Oxidation Number (O.N.) = a parameter which enables one to determine if a species is oxidized or reduced as the species undergoes chemical reaction.

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1 Page 1 University of Pennsylvania Department of Chemistry Chemistry 101 Engineering Sections Balancing OxidationReduction (Redox) Reactions I. The basic process: OxidationReduction Reaction A process which involves a "transfer of electrons (e ) " when species react chemically to form products. II. Definition of terms: Oxidation = when a species "loses" or supplies electrons to another species. This "supplying" species is said to be oxidized. It is also referred to as the reducing agent or reductant. Reduction = when a species accepts electrons from another species. This "accepting" species is said to be reduced. It is also referred to as the oxidizing agent. Thus, the species that undergoes oxidation is the reducing agent because it supplies the electrons to the species that is reduced. Alternatively, the species that undergoes reduction is the oxidizing agent or oxidant because it accepts electrons from the species that is oxidized. Oxidation Number (O.N.) = a parameter which enables one to determine if a species is oxidized or reduced as the species undergoes chemical reaction. III. Oxidation Numbers (ON) [or Oxidation States (OS)] Some rules for determining Oxidation Numbers (ON) (See the text, for a more detailed list.) 1. Free elements: ON of each atom is 0 (zero). Ex.: Zn(s) ON of Zn is 0 ; H 2 (g) ON of each H atom is 0 (zero) ; S 8 (s) ON of each S is 0 (zero). 2. Ions consisting of one (1) atom: ON = charge on ion. Ex.: Cl ON of Cl is 1 ; Al +3 ON of Al is Hydrogen, in compounds with other elements: ON = +1, except in hydrides: LiH, NaH, CaH 2, etc. In these cases, ON = Oxygen, in compounds with other elements: ON =, except in peroxides: H 2 O 2, Na 2 O 2, etc. In these cases, ON = 1. In KO 2 ON of O is 1/2. 5. Algebraic sum of ON's in a neutral compound must be zero (0) ; in a polyatomic ion the sum of the ON's equals the charge on the ion. Ex.: Li 2 O : ON of each Li = +1; ON of O = ; sum = (2)(+1) + () = 0. H 3 PO 4 : Since the molecule is neutral, sum of ON's = 0 (zero). \ 0 = (3)(+1) + (4)() + (ON of P) ; thus, ON of P = +5. Show: ON of Mn in MnO 1 4 is +7; and ON of Cr (each one), in Cr 2 O 7 is ON of each F in F 2 = 0 (zero). ON of F in compounds with other atoms = 1. "Always"!! We will see why when we discuss atomic structure and chemical bonding.

2 Page 2 IV. Balancing Redox Equations: A. Oxidation States Method This method will be presented by way of an example. The sample reaction is the combustion of methanol to carbon dioxide and water. Unbalanced Reaction: CH 3 OH(l) + O 2 (g) > CO 2 (g) + H 2 O(l). Step 1 Determine if the reaction is a redox reaction. Using the rules, we have oxidation state (OS) of H in methanol is +1 and the OS of O is. So, the OS of C must be by the sum rule. The OS of O in O 2 is 0. In CO 2 the OS of O is. From the sum rule, the OS of C is +4. The OS of O in H 2 O is and the OS of H in water is +1. Consolidating this information, we conclude the following: C : The OS changes from in methanol to +4 in carbon dioxide. C is oxidized and so methanol is the reducing agent or reductant. O : For each O, the OS changes from 0 in O 2 to in water (or carbon dioxide). O is reduced and so O 2 is the oxidizing agent or oxidant. Thus, this is a redox reaction. Step 2 Balance INCREASE in OS (of C ( of 6)) of oxidized species by the DECREASE in OS of reduced species (of each O (Ø of 2 each)). Need 3x as much O in O 2 as have C in methanol to balance the change. CH 3 OH(l) O 2(g) > CO 2 (g) + 2 H 2 O(l). The 2 in front of H 2 O is finally inserted to balance atoms of each element on both sides of the reaction. B. Balancing Redox Reactions Using the HalfReaction Method This method will be presented by way of an example. This example will contain another common situation oxidationreduction processes often occur in either (aqueous) acidic conditions or (aqueous) basic conditions. The following example reaction will be the oxidation of ethanol (C 2 H 5 OH) by dichromate anion (Cr 2 O 7 ) in acidic aqueous solution. The products are acetic acid (CH 3 CO 2 H) and the Cr +3 cation. The physical states of all species will be included in the initial unbalanced reaction and in the final balanced reaction. To avoid clutter we will leave out the physical states in the intermediate steps. Unbalanced Rxn: C 2 H 5 OH (aq) + Cr 2 O 7 (aq) > CH 3 CO 2 H (aq) + Cr +3 (aq) (orangered) (green) Step 1 What is oxidized/reduced? Using the rules, we have oxidation state (OS) of H in ethanol and in acetic acid is +1. The OS of O is in ethanol, in the dichromate anion, and in acetic acid. So, the OS of each C must be in ethanol by the sum rule. Also from the sum rule, the OS of each Cr is +6 in the dichromate anion. The OS of each carbon in the acetic acid is 0 by the sum rule. Finally, the OS of Cr in the Cr +3 cation is +3. Consolidating this information, we conclude the following: C : The OS changes from for each C in ethanol to 0 in acetic acid. C is oxidized and so ethanol is the reducing agent or reductant. Cr : For each Cr, the OS changes from +6 in Cr 2 O 7 to +3 in the Cr +3 cation. Cr is reduced and so Cr 2 O 7 is the oxidizing agent or oxidant.

