REDOX A guide for A level students KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

INTRODUCTION This 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... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING REDOX

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 REDOX

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

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

... 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 OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H OXIDATION & REDUCTION - Definitions

... 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 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

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 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 STATES

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 construct half equations and balance redox equations 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 STATES 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 construct half equations and balance redox equations 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 STATES 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

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 STATES 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

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 STATES 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

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 STATES 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

OXIDATION STATES in SO 4 2- the oxidation state of S = +6there is ONE S O = -2there are FOUR O’s (-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

OXIDATION STATES 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 (-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

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 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 values

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 STATE - 1 Many elements can exist in more than one oxidation state In compounds, certain elements are used as benchmarks to work out other values

OXIDATION STATES 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)

METALS 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,+7 NON-METALS mostly negative based on their usual ionCl usually -1 can have values up to their Group No. Cl or +7 OXIDATION STATES CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide

OXIDATION STATES Q. What is the theoretical maximum oxidation state of the following elements? NaPBaPbSMnCr What will be the usual and the maximum oxidation state in compounds of? LiBrSrOBN+1 Q. What is the theoretical maximum oxidation state of the following elements? NaPBaPbSMnCr What will be the usual and the maximum oxidation state in compounds of? LiBrSrOBN+1 METALS 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,+7 NON-METALS mostly negative based on their usual ionCl usually -1 can have values up to their Group No. Cl or +7 CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide

OXIDATION STATES CALCULATING OXIDATION STATE - 2 The position of an element in the periodic table can act as a guide A. What is the theoretical maximum oxidation state of the following elements? NaPBaPbSMnCr What will be the usual and the maximum oxidation state in compounds of? LiBrSrOBN USUAL or +5 MAXIMUM

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

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 = +2C = +4O = -2 SOCl 2 S = +4Cl = -1O = -2

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

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

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

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

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 C 2 O 4 2- —> CO 2 H 2 O 2 —> O 2 H 2 O 2 —> H 2 O Cr 2 O 7 2- —> Cr 3+ Cr 2 O 7 2- —> CrO 4 2- SO 4 2- —> SO 2

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 C 2 O 4 2- —> CO 2 O +3 to +4 H 2 O 2 —> O 2 O -1 to 0 H 2 O 2 —> H 2 O R -1 to -2 Cr 2 O 7 2- —> Cr 3+ R +6 to +3 Cr 2 O 7 2- —> CrO 4 2- N +6 to +6 SO 4 2- —> SO 2 R +6 to +4

1 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 side BALANCING REDOX HALF EQUATIONS

1 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 side Example 1 Iron(II) being oxidised to iron(III) Step 1Fe 2+ ——> Fe 3+ Step Step 3Fe 2+ ——> Fe 3+ + 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 5 BALANCING REDOX HALF EQUATIONS

1 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 side Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution

No need to balance Mn; equal numbers BALANCING REDOX HALF EQUATIONS 1 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 side Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+

BALANCING REDOX HALF EQUATIONS Overall charge on MnO 4 ¯ is -1; sum of the OS’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 + (4x -2) = -1] 1 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 side Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step

BALANCING REDOX HALF EQUATIONS 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 1 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 side Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step Step 3 MnO 4 ¯ + 5e¯ ———> Mn 2+

BALANCING REDOX HALF EQUATIONS 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 1 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 side Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step Step 3 MnO 4 ¯ + 5e¯ ———> Mn 2+ Step 4MnO 4 ¯ + 5e¯ + 8H + ———> Mn 2+

Example 2 MnO 4 ¯ being reduced to Mn 2+ in acidic solution Step 1 MnO 4 ¯ ———> Mn 2+ Step Step 3 MnO 4 ¯ + 5e¯ ———> Mn 2+ Step 4MnO 4 ¯ + 5e¯ + 8H + ———> Mn 2+ Step 5MnO 4 ¯ + 5e¯ + 8H + ———> Mn H 2 O now balanced BALANCING REDOX HALF EQUATIONS Everything balances apart from oxygen and hydrogenO LHS = 4RHS = 0 H LHS = 8RHS = 0 You must ADD 4 WATER MOLECULES to the RHS; the equation is now balanced 1 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 side

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 Cr 2 O 7 2- being reduced to Cr 3+ in acidic solution Step 1 Cr 2 O 7 2- ———> Cr 3+ there are two Cr’s on LHS Cr 2 O 7 2- ———> 2Cr 3+ both sides now have 2 Step both Cr’s are reduced Step 3 Cr 2 O e¯ ——> 2Cr 3+ each Cr needs 3 electrons Step 4 Cr 2 O e¯ + 14H + ——> 2Cr 3+ Step 5 Cr 2 O e¯ + 14H + ——> 2Cr H 2 O now balanced BALANCING REDOX HALF EQUATIONS

