Electrolytic CellsElectrolysis, electroplating: similar processes

From Pink Monkey

ElectrochemistryThe Two Types

▪Electrochemical Cells:▸Spontaneous flow of electrons from anode

through a meter or “work” to the cathode. Energy is produced by the cell and used to do work

▪Electrolytic Cells:▸Forced flow of electrons from anode to the

cathode by an external battery or generator. Energy is used by the electrochemical cell to make high-energy elements or compounds

ExampleThe Lead-Acid Battery (Pb/H

2SO

4/PbO

2)

▪Charged battery:Pb in contact with H

2SO

4: Pb oxidized to form PbSO

4.

This oxidation generates electrons (cathode)

PbO2 in contact with H

2SO

4: PbO

2 is reduced by adding

electrons to form PbSO4.

Discharged battery: PbSO4 coats both electrodes,

surrounded by depleted sulfuric acid

ExampleThe Lead-Acid Battery (Pb/H

2SO

4/PbO

2)

▪Discharged battery is recharged:PbSO

4 coats both electrodes, surrounded by depleted sulfuric acid

▸Electrons are forced in the opposite direction by the automobile generator

PbSO4 is oxidized to PbO

2 (+2 to +4)

PbSO4 is reduced to Pb at the other electrode

Now the Pb and PbO2 are restored at their proper electrodes so they

can start your car the next time you need some “juice”

SummaryLead-Acid Battery

▪Discharging: producing of electronsPb + H

2SO

4 → PbSO

4 + 2H+ + 2e-

PbO2 + 4H+ + SO

42- + 2e- → PbSO

4 + 2H

2O

▪Charging: restoring original situationPbSO

4 + 2H+ + 2e- → Pb + H

2SO

4

PbSO4 + 2H

2O → PbO

2 + 4H+ + SO

42- + 2e-

Batteries “die” when enough PbSO4 falls off one or the other

electrode so that it can’t “hold a charge” in the electrolysis cycle

What Reaction Happens? If a voltage is supplied to a mixture of cations,

which one will plate out? Since this would cause reduction, look at reduction

potentials

Ag+ + e- → Ag (s) E° = 0.80 V

Cu2+ + 2e- → 2Cu (s) E° = 0.34 V

Zn2+ + 2e- → 2Zn (s) E° = -0.76 V

The more positive voltage has a more negative G Thus Ag+ >Cu2+ >Zn2+

Chlorine & NaOHElectrolysis of Brine (concentrated NaCl)

▪Cathode: reduction--electrons provided2 H

2O + 2e- → H

2 (g) + 2 OH- (aq)

▪Anode: oxidation--electrons taken away2 Cl- (aq) → Cl

2 (g) + 2e-

▪Valuable stuff produced:▸Hydrogen gas▸Sodium hydroxide (lye)▸Chlorine gas

Hydrogen & OxygenElectrolysis of dilute sulfuric acid

▪Cathode: reduction--electrons provided2 H

2O + 2e- → H

2 (g) + 2 OH- (aq)

▪Anode: oxidation--electrons taken away2 H

2O → O

2 (g) + 4 H-(aq) + 4e-

▪Valuable stuff produced:▸Hydrogen gas▸Oxygen gas

▪Oxygen is more cheaply produced by liquefying air

Downs CellElectrolysis of molten sodium chloride (m.p. = 1074K)

From voltaicpower.com

Sodium & ChlorineElectrolysis of molten sodium chloride (m.p. = 1074K)

▪Anode: oxidation2 Cl- (l) → Cl

2 (g) + 2e-

▪Cathode: reduction2 Na+ (l) + 2e- → 2 Na (l)

▪This process can also be used to produce K from KCl, Li from LiCl, etc.

ElectroplatingCommercial use for electrolysis

▪Silver platingAg+ + e- → Ag (s)

▸Should we put the item to be plated at – the cathode or the anode?

▪Electrorefining:▸Impure metal at the anode; pure metal appears at the

cathode▪How would we produce the selectively plated Toyota

logo at top right?

Electrolysis StoichiometryCommercial use for electrolysis

We can predict amount of product at either electrode Current x time = coulombs of charge Coulombs x 1 mol e- = mol electrons

96,485 Coulombs Moles product = mol e- x Stoichiometric ratio Grams product = mol product x molar mass

1 mol electrons = 96,485 C = 1 Faraday

Electrolysis StoichiometryProblem to work

Current x time = coulombs of charge Coulombs x 1 mol e- = mol electrons

96,485 Coulombs

Mol product = mol e- x Stoichiometric ratio

Grams product = mol product x molar mass

How much silver can we plate out with a 12.0 A current running 3.5 hours?

Coulombs = 12.0 A x 3.5 hr x 60 min/hr = 2520 CMol e- = 2520 C x 1 mol e- = 2.6 x 10-2 mol e- 96,485 CMol Ag = 2.6 x 10-2 mol e- x 1 mol Ag = 2.6 x 10-2 mol Ag 1 mol e-Mass Ag = 2.6 x 10-2 mol Ag x 107.9 g Ag 1 mol Ag = 2.8 g Ag