3 Step 2 Write out each halfreaction (reductant & associated product(s) and oxidant & associated product(s)). Deal with stoichiometry of species changing their OS. Material balance each reaction first with respect to species undergoing OS change. Oxidation: C 2 H 5 OH > CH 3 CO 2 H (already balanced for C) Reduction: Cr 2 O 7 > 2 Cr +3 (now balanced for Cr) Balance OS change by appropriate addition of electrons. Each C atom changes OS form to 0. This gives rise to an overall loss of 4 electrons, (i.e., (2)() > 0). Thus we add 4 electrons to the products side of the oxidation halfreaction. Each Cr atom changes OS form +6 to +3. This gives rise to an overall gain of 6 electrons, (i.e., (2)(+6) > (2)(+3)). Thus we add 6 electrons to the reactants side of the reduction halfreaction. Oxidation: C 2 H 5 OH > CH 3 CO 2 H + 4 e Reduction: 6 e + Cr 2 O 7 > 2 Cr +3 Step 3 Balance each halfreaction for: (i) Charge (ii) Atoms In acid: Add H + & H 2 O as needed. In base: Add OH & H 2 O as needed. Oxidation: (i) Add 4 H + to righthandside to balance charge. (ii) Add H 2 O to lefthandside to balance mass [H's & O's]. [A] C 2 H 5 OH + H 2 O > CH 3 CO 2 H + 4 e + 4 H + (Net charge of 0 on each side of reaction [A].) Reduction: (i) Add 14 H + to lefthandside to balance charge. (ii) Add 7 H 2 O to righthandside to balance mass [H's & O's]. [B] 6 e + Cr 2 O H + > 2 Cr H 2 O (Net charge of +6 on each side of reaction [B].) Step 4 Multiply halfreactions by the smallest integers to balance the # of electrons... then add reactions and simplify. Here, to balance # of electrons: [A] x 3 & [B] x 2 Add and simplify. [A] x 3: 3 C 2 H 5 OH + 3 H 2 O > 3 CH 3 CO 2 H + 12 e + 12 H + [B] x 2: 12 e + 2 Cr 2 O H + > 4 Cr H 2 O Page 3 3 C 2 H 5 OH(aq) + 2 Cr 2 O 7 (aq) + 16 H + (aq)>3 CH 3 CO 2 H(aq) + 4 Cr +3 (aq) + 11 H 2 O(l) Doublecheck for yourself that atoms and net charge are balanced on each side of the reaction.

4 Page 4 Another Example Disproportionation Reactions A redox reaction where just a single substance acts as both oxidant and reductant. The example is the disproportionation of the aqueous manganate anion (MnO 4 ) into solid manganese (IV) oxide (MnO 2 ) and the aqueous permangante anion (MnO 4 ) in aqueous acid solution. This time we will display the steps without detailed comment. Unbalanced Reaction: MnO 4 (aq) > MnO 2 (s) + MnO 4 (aq). Step 1 What is oxidized/reduced? The OS of O is throughout. From the sum rule, we have the following. Mn in MnO 4 has an OS of +6, the Mn in MnO 2 has an OS of +4, and the Mn in MnO 4 has an OS of +7. As revealed above, Mn is both oxided and reduced. MnO 4 / MnO 4 : The OS of Mn changes from+6 to +7. Mn is oxidized in this case. MnO 4 / MnO 2 : The OS of Mn changes from+6 to +4. Mn is reduced in this case. Step 2 Write out each halfreaction. Deal with stoichiometry of species changing their OS. Material balance each reaction first with respect to species undergoing OS change. Oxidation: MnO 4 > MnO 4 (already balanced for Mn) Reduction: MnO 4 > MnO 2 (already balanced for Mn) Balance OS change by appropriate addition of electrons. In the oxidation halfreaction, the change in OS of Mn is +1, i,e., an overall loss of 1 electron. Thus we add 1 electron to the products side of the oxidation halfreaction. In the reduction halfreaction, the change in OS of Mn is, i,e., an overall loss of 2 electrons. Thus we add 2 electrons to the reactants side of the reduction halfreaction. Oxidation: MnO 4 > MnO 4 + e Reduction: 2 e + MnO 4 > MnO 2 Step 3 Balance each halfreaction for: (i) Charge (ii) Atoms In acid: Add H + & H 2 O as needed. Oxidation: (i) No H + needed to balance charge. (ii) No H 2 O needed to balance mass [H's & O's]. [A] MnO 4 > MnO 4 + e (Net charge of on each side of reaction [A].)

5 Page 5 Step 3 Continued: Reduction: (i) Add 4 H + to lefthandside to balance charge. (ii) Add 2 H 2 O to righthandside to balance mass [H's & O's]. [B] 2 e + MnO H + > MnO H 2 O (Net charge of zero (0) on each side of reaction [B].) Step 4 Multiply halfreactions by the smallest integers to balance the # of electrons... then add reactions and simplify. Here, to balance # of electrons: [A] x 2 & [B] No change. Add and simplify. [A] x 2: 2 MnO 4 > 2 MnO e [B] unchanged: 2 e + MnO H + > MnO H 2 O 3 MnO 4 (aq) + 4 H + (aq) > 2 MnO 4 (aq) + MnO 2 (s) + 2 H 2 O(l) Doublecheck for yourself that atoms and net charge are balanced on each side of the reaction.

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