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 Q. Balance the following half equations... Na —> Na + Fe 2+ —> Fe 3+ I 2 —> I¯ C 2 O 4 2- —> CO 2 H 2 O 2 —> O 2 H 2 O 2 —> H 2 O NO 3 - —> NO NO 3 - —> NO 2 SO 4 2- —> SO 2

BALANCING REDOX HALF EQUATIONS Q. Balance the following half equations... Na —> Na + + e - Fe 2+ —> Fe 3+ + e - I 2 + 2e - —>2I¯ C 2 O 4 2- —> 2CO 2 + 2e - H 2 O 2 —> O 2 + 2H + + 2e - H 2 O 2 + 2H + + 2e - —> 2H 2 O NO H + + 3e - —> NO + 2H 2 O NO H + + e - —> NO 2 + H 2 O SO H + + 2e - —> SO 2 + 2H 2 O

COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

COMBINING HALF EQUATIONS A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides The reaction between manganate(VII) and iron(II)

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 OReduction A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 OReduction Step 2 5Fe 2+ ——> 5Fe e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 Omultiplied by 1 A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 OReduction Step 2 5Fe 2+ ——> 5Fe e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 Omultiplied by 1 Step 3MnO 4 ¯ + 5e¯ + 8H + + 5Fe 2+ ——> Mn H 2 O + 5Fe e¯ A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

COMBINING HALF EQUATIONS The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 OReduction Step 2 5Fe 2+ ——> 5Fe e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 Omultiplied by 1 Step 3MnO 4 ¯ + 5e¯ + 8H + + 5Fe 2+ ——> Mn H 2 O + 5Fe e¯ Step 4 MnO 4 ¯ + 8H + + 5Fe 2+ ——> Mn H 2 O + 5Fe 3+ A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

The reaction between manganate(VII) and iron(II) Step 1Fe 2+ ——> Fe 3+ + e¯Oxidation MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 OReduction Step 2 5Fe 2+ ——> 5Fe e¯multiplied by 5 MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 Omultiplied by 1 Step 3MnO 4 ¯ + 5e¯ + 8H + + 5Fe 2+ ——> Mn H 2 O + 5Fe e¯ Step 4 MnO 4 ¯ + 8H + + 5Fe 2+ ——> Mn H 2 O + 5Fe 3+ COMBINING HALF EQUATIONS SUMMARY A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides

A combination of two ionic half equations, one involving oxidation and the other reduction, produces a REDOX equation. The equations are balanced as follows... Step 1 Write out the two half equations Step 2 Multiply the equations so that the number of electrons in each is the same Step 3 Add the two equations and cancel out the electrons on either side Step 4 If necessary, cancel any other species which appear on both sides COMBINING HALF EQUATIONS Q. Construct balanced redox equations for the reactions between... Mg and H + Cr 2 O 7 2- and Fe 2+ H 2 O 2 and MnO 4 ¯ C 2 O 4 2- and MnO 4 ¯ S 2 O 3 2- and I 2 Cr 2 O 7 2- and I¯

Mg ——> Mg e ¯ (x1) H + + e ¯ ——> ½ H 2 (x2) Mg + 2H + ——> Mg 2+ + H 2 Cr 2 O H + + 6e ¯ ——> 2Cr H 2 O (x1) Fe 2+ ——> Fe 3+ + e ¯ (x6) Cr 2 O H + + 6Fe 2+ ——> 2Cr Fe H 2 O MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 O (x2) H 2 O 2 ——> O 2 + 2H + + 2e ¯ (x5) 2MnO 4 ¯ + 5H 2 O 2 + 6H + ——> 2Mn O 2 + 8H 2 O MnO 4 ¯ + 5e¯ + 8H + ——> Mn H 2 O (x2) C 2 O 4 2- ——> 2CO 2 + 2e ¯ (x5) 2MnO 4 ¯ + 5C 2 O H + ——> 2Mn CO 2 + 8H 2 O 2S 2 O 3 2- ——> S 4 O e¯ (x1) ½ I 2 + e¯ ——> I ¯ (x2) 2S 2 O I 2 ——> S 4 O I ¯ Cr 2 O H + + 6e ¯ ——> 2Cr H 2 O (x1) ½ I 2 + e¯ ——> I ¯ (x6) Cr 2 O H + + 3I 2 ——> 2Cr I ¯ + 7H 2 O BALANCING REDOX EQUATIONS ANSWERS

REVISION CHECK What should you be able to do? Recall the definitions for oxidation and reduction in terms of oxygen, hydrogen and electrons Write balanced equations representing oxidation and reduction Know the trend in electronegativity across periods Predict the oxidation state of elements in atoms, simple ions, compounds and complex ions Recognize, in terms of oxidation state, if oxidation or reduction has taken place Balance ionic half equations Combine two ionic half equations to make a balanced redox equation CAN YOU DO ALL OF THESE? YES NO

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