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CHEM 126 Chapter 21

CHEM 126 Final

reducing agent - gives electrons to substance being reduced - substance being oxidized
oxidizing agent - takes electrons from substance being oxidized - substance being reduced
balance redox equations 1. separate into 1/2 reactions 2. balance atoms other than H & O 3. balance O by adding other H20 4. balance H by adding H+ ions 5. basic solutions: add OH- for every H+ present 6. balance charge by adding e- 7. multiply 1/2 rxn by integer
reaction at cathode - reduction - Voltaic: Y+ + e- -> Y - Electrolytic: B+ + e- -> B
reaction at anode - oxidation - Voltaic: X -> X+ + E- - Electrolytic: A- -> A + e-
direction of electron flow in voltaic cell anode to cathode
overall redox reaction for electrochemical cell A- + B- -> A + B when ∆G > 0
elements of voltaic cell - does work on surroundings - Electrolyte Y+ + - (anode) -> Electrolyte X- + +(cathode) - components: electrode, electrolytes, salt bridge (release energy to surroundings) - anode -> cathode (electrodes) vs. cathode -> anode (salt bridge)
spontaneous or non spontaneous? - voltaic cell: spontaneous, work on surroundings (∆G < 0) - electrolytic: non spontaneous, surroundings do work on system (∆G > 0)
characteristics of voltaic cell - spontaneous - work on surroundings - higher energy reactants, lower energy products * create electrical energy - EX: batteries
characteristics of electrolytic cell - non spontaneous reaction (absorb energy from surroundings) - does work on system - lower energy reactants, higher energy products - EX: electroplating
elements of electrolytic cell - one solution of 2 mixed electrolytic solutions - have anode (+) and cathode (-), flow from + to - - power supply gives energy to electrolytes
spontaneity vs. electrochemical work - non spontaneous - electrical energy from surroundings do work on system
shorthand notation for voltaic cell - LEFT: anode components - RIGHT: cathode components - single vertical line: phase boundary - comma: same phase, different substance - parenthesis: dissolved substance - EX: Zn(s)| Zn+2 (1 M) || Cu+2 (1M) | Cu (s)
determine standard cell potential - Ecell: difference in electrical potential for an electrochemical cell - + = spontaneous rxn, more positive -> more can do -> farther rxn proceeds - - = nonspontaneous cell rxn - Ecell = 0: at equilibirium
determine free energy - 0 > ∆G for ) < Ecell - ∆G = -nFEcell * n: number of moles of e-, F: 9.65x10^4J, Ecell: number in V
determine max amount of work w max = -Ecell * charge - same for electrolytic & voltaic cell
determine equilibrium constant log K = (nE°cell)/.0592V - K = 10^ (log K)
describe cell behavior with respect to K - Q/K < 1: E cell is +, the smaller Q/K is -> greater E cell is - Q/K >1: E cell is -, reverse reaction will take place & cell will do work until Q/k = 1 - Q/K = 1: cell has reached equilibrium
determine how ∆G°is different from ∆G - ∆G = -nFEcell (at any point) vs. - ∆G° = -nFE°cell (all components in standard state) **E°cell = (0.0529V/n)*logQ at 298.15K
cell concentration changes during opporation - Q > 1 -> [reactant]<[product] so Ecell < E°cell - Q < 1 -> [reactant] > [product] so Ecell > E°cell - Q = 1 -> [reactant] = [product] so Ecell = E°cell
concentration cell - mix concentrated solution (C) of substance with dilute solution (A) in it - spontaneous rxn -> electrical energy - 2 solutions in own 1/2 cells, concentrations become equal as cell operates - A-> gives e-, becomes concentrate - C-> lose e-, more dil
Calculate cell potential - Ecell = E°cell - ((o.o529)V/n)* logQ at 298.15K - Q = [(dilute)(anode)]/[(concentrated)(cathode)]
lead acid rechargeable batteries - 6 cells connected in series - each with 2 lead grids, high SA of Pb in A vs high SA of PbO2 in C - grids immersed in H2SO4 - discharge: creates energy, recharge: use electrical energy
Hydrogen fuel cell (anode) - H2 molecules adsorb onto catalyst - 2e- travel through wire to cathode - 2 H+ are hydrated & go through electrolytes as H30+
Hydrogen fuel cell (cathode) - O2 molecules adsorb onto catalyst, - provide e- to form 02- + 2H+ from H30+ - allow H20 to form - H20 leaves the cell
reactions for hydrolysis of h20 - anode: 2H20 -> 4H+ + 4e- + 02 - cathode: 2H20 + 2e- -> H2 + 2OH-
faraday's law of electrolysis - amount of substance produced at electrode is directly prop to quantity of charge flowing through the - current x time = charge (1 ampere = 1 columb/sec)
how does iron rust? 1. Loss of Fe Fe -> Fe+2 + 2e- * pit formed in A region where there is loss of e-, e- move towards high region of 02 2. Rusting process 2 Fe+2 + 1/2 O2 + (2 + n)H20 -> Fe2O3 + 4H+ * Fe+2 formed in a region disperse through h20 & reacts with 02
Created by: ccottrel



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