Electrochemistry

Electrochemistry Textbook Chapter 20 IB Topics 9 and 19

Oxidation Numbers • IB Topic 9.1.1- 9.1.4 • Text p. 843-846 • Oxford p. 54. 55

LEO

LEO says GER!

GER! LEO says Loss of Electrons = Oxidation Gain of Electrons =Reduction

Reduction Is Gain (of electrons) • Oxidation Is Loss (of electrons)

Oxidation Numbers • Oxidation/Reduction (ReDox)-- Whenever a substance loses electrons and another substance gains electrons • Oxidation Numbers--system used to keep track of electron transfers

Oxidation Numbers Rules

Practice Assigning Oxidation Numbers

Using Oxidation Numbers • Careful examination of the oxidation numbers of atoms in an equation allows us to determine what is oxidized and what is reduced in an oxidation-reduction (Redox) reaction

An increase in the oxidation number indicates that an atom has lost electrons and therefore oxidized. A decreasein the oxidation number indicates that an atom has gained electrons and therefore reduced Example Zn + CuSO4 ZnSO4 + Cu 0 +2 +6-2 +2+6-2 0 Zn: 0+ 2Oxidized Cu: +20Reduced Using Oxidation Numbers

Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 + KBr  KCl + Br2 Cu + HNO3 Cu(NO3)2+ NO2 + H2O HNO3 + I2 HIO3 + NO2

Exercise For each of the following reactions find the element oxidized and the element reduced Cl2 + KBr  KCl + Br2 0 +1-1 +1-10 Br increases from –1 to 0 -- oxidized Cl decreases from 0 to –1 -- Reduced K remains unchanged at +1

Exercise For each of the following reactions find the element oxidized and the element reduced Cu + HNO3 Cu(NO3)2+ NO2 + H2O 0 +1+5-2 +2 +5-2 +4 –2 +1-2 • Cu increases from 0 to +2. It is oxidized • Only part of the N in nitric acid changes from +5 to +4. It is reduced • The nitrogen that ends up in copper nitrate remains unchanged

Exercise For each of the following reactions find the element oxidized and the element reduced HNO3 + I2 HIO3 + NO2 1 +5 -2 0 +1+5-2+4-2 • N is reduced from +5 to +4. It is reduced • I is increased from 0 to +5 It is oxidized • The hydrogen and oxygen remain unchanged.

Assignment • Oxidation Numbers/Naming Review packet

Balancing Redox Reactions • IB Topics 9.2.1-9.2.4 • Text p. 846-850 • Oxford p. 55

Balancing Redox Reactions • Many Redox rxns are complex and difficult to balance . • A systematic approach to balancing these reaction is required.

MnO41- (aq) + C2O42- (aq)  Mn2+(aq) + CO2(aq) Balancing Redox Equations 1 • Divide the equation into 2 half reactions—one for oxidation, one for reduction. • Balance each half reaction • Balance elements other than H and O • Balance O by adding H2O as needed • Balance H by adding H+ as needed.(acidic solution) • Balance charge by adding e- as needed. • Multiply half reactions by integers so that the # of e- lost in one reaction = # of e- gained in the other reaction. • Add the two half reactions. Simplify by canceling species that appear on both sides of the arrow. • Check your work. Make sure that both the atoms and charges balance

Cr2O72- (aq) + Cl1- (aq)  Cr 3+ (aq) + Cl2 (g) • Divide the equation into 2 half reactions—one for oxidation, one for reduction. • Balance each half reaction • Balance elements other than H and O • Balance O by adding H2O as needed • Balance H by adding H+ as needed. .(acidic solution) • Balance charge by adding e- as needed. • Multiply half reactions by integers so that the # of e- lost in one reaction = # of e- gained in the other reaction. • Add the two half reactions. Simplify by canceling species that appear on both sides of the arrow. • Check your work. Make sure that both the atoms and charges balance Balancing Redox Equations 2

Divide the equation into 2 half reactions—one for oxidation, one for reduction. • Balance each half reaction • Balance elements other than H and O • Balance O by adding H2O as needed • Balance H by adding H+ as needed. .(acidic solution) • Balance charge by adding e- as needed. • Multiply half reactions by integers so that the # of e- lost in one reaction = # of e- gained in the other reaction. • Add the two half reactions. Simplify by canceling species that appear on both sides of the arrow. • Check your work. Make sure that both the atoms and charges balance Balancing Redox Equations 3 Cu (s) + NO31- (aq)  Cu 2+(aq) + NO2(aq)

Divide the equation into 2 half reactions—one for oxidation, one for reduction. • Balance each half reaction • Balance elements other than H and O • Balance O by adding H2O as needed • Balance H by adding H+ as needed. .(acidic solution) • Balance charge by adding e- as needed. • Multiply half reactions by integers so that the # of e- lost in one reaction = # of e- gained in the other reaction. • Add the two half reactions. Simplify by canceling species that appear on both sides of the arrow. • Check your work. Make sure that both the atoms and charges balance Balancing Redox Equations 4 Mn2+ (aq) + NaBiO3 (s)  Bi 3+(aq) + MnO4 1- (aq) + Na1+(aq)

Text p. 884 • Text p. 884 • Problems # 20a, b, c, d 22 a, b, c, d “acidic solution only”

Electrochemistry