species has been OXIDISED OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Q. State if the changes involve oxidation (O) or reduction (R) or neither (N) Fe 2+ —>Fe 3+ O +2 to +3 I 2 —>I¯ R 0 to -1 F 2 —> F 2 OR 0 to -1

28 Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) Redox reactions of metals with acids

29 Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) Redox reactions of metals with acids We can assign oxidaton numbers to each atom in any equation in order to –  Identify whether a redox reaction has taken place  Work out what has been oxidised and what has been reduced.

30 Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) Redox reactions of metals with acids The metal is oxidised, forming positive metal ions The hydrogen ion in the acid is reduced, forming the element hydrogen, as a gas We can write the above equation to show the role of the hydrogen ion, H + H + (aq)  Mg 2+ +H 2 (g) Mg (s) + 2H + (aq)  Mg 2+ +H 2 (g) We can assign oxidaton numbers to each atom in any equation in order to –  Identify whether a redox reaction has taken place  Work out what has been oxidised and what has been reduced.


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