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CHEM 126 Chapter 21
CHEM 126 Final
| Question | Answer |
|---|---|
| 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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