Oxidation-Reduction Reactions LESSON OBJECTIVES You will learn: 1) The purpose for oxidation numbers 2) The rules for oxidation numbers 3) How to find oxidation numbers, example questions 4) How to use oxidation numbers to determine if a reaction is oxidation or reduction

MOLECULES: SO VERY HARD TO FOLOW THE e- Compare and ionic reaction to a molecular one: Ionic: 2Na(s) + Cl2(g)  2 NaCl(s) Following the electrons we see Na (s) has picked up an e- and Cl has gained it. Molecular: 2 H2(s) + O2(g)  2H2O(g) Ok, so were did the electrons go? You cannot tell. So we need a way to FOLLOW the e- The way we “follow” electrons is to use oxidation numbers, or rules invented to help us to “see” the electrons when they are invisible!

Oxidation Numbers for Bookkeeping Hydrogen has "lost" 1 electron to oxygen. Oxygen has "gained" 2 electrons Partial charges given to atoms in a molecule in this way are called oxidation numbers. We can use oxidation numbers to keep track of where electrons are in a molecule Oxygen has an oxidation number = -2 because the oxygen atom has "gained" 2 e- Each H still has an oxidation number = +1 . Oxygen, however, now has an oxidation # of -1 because each oxygen gains 1 electron from each hydrogen (it equally shares the other so it does not “go” to anyone in particular) 

Oxidation Number Rules The charge the atom would have in a molecule if electrons were completely transferred. Free elements (uncombined state); oxidation number = 0. Na, Be, K, Pb, H2, O2, P4 = 0 In monatomic ions, the oxidation # is equal to the charge. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 The oxidation # of oxygen is usually - 2. In H2O2 , oxygen = -1. The oxidation number of hydrogen is +1 except when it is bonded to metals (as in metal hydrides) in binary compounds. In these cases, its oxidation number is : -1. Ex: (LiH4)

Oxidation Numbers The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge. Ex: Cr2O72- all the ox #’s must add to equal -2 thus 2 Cr + 7(-2) = -2 2 Cr + -14 = -2 2 Cr = -12 Cr = +6 If all else fails play the electronegativity's game, whichever has the higher electronegativity, wins its charge. Ex: PCl3 notice we have no rule for either P or Cl Thus we look up their electronegativity's, P= 2.1 and Cl =3.0 Cl “wins” or it has a higher electronegaivirty thus we give it whatever charge it would have if it were ionic thus: Cl = -1 and P has to be -3 in PCl3

CALCULATING OXIDATION STATE OXIDATION STATES CALCULATING OXIDATION STATE Q. 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

CALCULATING OXIDATION STATE OXIDATION STATES CALCULATING OXIDATION STATE Q. 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

THE ROLE OF OXIDATION STATE IN NAMING SPECIES 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 manganese(IV) oxide shows that Mn is in the +4 oxidation state in MnO2 sulphur(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 Name the following... PbO2 Lead (IV) Oxide SnCl2 Tin (II) Chloride SbCl3 Antimony (III) Chloride TiCl4 Titanium (IV) Chloride

REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS Oxidation and reduction are not only defined as changes in O and H REDOX When reduction and oxidation take place OXIDATION LEO REDUCTION GER 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)

GER! LEO says Loss of Electrons = Oxidation Gain of Electrons = Reduction

OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species has been OXIDISED Q. State if the changes involve oxidation (O) or reduction (R) or neither (N) Fe2+ —> Fe3+ I2 —> I¯ F2 —> F2O C2O42- —> CO2 H2O2 —> O2 H2O —> H2O Cr2O72- —> Cr3+ Cr2O72- —> CrO42- SO42- —> SO2

OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDOX REACTIONS OXIDATION AND REDUCTION IN TERMS OF ELECTRONS REDUCTION in O.S. INCREASE in O.S. Species has been REDUCED Species has been OXIDISED O = Oxidation , R = Reduction 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 H2O —> H2O N -1 to -1 for H and -2 to -2 for O Cr2O72- —> Cr3+ R +6 to +3 Cr2O72- —> CrO42- N +6 to +6 SO42- —> SO2 R +6 to +4

Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 + KBr  KCl + Br2  

So this is a reduction reaction Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 + KBr  KCl + Br2  0 +1-1 +1-1 0 Br increases from –1 to 0 -- oxidized Cl decreases from 0 to –1 -- Reduced So this is a reduction reaction K remains unchanged at +1

Exercise For each of the following reactions find the element oxidized and the element reduced   Cu + HNO3  Cu(NO3)2 + NO2 + H2O

Exercise For each of the following reactions find the element oxidized and the element reduced Cu + HNO3  Cu(NO3)2 + NO2 + H2O 0 +1+5-2 +2 +5-2 +4 –2 +1-2 Cu increases from 0 to +2. It is oxidized Only part of the N in nitric acid changes from +5 to +4. It is reduced The nitrogen that ends up in copper nitrate remains unchanged

Exercise For each of the following reactions find the element oxidized and the element reduced HNO3 + I2  HIO3 + NO2

Exercise For each of the following reactions find the element oxidized and the element reduced HNO3 + I2  HIO3 + NO2 1 +5 -2 0 +1+5-2 +4-2 N is reduced from +5 to +4. It is reduced I is increased from 0 to +5 It is oxidized The hydrogen and oxygen remain unchanged.

