Chapter 20: Electrochemistry Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor.

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Chapter 20: Electrochemistry Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor

Oxidation-Reduction reactions Oxidation-reduction (redox) reaction: transfer of electrons from one species to another H 3 O + becomes simply H + when dealing with redox reactions to simplify balancing –(still the same species, just different notation) Skeleton oxidation-reduction equation: involves only the species being oxidized and reduced. –Write oxidation numbers above each species. –No spectator ions, no balancing Half reaction: shows only one oxidation OR one reduction –Most redox reactions are split into an oxidation half- reaction and a reduction half-reaction –LEO, GER

Balancing redox equations in acidic solutions For each half reaction… –Balance everything except H or O –Balance O by adding H 2 O to one side –Balance H by adding H + to one side –Balance charge by adding e - to one side Multiply each half reaction by a factor so that the electrons cancel when the two half reactions are added together (e - cannot appear in the final equation) Add the reactions, cancel anything that appears on the left and right, and simplify the coefficients to the smallest integers

Practice balancing acidic redox reactions Balance I 2 (s) + NO 3 - (aq)  IO 3 - (aq) + NO 2 (g) in acidic solution Half reactions Cancel electrons Add half-reactions Simplify

Voltaic cells A voltaic cell consists of two half-cells –Each half-cell contains a metal rod dipped in a solution containing that metal ion –Anode: a species is being oxidized –Cathode: a species is being reduced Cell reaction: redox reaction for entire voltaic cell Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Zn(s)  Zn 2+ (aq): oxidation half-reaction, anode Cu 2+ (aq)  Cu(s): reduction half-reaction, cathode

Cell notation Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s) Cell notation: Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) –Anode || Cathode Write half reactions and cell reactions for the following cell: Tl(s) | Tl + (s) || Sn 2+ (aq) | Sn(s)

emf, Standard Cell emf, Standard electrode potential Electromotive force, emf, E cell = electrical pressure across the conductors of an electrochemical cell –Unit: Volt, V –Measure of the driving force of a cell reaction Standard cell emf, E o cell = solutes are 1 M, gases are 1 atm, temperature is 25 o C Standard electrode potenital –By convention, the standard hydrogen electrode has an emf of 0 V –All reactions shown as reductions E cell = E cathode – E anode E cell is positive for spontaneous reactions as written

Practice calculating E cell Using standard potentials, calculate E cell for Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) Standard cell potentials are an intensive property –Do not depend on quantity! –If you have to multiply a half-reaction to cancel electrons, do not multiply the E o for that half- reaction

Free energy and K from E cell ΔG o = -nFE cell n = moles of electrons transferred F = Faraday’s constant, 96,500 C/mol e - This gives an answer in J, since 1 J = 1 C·V Convert to kJ since that’s what ΔG o is usually expressed in E cell = ( / n) log K (Nernst equation)

Practice with ΔG o and K Calculate ΔG o and K for the following cell: Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) ΔG o = -nFE cell E cell = ( / n) log K