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Chapter 18: Electrochemistry I Chem 102 Dr. Curtis Electrochemistry • The study of the relationships between electrical processes and chemical processes • Batteries, electroplating, fuel cells, hydrogen production, biology 2 Electrochemistry Interconversion Electrical energy Chemical energy • Electrochemical processes • Oxidation-reduction processes, which involve electron transfers from one substance to another • Energy released by a spontaneous chemical reaction is converted into electricity (e.g. battery) • Electrical energy can be used to force a non-spontaneous reaction to occur (e.g. electrolysis) 3 Voltage (E or V), Current (I), Resistance (R) • Voltage is a potential energy • work needed or can be done when moving an electric charge • Work (Joules) = E (Volts) x q (Coulombs) • Current is a kinetic energy • the actual movement of the electrons • units: Current (amperes), Resistance (ohms) • Ohm’s Law: E = IxR • Power (Watts): P = ExI 4 Oxidation-reduction reactions (redox reactions) • Redox reactions involve the transfer of electrons from one atom to another • Example: 4 Fe(s) + 3 O2(g) → 2 Fe2O3 (s) • Electrons are transferred from the iron to the oxygen 5 Combustion as a redox reaction 6 Redox reactions • Reaction atoms gain or lose electrons • if one loses one (or more) electron, another must gain one (or more) electron • Atoms that lose electrons are being oxidized • Atoms that gain electrons are being reduced • LEO GER • Loss of electrons is oxidation, gain of electrons is reduction • OIL RIG • Oxidation is loss of electrons, reduction is gain of electrons 7 Redox reactions 8 How can we keep track of electron transfer • Oxidation state • Oxidation states are not ion charges • Oxidation states are imaginary charges based on a set of rules • Ion charges are real and measurable • Oxidation states are written -1, -2, +2, etc. • Ion charges are written 1-, 2-, 2+, etc. 9 Oxidation state rules 10 Oxidation state rules 11 Practice: Assign oxidation states to each element • Br2 • K+ • LiF • CO2 • SO42- 12 Common oxidation states These are the most common oxidation states. However, other oxidation states are also possible. +1 +2 -2 -1 13 Oxidation and reduction - another definition • Oxidation: an atom’s oxidation state increases • Reduction: an atom’s oxidation state decreases 14 Oxidizing and reducing agents • Oxidation and reduction must occur simultaneously • The reactant that reduces an atom is called the reducing agent • the reducing agent contains the element that is oxidized • The reactant that oxidizes an atom is called the oxidizing agent • the oxidizing agent contains the element that it reduced 15 Oxidation and reduction half-reactions 0 0 2+ 2- 2Mg (s) + O2 (g) 2MgO (s) This reaction can be split into two half-reactions Oxidation half-reaction reactant (= reducing agent) loses e2 Mg → 2 Mg2+ + 4 e- Reduction half-reaction Reactant (= oxidizing agent) gains eO2 + 4 e- → 2 O2- 16 Balancing Redox Equations • “Half-reaction method” Tro: 7 steps Al (s) + Cu2+ (aq) → Al3+ (aq) + Cu (s) 1. Assign oxidation states to all atoms and identify the substances being oxidized and reduced 17 Balancing Redox Equations 2. Separate the overall reaction into two half-reactions: One for oxidation, one for reduction Oxidation: Al (s) → Al3+ (aq) Reduction: Cu2+ (aq) → Cu (s) 3. Balance each half-reaction with respect to mass in the following order A. Balance all elements other than H and O B. Balance O by adding H2O C. Balance H by adding H+ 18 Balancing Redox Equations 4. Balance each half-reaction with respect to charge by adding electrons Al (s) → Al3+ (aq) + 3 eCu2+ (aq) + 2 e- → Cu (s) 19 Balancing Redox Equations 5. Make the number of electrons in both half-reactions equal by multiplying one or both half-reactions by a small whole number 2 [Al (s) → Al3+ (aq) + 3 e-] 2 Al (s) → 2 Al3+ (aq) + 6 e3 [Cu2+ (aq) + 2 e- → Cu (s)] 3 Cu2+ (aq) + 6 e- → 3 Cu (s) 20 Balancing Redox Equations 6. Add the two half-reactions together, canceling electrons and other species as necessary 2 Al (s) → 2 Al3+ (aq) + 6 e3 Cu2+ (aq) + 6 e- → 3 Cu (s) 2 Al (s) + 3 Cu2+ (aq) → 2 Al3+ (aq) + 3 Cu (s) 21 Balancing Redox Equations 7. Verify that the reaction is balanced both with respect to mass and with respect to charge 2 Al (s) + 3 Cu2+ (aq) → 2 Al3+ (aq) + 3 Cu (s) Reactants Products 2 Al 2 Al 3 Cu 3 Cu +6 Charge +6 Charge 22 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 1. Assign oxidation states to all atoms and identify the substances being oxidized and reduced 23 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 2. Separate the overall reaction into two half-reactions: One for oxidation, one for reduction 24 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 3. Balance each half-reaction with respect to mass in the following order A. Balance all elements other than H and O B. Balance O by adding H2O C. Balance H by adding H+ 25 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 4. Balance each half-reaction with respect to charge by adding electrons 26 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 5. Make the number of electrons in both half-reactions equal by multiplying one or both half-reactions by a small whole number 27 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 6. Add the two half-reactions together, canceling electrons and other species as necessary 28 Balancing redox reactions in acidic solution Fe2+ (aq) + Cr2O72- (aq) → Fe3+ (aq) + Cr3+ (aq) 7. Verify that the reaction is balanced both with respect to mass and with respect to charge 29 Balancing redox reactions in basic solution 3. Balance each half-reaction with respect to mass in the following order A. Balance all elements other than H and O B. Balance O by adding H2O C. Balance H by adding H+ D. Neutralize H+ by adding enough OH- to neutralize each H+. Add the same number of OH- ions to each side of the equation. H+ + OH- → H2O (l) 30 Example • Balance the following reaction in basic solution MnO4- (aq) + Br- (aq) → MnO2 (s) + BrO3- (aq) H2O (l) + 2 MnO4- (aq) + Br- (aq) → 2 MnO2 (s) + BrO3- (aq) + 2 OH- (aq) 31