Electrochemical Reactions In electrochemical reactions, electrons are transferred from one species to another. Mg (s) + HCl (aq) →

Oxidation Numbers In order to keep track of what loses electrons and what gains them, we assign oxidation numbers.

Oxidation and Reduction A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ ion.

Oxidation and Reduction A species is reduced when it gains electrons. Here, each of the H+ gains an electron, and they combine to form H2.

Oxidation and Reduction What is reduced is the oxidizing agent. H+ oxidizes Zn by taking electrons from it. What is oxidized is the reducing agent. Zn reduces H+ by giving it electrons.

In a spontaneous Oxidation-Reduction (redox) Reaction electrons are transferred and energy is released as the reaction proceeds: Zn strip dissolves… The blue color due to Cu2+ fades. Copper metal is deposited.

Oxidation and Reduction Half-Reactions A reaction represented by two half-reactions. Oxidation: Zn(s) → Zn2+(aq) + 2 e- Reduction: Cu2+(aq) + 2 e- → Cu(s) Overall: Cu2+(aq) + Zn(s) → Cu(s) + Zn2+(aq) Prentice-Hall © 2007 General Chemistry: Chapter 5

Assigning Oxidation Numbers, ON Elements in their elemental form have an oxidation number of 0. Mg Cl2 Pb The oxidation number of a monatomic ion is the same as its charge. Zn2+ F- Na+

HF NaH Nonmetals tend to have negative oxidation numbers In combination with other elements: Nonmetals tend to have negative oxidation numbers Oxygen has an oxidation number of −2, except in the peroxide ion, which has an oxidation number of −1. MgO Na2O Hydrogen is −1 when bonded to a metal and +1 when bonded to a nonmetal. HF NaH

Assigning Oxidation Numbers Fluorine always has an oxidation number of −1. HF NaF The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions. ZnBr2 CuCl2

Assigning Oxidation Numbers The sum of the oxidation numbers in a neutral compound is 0. HCl H2O K2O 5. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion. CO32- ClO4-

Examples Determine the oxidation number of sulfur in the following compounds: H2S SO42- S8 S2O32-

Electrochemical Reactions

Corrosion and…

Figure: 20-20 Title: Protection of iron against electrolytic corrosion Caption: In the anodic reaction, the metal that is more easily oxidized loses electrons to produce metal ions: in (a), this is iron; in (b), it is zinc. In the cathodic reaction, oxygen gas, which is dissolved in a thin film of water on the metal, is reduced to OH-. Rusting of iron occurs in (a), but not in (b). When iron corrodes, Fe2+ and OH- ions from the half-reactions initiate these further reactions: Fe2+ + 2OH--> Fe(OH)2(s) 4Fe(OH)2(s) + O2 + 2H2O --> 4Fe(OH)3(s) 2Fe(OH)3(s) --> Fe2O3 • H2O(s) + 2H2O(l)

Figure: 20-20-02UN Title: Magnesium sacrificial anodes Caption: The small cylindrical bars of magnesium attached to the steel ship provide cathodic protection against corrosion. Galvanization is another type of cathodic protection, the use of magnesium here is more cost effective.

Oxidation and Reduction O.S. of some element increases in the reaction. Electrons are on the right of the equation Reduction O.S. of some element decreases in the reaction. Electrons are on the left of the equation. Prentice-Hall © 2007 General Chemistry: Chapter 5

Oxidation State Changes Assign oxidation states: Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) D Fe3+ is to metallic iron. CO(g) is to carbon dioxide. Prentice-Hall © 2007 General Chemistry: Chapter 5

Balancing Oxidation-Reduction Equations Few can be balanced by inspection. Systematic approach required. The Half-Reaction (Ion-Electron) Method Prentice-Hall © 2007

Balancing in Acid Write the equations for the half-reactions. Balance all atoms except H and O. Balance oxygen using H2O. Balance hydrogen using H+. Balance charge using e-. Equalize the number of electrons. Add the half reactions. Check the balance.

SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) EXAMPLE 5-6 Balancing the Equation for a Redox Reaction in Acidic Solution. The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution.. SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) Prentice-Hall © 2007

EXAMPLE 5-6 Determine the oxidation states: 4+ 6+ 7+ 2+ SO32-(aq) + MnO4-(aq) → SO42-(aq) + Mn2+(aq) Write the half-reactions: SO32-(aq) → SO42-(aq) + 2 e-(aq) 5 e-(aq) +MnO4-(aq) → Mn2+(aq) Balance atoms other than H and O: Already balanced for elements. General Chemistry: Chapter 5 Prentice-Hall © 2007

EXAMPLE 5-6 Balance O by adding H2O: H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq) 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Balance hydrogen by adding H+: H2O(l) + SO32-(aq) → SO42-(aq) + 2 e-(aq) + 2 H+(aq) 8 H+(aq) + 5 e-(aq) +MnO4-(aq) → Mn2+(aq) + 4 H2O(l) Check that the charges are balanced: Add e- if necessary. Prentice-Hall © 2007

EXAMPLE 5-6 Multiply the half-reactions to balance all e-: 5 H2O(l) + 5 SO32-(aq) → 5 SO42-(aq) + 10 e-(aq) + 10 H+(aq) 16 H+(aq) + 10 e-(aq) + 2 MnO4-(aq) → 2 Mn2+(aq) + 8 H2O(l) 6 3 Add both equations and simplify: 5 SO32-(aq) + 2 MnO4-(aq) + 6H+(aq) → 5 SO42-(aq) + 2 Mn2+(aq) + 3 H2O(l) Check the balance! Prentice-Hall © 2007