Oxidation half-reaction (lose e-) Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 2+ 2- 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-)

Balancing Redox Equations LESSON OBJECTIVES You will learn: How to balance redox equations with acids

Balancing Redox Equations Balance the oxidation of Fe2+ to Fe3+ by Cr2O72- in acid solution Write the unbalanced equation for the reaction in ionic form. Fe2+ + Cr2O72- Fe3+ + Cr3+ Separate the equation into two half-reactions. Fe2+ Fe3+ +2 +3 Oxidation: Cr2O72- Cr3+ +6 +3 Reduction: Balance the atoms other than O and H in each half-reaction. Cr2O72- 2Cr3+

Balancing Redox Equations For reactions in acid, add H2O to balance O atoms and H+ to balance H atoms. Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- 2Cr3+ + 7H2O Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe2+ Fe3+ + 1e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O If necessary, equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 6Fe2+ 6Fe3+ + 6e- 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O

Balancing Redox Equations Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. Oxidation: 6Fe2+ 6Fe3+ + 6e- Reduction: 6e- + 14H+ + Cr2O72- 2Cr3+ + 7H2O 14H+ + Cr2O72- + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 For reactions in basic solutions, add OH- to both sides of the equation for every H+ that appears in the final equation.

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 Q. Construct balanced redox equations for the reactions between... Mg and H+  H2 and Mg2+ Cr2O72- and Fe2+  Fe3+ + 2Cr3+ H2O2 and MnO4¯  Mn2+ + O2 C2O42- and MnO4¯.  CO2 + Mn2+ S2O32- and I2.  S4O62- + I¯ Cr2O72- and I¯

ANSWERS Mg ——> Mg2+ + 2e¯ (x1) H+ + e¯ ——> ½ H2 (x2) Mg + 2H+ ——> Mg2+ + H2 Cr2O72- + 14H+ + 6e¯ ——> 2Cr3+ + 7H2O (x1) Fe2+ ——> Fe3+ + e¯ (x6) Cr2O72- + 14H+ + 6Fe2+ ——> 2Cr3+ + 6Fe2+ + 7H2O 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) ½ I2 + e¯ ——> I¯ (x6) Cr2O72- + 14H+ + 3I2 ——> 2Cr3+ + 6I ¯ + 7H2O

Balancing redox reactions by the half-reaction method PROBLEM: Permanganate ion is a strong oxidizing agent and its deep purple color makes it useful as an indicator in redox titrations. It reacts in basic solution with the oxalate anion to form carbonate anion and solid manganese dioxide. Balance the skeleton ionic reaction that occurs between KMnO4 and Na2C2O4 in basic solution. MnO4-(aq) + C2O42-(aq) MnO2(s) + CO32-(aq) PLAN: Proceed in acidic solution and then neutralize with base. SOLUTION: Red Ox MnO4- MnO2 C2O42- CO32- +7 +4 +3 +4 MnO4- MnO2 C2O42- CO32- 4H+ + MnO4- MnO2 + 2H2O C2O42- 2CO32- + 2H2O + 4H+ + 3e- + 2e-

(continued) x2 x3 C2O42- + 2H2O 2CO32- + 4H+ + 2e- 4H+ + MnO4- +3e- MnO2+ 2H2O 8H+ + 2MnO4- + 6e- 2MnO2+ 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 8H+ + 2MnO4- + 6e- 2MnO2 + 4H2O 3C2O42- + 6H2O 6CO32- + 12H+ + 6e- 2MnO2-(aq) + 3C2O42-(aq) + 2H2O(l) 2MnO2(s) + 6CO32-(aq) + 4H+(aq) + 4OH- + 4OH- 2MnO2-(aq) + 3C2O42-(aq) + 4OH-(aq) 2MnO2(s) + 6CO32-(aq) + 2H2O(l)

Disproportionation Disproportionation is the simultaneous oxidation and reduction of the same species. 3 Cl2 + 6 OH− → 5 Cl− + ClO3− + 3 H2O The chlorine gas reactant is in oxidation state 0. In the products, the chlorine in the Cl− ion has an oxidation number of −1, having been reduced. Whereas the oxidation number of the chlorine in the ClO3− ion is +5, indicating that it has been oxidized.