Electrochemistry

Define oxidation and reduction. Determine oxidation numbers for atoms. Identify the oxidizing agent, the reducing agent. Distinguish between redox and non-redox reactions.

Burning and corrosion needs oxygen – oxidation. Oxidation-reduction reactions – (redox) Chemical changes when electrons are transferred from one reactant to another. Oxidation - an atom loses one or more electrons. Reduction - an atom gains one or more electrons. "LEO says GER” 
Losing Electrons is Oxidation, Gaining Electrons is Reduction

Magnesium is oxidized. Oxygen is reduced. 2 Mg (s) + O2 (g) → 2 MgO (s) Mg – neutral Mg2+ ion O – neutral O2– ion Magnesium is oxidized. Oxygen is reduced. Mg → Mg2+ 2e- O + 2e- → O2-

Written difference between ion and oxidation: Mg (s) + Cl2 (g) → MgCl2 (s) Written difference between ion and oxidation: Chlorine ion – Cl1- ion charge = 1-
 oxidation number = -1 Mg → Mg2+ 2e- Cl + 1e- → Cl1- Sometimes these numbers are the same (like above) sometimes they are very different – which is why we write them differently.

**You assign them to EACH atom** Oxidation number represents the charge the atom would have if every bond were ionic. **You assign them to EACH atom** 1. Assign known numbers first (below). Then calculate the others. All uncombined elements (diatomic) – zero. Monatomic ions in ionic bond - equals ion charge.

In compounds: Alkali metals - always +1. Earth metals - always +2 Al: +3, F: -1, H: +1*, O: –2* Neutral compound: Sum of ox.numbers for each atom must be zero. Polyatomic ion: Sum of ox.numbers must be the charge of that ion. Assign oxidation numbers to each atom in SO2. S = +4 O = –2

Assign ox.numbers for each atom in K2Cr2O7 Step 1: Start with atoms which are known. O: –2
 K: +1 Step 2: Solve for other atoms. All O atoms: –2 × 7 = –14
 All K atoms: +1 × 2 = +2 The total for the compound must be zero. All Cr atoms: -14 + 2 + ?? = 0 Two Cr atoms - (+12) ÷ 2 = +6 each K = +1 Cr = +6 O = –2.

Assign ox. numbers for each atom in Fe(NO3)3 Ionic bond between Fe3+ and NO3– Fe: +3 Sum of ox.numbers for the compound must be 0. All O atoms: -2 × 9 = –18 +3 (-18) + ?? = 0 All N atoms: 15 ÷ 3 = Fe = +3 N = +5 O = –2.

Ox.numbers do not change – no e- transferred – NOT a redox reaction. Use ox.numbers to determine if reaction is a redox reaction. +4 -2 +1 -2 +1 +4 -2 SO2 + H2O → H2SO3 +4 -4 +2 -2 +2 +4 -6 Ox.numbers do not change – no e- transferred – NOT a redox reaction.

Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) Is the following reaction a redox reaction? +1 +5 -2 +1 +5 -2 Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) +1 +5 -6 +1 +5 -6 • Cu – oxidized (loss of electrons). 
 • Ag – reduced (gain of electrons). Oxidation cannot occur without reduction.

Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) Oxidizing agent - causes the oxidation of another substance. AgNO3 is the oxidizing agent Reducing agent - causes the reduction of another substance. Cu is the reducing agent Oxidizing agent becomes reduced and the reducing agent becomes oxidized. +1 Cu(s) + 2 AgNO3(aq) → CuNO3(aq) + 2 Ag(s) +1

2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) Identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent. +1 +5 -2 +1 -2 +2 -2 +1 -2 2 HNO3(aq) + 3 H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) +1 +5 -6 +2 -2 +2 -2 +2 -2 S – oxidized N – reduced H2S – reducing agent HNO3 – oxidizing agent

How many electrons are transferred in the reaction below: gains 3e- +1 +5 -2 +1 -2 +2 -2 +1 -2 HNO3(aq) + H2S(g) → 2 NO(g) + 3 S(s) + 4 H2O(l) 2 3 loses 2e- S: (3 atoms) x (2e- lost) = 6 electrons lost N: (2 atoms) x (3e- gained) = 6 electrons gained Stoichiometry used to determined total electrons transferred.

Strong oxidizing agents: Are very reactive – will take from anything Oxidizing Agents Reaction Products O2 O2–, H2O, CO2 F2, Cl2, Br2, I2 F–, Cl–, Br–, I– MnO4– Mn2+ Cr2O72– Cr3+ HNO3 NO, NO2 H2O2 O2, H2O Strong reducing agents: Are very reactive – will give to anything. Metals, substances that burn easily – H2, CxHy