Electrochemistry Oxidation Reduction and Oxidation Numbers. Many important chemical reactions involve the transfer of electrons.

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1 Electrochemistry Oxidation Reduction and Oxidation Numbers Many important chemical reactions involve the transfer of electrons. e.g. 2Na + Cl 2 2 NaCl Many do not. e.g. Pb(NO 3 ) 2 (aq) + 2KI(aq) PbI 2 (s) + 2KNO 3 (aq) Chemists have devised a useful bookkeeping method to determine if electron transfer is involved in a chemical reaction. It is referred to as the oxidation state method. If an atom loses electrons, it is oxidized. If an atom gains electrons, it is reduced. The oxidation state or equivalently, the oxidation number, of an atom is the charge assigned to an atom using the oxidation state method.

2 Rules for assigning oxidation numbers: 1. Elements in their most abundant naturally occurring form are assigned an oxidation number of zero. e.g. Na, Fe, Cl 2, O 2 2. The sum of the oxidation numbers for a compound or formula unit is zero. 3. For a polyatomic ion, the oxidation numbers of the constituent atoms sums to the ion charge. 4. Monatomic ions are assigned oxidation numbers equal to the ion charge. e.g., Na + = +1, Al 3+ = +3, Cl - = -1, N 3- = Oxygen in a compound or ion usually has an oxidation state of 2. (Peroxides are the exception, in which case the oxidation number is 1.) 6. Hydrogen in a compound or ion usually has an oxidation state of +1. (Hydrides are the exception, in which case the oxidation number is 1.) 7. For covalently bonded substances, shared electrons are assigned to the more electronegative element. 8. When two atoms of the same element share electrons, they are divided equally between the atoms (each atom gets half.)

3 Determining oxidation numbers from a formula is usually an exercise in solving for the unknown oxidation number of one if the elements in the formula. Procedure for assigning an oxidation number: 1. List the known oxidation numbers of all atoms in the compound or ion. 2. Multiply each oxidation number by the relevant number of atoms. 3. Determine the difference between the sum of the oxidation numbers of the known atoms and the total charge of the molecule or ion. 4. Divide the difference by the number of atoms of the unknown oxidation state. e.g. K 2 Cr 2 O 7 O = -2; K = +1; solve for Cr 0 = 7(-2) + 2(+1) + 2(oxidation number of Cr) +12 = 2(oxidation number of Cr) +6 = oxidation number of Cr e.g NaNO 3 O = -2; Na = +1; solve for N 0 = 3(-2) + 1(+1) + (oxidation number of N) +5 = oxidation number of N

4 Acid-Base reactions: H + transfer Acid A + Base B Base A + Acid B Acid A,Base A are protonated and deprotonated forms of A. They constitute a conjugate acid-base pair. (The same holds for Base B, Acid B.) Oxidation-Reduction reactions: e - transfer A OX + B RED A RED + B OX A OX, A RED are oxidized and reduced forms of A. B OX, B RED are oxidized and reduced forms of B. A OX and B OX are oxidizing agents A RED and B RED are reducing agents A OX oxidizes B RED and B RED reduces A OX In acid-base chemistry we speak in terms of strong acids and bases and weak acids and bases. We also speak of strong and oxidizing agents and strong and weak reducing agents. Strong oxidizing agents have high electron affinities. Strong reducing agents readily donate electrons.

5 Balancing oxidation/reduction reactions: as always, the conservation laws must be obeyed the half reaction method (two examples: acid solution, base solution) In acid solution: consider the unbalanced net reaction: MnO C 2 O 4 Mn 2+ + CO 2 Step 1: write the half reactions (for each redox pair) (i.e. A OX A RED ; B RED B OX ) - MnO 4 C 2 O 4 Mn 2+ CO 2 Step 2: balance each half reaction using H 2 O, H +, and e -, as needed MnO H + + 5e - Mn H 2 O C 2 O 4 2CO 2 + 2e -

6 Step 3: multiply each half reaction as needed to balance the number of electrons on each side and add the half reactions 2(MnO H + + 5e - Mn H 2 O) 5(C 2 O 4 2CO 2 + 2e - ) 2MnO H + + 5C 2 O 4 2Mn H 2 O + 10CO 2 In base solution: MnO C 2 O 4 Mn 2+ + CO 2 Step 1: write the half reactions (for each redox pair) (i.e. A OX A RED ; B RED B OX ) - MnO 4 C 2 O 4 Mn 2+ CO 2 Step 2: balance each half reaction using H 2 O, OH -, and e -, as needed MnO H 2 O + 5e - Mn OH - C 2 O 4 2CO 2 + 2e -

7 Step 3: multiply each half reaction as needed to balance the number of electrons on each side and add the half reactions 2(MnO H 2 O + 5e - Mn OH - ) 5(C 2 O 4 2CO 2 + 2e - ) 2MnO H 2 O + 5C 2 O 4 2Mn OH CO 2 n = #electrons transferred = 10

8 Ex. 2 H 3 AsO 3 + I 2 H 3 AsO 4 + I - In acid: H 3 AsO 3 + H 2 O H 3 AsO 4 +2H + + 2e - I 2 + 2e - 2I - H 3 AsO 3 + I 2 + H 2 O H 3 AsO 4 + 2I - + 2H + In base: H 3 AsO 3 + 2OH - H 3 AsO 4 + H 2 O + 2e - I 2 + 2e - 2I - H 3 AsO 3 + I 2 + 2OH - H 3 AsO 4 + 2I - + H 2 O n = #electrons transferred = 2

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