2.7: Demonstrate understanding of oxidation-reduction

Electrochemistry: Study of relationships between electrical and chemical energy. Electrochemistry is concerned with the movement of electrons from one species to another

Electrochemistry: Voltaic cells / Galvanic Cells Devices that use chemical reaction to produce an electric current Named after Count Alessandro Volta (1745-1827) and Luigi Galvani (1737-1789)

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

Oxidation Numbers “Oxidation States” Bookkeeping method of electrons. All molecules are treated as completely ionic in nature, with all shared electrons assigned to the more electronegative atom in the bonding pair

Oxidation Numbers Rules for Assigning Oxidation Numbers: The oxidation number of an element in its elemental form is zero “Genine” … or “7 starting at 7” The oxidation number of a monatomic ion is the same as its charge. The sum of the oxidation numbers of all atoms in a species must equal the overall charge on the species

Oxidation Numbers Empirical Observations on Oxidation Numbers for Elements in Compounds: Group I metals are always +1 Group II metals are always +2 Fluorine is always in -1 oxidation state Oxygen is usually in -2 oxidation state (except H2O2)

Oxidation Numbers Assign Oxidation Numbers to Each Element in the Compounds: CrO4-  Oxygen (-2) so Cr (+7) Cl2O3  Oxygen (-2) so Cl (+3) K2SO4  2K+ + SO4-2  Oxygen (-2), Sulfur (+6), Potassium (+1)

Oxidation and Reduction Process of losing electrons Reduction- Process of gaining electrons

Balancing Oxidation-Reduction Reactions Perhaps the easiest way to balance the equation of an oxidation-reduction reaction is via the half-reaction method. This involves treating (on paper only) the oxidation and reduction as two separate processes, balancing these half reactions, and then combining them to attain the balanced equation for the overall reaction

Balancing Oxidation-Reduction Reactions Basic Method: Divide equation into 2 half-reactions (one oxidation and one reduction) Balance each half-reaction Meaning to balance the mass (balance the elements) Balance the charge on each side of the half-reaction by adding electrons. Multiply each half-reaction by an integer number in order to cancel electrons Add and simpify

Balancing Oxidation-Reduction Reactions Example: The redox reaction of copper metal and silver ion: Cu + Ag+  Cu+2 + Ag (not balanced) Break into 2 half reactions: Oxidation: Cu  Cu+2 Reduction: Ag+  Ag

Balancing Oxidation-Reduction Reactions Already balanced, so add electrons Oxidation: Cu  Cu+2 + 2 e- Reduction: Ag+ + e-  Ag Multiply so the electrons cancel Reduction: 2Ag+ + 2e-  2Ag Add and Simplify: Cu + 2Ag+  Cu+2 + 2Ag

Balancing Oxidation-Reduction Reactions in Aqueous Solutions Redox reactions in solution can be very complex. H2O, H+(aq), and OH-(aq) are sometimes necessary to balance the equations We will discuss balancing redox reactions in both acidic and basic solutions.

Balancing Redox in Acidic Solutions Balance the following redox equations in an acidic solution: NO2− + Cr2O72-  Cr3+ + NO3-

Balancing Redox in Acidic Solutions Before we begin, assign oxidation numbers so that we can identify the half-reactions: NO2− + Cr2O72-  Cr3+ + NO3- +3 +6 +5

Balancing Redox in Acidic Solutions Divide equations into 2 half-reactions (one for oxidation and one for reduction) NO2−  NO3- oxidation Cr2O72−  Cr+3 reduction

Balancing Redox in Acidic Sol’ns Balance each half-reaction: 1st balance elements other than H or O NO2−  NO3- Cr2O72−  2Cr+3 Balance oxygen by adding H2O to the side of the reaction that needs oxygen H2O + NO2−  NO3- Cr2O72−  2Cr+3 + 7H2O Balance Hydrogen by adding H+ (in acidic solution) H2O + NO2− NO3- + 2H+ 14H+ + Cr2O72− 2Cr+3 + 7H2O Balance charge on each side of each half-reaction by adding electrons H2O + NO2−  NO3- + 2H+ + 2e- 6e- + 14H+ + Cr2O72−  2Cr+3 + 7H2O

Balancing Redox in Acidic Sol’ns Multiply each half-reaction by an integer number in order to cancel electrons ( H2O + NO2−  NO3- + 2H+ + 2e-) x 3 (6e- + 14H+ + Cr2O72−  2Cr+3 + 7H2O) x 1 Add and simplify 3H2O + 3NO2− + 6e- + 14H+ + Cr2O72−  3NO3- + 6H+ + 6e- + 2Cr+3 + 7H2O 3NO2− + 8H+ + Cr2O72−  3NO3- + 2Cr+3 + 4H2O Check – same number of each type of atom on both sides; same total charge on each side of equation

Balancing Redox in Acidic Sol’ns Consider the reaction between MnO4− and C2O42− : MnO4− (aq) + C2O42− (aq)  Mn2+ (aq) + CO2 (aq)

Balancing Redox in Acidic Sol’ns 5 C2O42−  10 CO2 + 10 e− 10 e− + 16 H+ + 2 MnO4−  2 Mn2+ + 8 H2O Add and simplify: 16 H+ + 2 MnO4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2

Oxidation and Reduction Process of losing electrons Reduction- Process of gaining electrons Oxidizing Agent (Oxidant) Substance which causes the oxidation of another compound It is reduced Reducing Agent (Reductant) Substance which causes reduction in another compound It is oxidized

Oxidation and Reduction Which species is being oxidized? Which species is being reduced? Which species is the oxidizing agent? Which species is the reducing agent?

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. It is the reducing agent

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. It is the oxidizing agent

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.