Flashcards: H2 Chemistry 21 - Electrolytic cells

state ALL species present in electrolyte >

which species reduced @ cathode, oxidised @ anode? >

to determine substance discharged @ each electrode, 3 factors affect selective discharge of species;

1. compare e nought values; > -ve gets oxidised, > +ve gets reduced

2. [ion] affects reduction potential

3. nature of electrode; reactive?

cathode(-ve) -> anode(+ve)

not like galvanic cells where anode(-ve) -> cathode(+ve)

identify all species present >

consider possible species for each electrode(metals only [O]); >1 species for electrode, compare e nought values to predict species discharged

template:

@ -ve cathode, ___ > easily [R] than ___ as e nought() is >+ve/<-ve than e nought()

@ +ve anode, ___ > easily [O] than ___ as e nought() is >-ve/<+ve than e nought()

always use full forward arrow

total charge by 1 mol of e^- = (6.02x10²³) x (1.60x10^-19)

= 96500C

total charge passed, Q = amount of e^- passed in circuit x Faraday's constant

Q = n_eF

current(rate of flow of electric charge in s), I=Q/t

total charge passed, Q = current x time; Q=It

current in amps(A), time in seconds(t)

mass/volume of substance formed proportional to total charge Q=It

purpose: form stronger, thicker layer of aluminium oxide to improve corrosion resistance

components in cell set up

electrolyte: dilute H₂SO₄(aq)

cathode: Pt/graphite electrode

anode: Al electrode

species present: H⁺, SO₄^2-, H₂O

@ anode, H₂O oxidised; 2H₂O(l) -> O₂(g) + 4H⁺ + 4e^-

O₂(g) reacts w/ Al anode to form thick porous layer of oxide

2Al(s) + 3/2O₂(g) -> Al₂O₃(s)

OVERALL anode eqn

2Al(s) + 3H2O(l) -> Al₂O₃(s) + 6H⁺ + 6e^-

@ cathode, H+ reduced; 2H⁺ + e^- -> H₂(g)

OVERALL eqn

2Al(s) + 3H₂O(l) -> Al₂O₃(s) + 3H₂(g)

metal corrosion resistance increased

porous surface of oxide layer can be dyed to diff colours

purpose: extract pure Cu from manufactured product

components in cell set up

electrolyte: CuSO₄(aq)

cathode: pure Cu electrode

anode: impure Cu electrode

species present: Cu²⁺(aq), SO₄^2-(aq), H₂O(l), Cu&impurities

anode: Cu(s) -> Cu²⁺(aq) + 2e^- (impure Cu dissolves)

cathode: Cu²⁺(aq) + 2e^- -> Cu(s) (Cu deposited on pure Cu)

blue soln does not fade during electrolysis

impurities @ anode may not get oxidised & get collected as anode sludge,

some may get oxidised but not reduced @ Cu cathode as e nought value too

-ve wrt Cu²⁺

used for electrical wiring as Cu has high electrical conductivity, tensile strength, corrosion resistance, etc.