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 |