Electrolytic Cell Flashcards
Describe what happens to an ion being discharged in electrolytic cell
discharged: when ion lose charge, it is said to b discharged. An ion can lose charge by lose or gain e-, depending on initial charge
Describe set up of electrolytic cell
- current flow fr + to - of battery
- electrolyte is molten or aq
anode: - +ve electrode (attached to + terminal of battery)
- anion attracted to, migrate to anode
- oxidat n occur; anion r discharged
cathode:
- -ve electrode (attached to - terminal of battery)
- cation attracted, migrate to cathode
- reduct n occur; cation discharged
Describe what happens to electrode potential of ions at cathode and anode of electrolytic cell
cathode:
- cation w more +ve (less -ve) standard electrode potential is preferentially reduced (first)
anode:
- anion w less +ve (more -ve) standard electrode potential is preferentially oxidised (first)
What are factors affecting discharge of ions?
- electrode potential of ion
- state of electrolyte (molten vs aq)
- conc of ion
- nature of electrode
Describe what happens to electrolysis of molten ionic compound
- electrolyte is molten ionic cpd, contain oni 1 type cation, anion. So, no other choice for ion being discharged at anode, cathode
eg
molten lead (II) chloride, oni Pb2+(l) & Br-(l) present
at anode: - Br- anion migrate to anode
- Br- ion lose e-, oxidised form Br2(g)
*Br2 is gas as temp vv high for molten ionic cpd
at cathode: - Pb2+cation migrate to cathode
- Pb2+ ion gain e-, reduced to form Pb(s)
Describe what happens to electrolysis of aqueous solution
- aq electrolyte; water present in large amt as solvent, found at surface of both electrode (so reduct n, oxidat n of water shd oso b considered)
- water undergo vv slight self-dissociat n produce H+, OH-, thus in aq neutral medium, reduct n of H+ & oxidat n of OH- (half eqn in data book) not considered generally due to low conc
eg
NaBr(aq)
at anode: - Br- & H2O
- since Eθ Br2/Br- less +ve than Eθ O2/H2O, Br- will b preferentially oxidised give bromine
at cathode - Na+, H2O
- Since Eθ H2O/H2 more +ve than Eθ Na+/Na, H2O preferentially reduced to H2 gas
Describe how concentration of ions affect electrolysis
- the higher conc ion, easier for species b reduced/oxidised
*NOTE: conc factor may override predict n based on electrode potential value (usually for discharge of anion)
eg
NaCl(conc)/brine using Pt (inert) electrode give both H2, Cl2 gas
at anode - prefer discharge Cl- as it hv high conc
- effervescence of greenish-yellow Cl2 gas observed
at cathode - prefer discharge H2O since Eθ H2O/H2 more +ve than Eθ Na+/Na
- effervescence of H2 gas observed
Explain how concentration factor overrides Eθ factor
- recall Eθ value in data book measured against SHE as refer, under standard condit n
- As [Cl-] increase, eqm pos n of (Cl2 + 2e- ⇌ 2Cl-) shift left, Eθ Cl2/Cl- decrease to pt where less +ve than Eθ O2/H2O = +1.23V. (this cld occur as both Eθ similar in value)
- Since Eθ Cl2/Cl- is now less +ve than Eθ O2/H2O, Cl- prefer oxidised
NOTE: in syllabus, just know this expl n, not predict when conc factor override Eθ factor
Explain nature of electrodes during electrolysis
- inert electrode (eg Pt, graphite) no participate in rxn
- reactive electrode (eg Cu anode) might possibly b oxidised in rxn
eg
electrolysis of CuSO4 (aq) using reactive Cu electrode
at anode: - prefer oxidat n of reactive Cu anode as Eθ Cu2+/Cu is least +ve (vs use inert electrode, HwO will b oxidised due to hv least +ve Eθ)
- Cu anode decrease in mass
cathode:
- prefer discharge Cu2+ as Eθ Cu2+/Cu is more +ve than Eθ H2O/H2
- pink solid Cu formed
- Cu cathode increase in size
*Blue colour of CuSO4(aq) electrolyte does not fade since Cu2+ removed is replenished by oxidat n of Cu anode to Cu2+ (vs using inert electrode, where sol n fade/decolourise)
Define Faraday’s First Law
states that mass of substance produced at electrode during electrolysis directly proportional to quantity of electricity passed
Define Faraday constant F
charge of 1 mol e-
F = Le = 96500C mol-1
where
L is Avogadro’s const (6.02E23),
e is charge on 1 e-
Define Faraday’s second law
states no. of Faraday needed discharge 1 mol of ion = no. of charge on ion/ no of mol of e- transferred, n)
Give overall Faraday Law formula
amt of pdt formed
= Qsupplied/Qneeded form 1 mol pdt
= It/(nF)
where
n is no of mol e- transferred to form 1 mol pdt,
F is Faraday const,
I is current,
t is time
*Faraday’s laws r independent of temp, conc ion
Describe steps to use Faraday’s laws
- write balanced eqn for rxn at cathode &/or anode
- determine n, amt e- needed form 1 mol pdt
- calculate amt pdt by using formula,
amt of pdt formed
= Qsupplied/Qneeded form 1 mol pdt
= It/(nF) - calculate mass or vol of pdt formed
Name some industrial applications of electrolysis
- anodise Al
- electrolytic purificat n of Cu
- electroplating
