Electrochemistry Flashcards

1
Q

define electrolysis

A

break down of a compound into its elements using an electric current

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2
Q

3 uses of electrolysis

A
  • extract metals from ores if it cannot be done by heating with carbon
  • purify metals
  • produce non metals
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3
Q

components of electrolysis setup

A
  • 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
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4
Q

Anode is

Cathosde is

A

A +

C -

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5
Q

State what happens during electrolysis of molten electrolytes

A
  • 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
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6
Q

Ions discharged during electrolysis of aq sols depend on x and y `

A
  • relative electrode potential of ions

- conc of ions (cos H2O, weak electrolyte, splits into H+ and OH-)

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7
Q

what does relative electrode potential of ions describe

A

how easily its discharged during electrolysis

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8
Q

relate E0 values to relative electrode potential of cations and anions to determine which is discharged

A
  • 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
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9
Q

relate ion conc to discharge amount

A

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

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10
Q

state faraday’s law

A

amount of substance formed during electrolyis is proportional to:
-time a constant current passes
-electricity that passes through electrolyte (strength of current)
Q=It

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11
Q

what is the faraday unit

A

amount of electric charge carried by 1 mole of electrons

ie F=Le which is 96500

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12
Q

how to experimentally find the avogadro constant by electrolysis

A

L=charge on 1mol e-/charge on 1e-

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13
Q

how to find charge on 1mol e- and hence L

A

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)

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14
Q

what is electrode potential, and how are the half eqns written

A
  • 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
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15
Q

why do diff species have diff electrode potentials

A

diff equilibrium positions

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16
Q

relationship between elctrode potential and reduction-

A

more positive means more likely to undergo reduction- equilibrium lies more to right

17
Q

position of equilibrium and electrode potential depend on

A
  • temp
  • pressure of gases
  • conc of reagent
18
Q

std conditions used to compare electrode potentials (cos otherwise will affect comparison of diff species hence a reference)

A
  • ion conc 1mol/dm^3
  • temp 298K
  • 1atm
19
Q

std electrode potential is produced is …

A

the voltage produced when a std half cell is connected to a std hydrogen cell under std conditions

20
Q

when compared to the std hydrogen cell’s E0, if the half cell’s E0 is more positive relative to 0V it is …

A

more likely the species on left side of eqn is to get reduced

21
Q

what is the std cell potential

A

difference in E0 between any 2 half cells

22
Q

describe a redox equilibrium

A

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

23
Q

if a species redox equilibrium lies more to the right it means that

A

it is more easily reduced

viceversa tho ie lies to left cos oxidised easier

24
Q

in a redox equilibrium metal atomd and ions cause a voltage/potential. but it cant be measured directly. instead, …

A

potential difference between the metal ion system and another can be measured using a voltmeter

25
Q

what is the std hydrogen electrode

A

ref electrode of: H2 gas in equilibrium w H+ ions of conc 1mol/dm^3 at 1atm w inert platinum electrode
2H+2e-H2

26
Q

metal/metal ion half cell connected to a SHE

A
  • electrode, ion sol, and salt bridge used in addition to SHE
  • the more positive E0, thats the +ve electrode
  • reverse the equation and E0 of the less negative eqn (the -ve electrode/ where oxidation is happening) before adding
27
Q

nonmetal/nonmetal ion half cell connected to a SHE

A
  • inert platinum wire/foil used for electrical contact w sol along w a salt brdige
  • more E0, means +ve electrode
28
Q

ion/ion half cell connected to a SHE

A
  • platinum electrode used to form a half cell of ions in diff oxi states and ofc salt brdige
  • more negative E0 eqn is that of the -ve electrode
29
Q

how to tell electron flow from E0 values of eqns

A
  • the more positive the E0, the more readily the species is reduced/ gaining e-. this is the +ve pole.
  • the more negative the E0, the more readily its oxidised/ losing e-. this is the -ve pole.
  • flow is from -ve to +ve poles ie from where e- are lost to gained
30
Q

the less positive the E0 value, the easier it is to _ species on right

A

oxidise

31
Q

a reaction is feasible when

A

E0 is positive

32
Q

to get the overall eqn form 2 diff ones

A

make even no of electrons then subtract

33
Q

method for calculating mass deposited or gas liberated at electrode

A
  • write half eqn at electrode
  • determine no. of Coulombs needed to get 1mol of susbstance
  • calculate charge transferred during electrolysis
  • using simple proportion and relative atomic mass/ relationship 1mol gas occupies 24dm^3