Electrochemistry Flashcards
define electrolysis
break down of a compound into its elements using an electric current
3 uses of electrolysis
- extract metals from ores if it cannot be done by heating with carbon
- purify metals
- produce non metals
components of electrolysis setup
- electrolyte: molten form of compound conc aq sol of ions
- 2 electrodoes: metal’graphite rods that conduct electricity to and from the electrolyte
- power supply: direct current
Anode is
Cathosde is
A +
C -
State what happens during electrolysis of molten electrolytes
- cations move to cathode and gain e- ie reduction
- a metal layer/ molten layer may form on cathode
- bubbles if H2
- anions move to anode and lose e- ie oxidation
Ions discharged during electrolysis of aq sols depend on x and y `
- relative electrode potential of ions
- conc of ions (cos H2O, weak electrolyte, splits into H+ and OH-)
what does relative electrode potential of ions describe
how easily its discharged during electrolysis
relate E0 values to relative electrode potential of cations and anions to determine which is discharged
- a positive cation with most positive E0 is discharged at cathode cos its most easily reduced
- a negative anion with most negative E0 is discharged at anode cos its most easily oxidised
relate ion conc to discharge amount
more conc it is, the more of it will form at anode because less hydroxide ions from water are there to be discharged in comparison to what we have
state faraday’s law
amount of substance formed during electrolyis is proportional to:
-time a constant current passes
-electricity that passes through electrolyte (strength of current)
Q=It
what is the faraday unit
amount of electric charge carried by 1 mole of electrons
ie F=Le which is 96500
how to experimentally find the avogadro constant by electrolysis
L=charge on 1mol e-/charge on 1e-
how to find charge on 1mol e- and hence L
Method: pure Cu anode and cathode weighed, variable resistor kept at a constant current, electric current passed for certain time, anode and cathode removed washed w distilled water dried w propanone and reweighed
Result: Cu deposited on cathode so increased in mass, while anode lost mass ie as Cu2+ ions to sol. decreased mass of anode used in calc cos not all solid Cu formed sticks properly to cathode
Calculation: find charge (Q=It) and mass (n=m/M) via proportion to find charge for 1mol to be depositied. use half eqn to know mol of e- for 1 mol of Cu and divide the total charge by this to get charge per mol. divide this by charge of 1e- (formula)
what is electrode potential, and how are the half eqns written
- value that shows how easily a substance is reduced (in the sitch of a redox equilibrium ie related species in diff oxi states)
- electrons to left ie reduction
why do diff species have diff electrode potentials
diff equilibrium positions
relationship between elctrode potential and reduction-
more positive means more likely to undergo reduction- equilibrium lies more to right
position of equilibrium and electrode potential depend on
- temp
- pressure of gases
- conc of reagent
std conditions used to compare electrode potentials (cos otherwise will affect comparison of diff species hence a reference)
- ion conc 1mol/dm^3
- temp 298K
- 1atm
std electrode potential is produced is …
the voltage produced when a std half cell is connected to a std hydrogen cell under std conditions
when compared to the std hydrogen cell’s E0, if the half cell’s E0 is more positive relative to 0V it is …
more likely the species on left side of eqn is to get reduced
what is the std cell potential
difference in E0 between any 2 half cells
describe a redox equilibrium
eg: metal in aq sol, atoms oxidised and ions enter sol. ions gain e- from metal and depositi metal on rod
- for equilibrium, rate of reduction=rate of oxi
if a species redox equilibrium lies more to the right it means that
it is more easily reduced
viceversa tho ie lies to left cos oxidised easier
in a redox equilibrium metal atomd and ions cause a voltage/potential. but it cant be measured directly. instead, …
potential difference between the metal ion system and another can be measured using a voltmeter
