electrochemistry

Electrochemistry

Guaranteed to give you a jolt

Electrochemical cells

A chemical system in which oxidation and reduction can occur – often a single displacement reaction

Zn + Cu+2 Zn+2 + Cu Oxidation reaction and reduction

reaction are physically separated so that useable work can be obtained from the reaction

Electrochemical cells Voltaic cells (galvanic cells) – redox

occurs spontaneously Anode – where oxidation takes place

(solid metal becomes aqueous positive ions)

Cathode – where reduction takes place (metal ions deposit as solid metal)

AN OX RED CAT

Electrochemical cells

anode cathode

Salt bridge

Electrons flow from anode to cathode through wire

Electrochemical cells

Salt bridge – usually KNO3 or a semipermeable membrane– completes circuit by providing mobile ions

Conditions for spontaneity – one metal must lose electrons more easily than another (metals can be identical if one is warmer) p. 288 (activity series)

Electric current

Rate of flow – amperes1 amp = 1 coulomb/sec

1 coulomb = 1/96490 of a mole of electrons (6.24x1018e-)

Faraday’s number = 96490coul/mole Potential – volts (joules/coulomb)

Cell potential

Half cell potential (E) – the voltage contribution of a half reaction to the cell

Standard half cell potential (Eº) - voltage when solutions are 1M, gases are 1atm and temp. is 25ºC.

Half cell potentials are given as reductions relative to the reduction of H+ to H2, which is assigned 0 volts.

Cell potential

Half cell potential is a measure of the tendency of a particle to gain electrons.

The cell potential is the sum of the two half cell potentials.

For oxidation, the sign of the reduction potential is reversed.

A positive cell potential means a spontaneous reaction, i.e. a galvanic cell.

Cell potential

Example. Find the cell potential for a cell with a mercury/mercury (II) cathode and a lead/lead (II) anode.

Standard reduction potentials:Hg+2 +2e- Hg Eº = 0.851V

Pb+2 + 2e- Pb Eº = -0.1262V Lead is the anode, so it is oxidized

(reduction equation is reversed)

Cell potential

Hg+2 +2e- Hg Eº = 0.851VPb Pb+2 + 2e- Eº = 0.1262V

Cell potential is the sum of the half cell potentials

Hg+2 + Pb Pb+2 + Hg E = 0.851V + 0.1262V = 0.977V

Potential is intensive, so it’s not affected by coefficients

Cell potential

A positive cell potential means a spontaneous reaction, i.e. a galvanic cell.

Galvanic cell calculations Cell notation

Zn|Zn+2║Cu+2|Cu anode salt bridge cathode

Galvanic cell calculations

Vertical lines represent the barrier between two different states of matter.

Two different materials in the same part of the cell are separated by commas.

H2,Pt|H+║Ag+|Ag+

Galvanic cell calculations

Calculate the voltage of this cell:Mg|Mg+2║Au+3|AuMg Mg+2 + 2e-

E º = +2.37Vcathode: Au+3 + 3e- Au

E º = +1.50V Total cell voltage = 3.87V

Current calculations

Extent of oxidation/reduction and amount of material oxidized or reduced is directly related to the number of electrons transferred

Moles e- = current x time / Faraday’s # = It/F F =

96485coul/mole

Current calculations A zinc anode with mass 2.30g is used

in a copper/zinc cell. The cell has a current of 0.00140 amps. How long will the electrode last?

Solution: 2.30 g Zn is 0.0352 mole zinc.

Each mole zinc requires 2e-, so 0.0704 mole e- are required to completely oxidize the anode.

Current calculations

There are 96,485 coulombs/mole, so 6790 coulombs are needed.

0.0704mol e- x 96485 coul/mol = 6790c At 0.00140 coulombs/sec, the

electrode will last 4.85x106 sec, or 1350 hours.

Types of cells Concentration

cell – uses identical electrodes – potential difference due to concentration differences in the cell

Nonstandard cells

Ecell = Eºcell – (RT/nF )lnQ Q = reaction quotient For the reaction aA + bB cC + dD,

Q = [C]c[D]d/[A]a[B]b

Find the voltage for a zinc/copper cell at 55ºC where the concentrations of zinc and copper are 0.10 and 0.75M respectively.

Types of cells

Leclanché cell (dry cell) MnO2, water, NH4Cl in a paste around a graphite cathode, with a zinc anode

Types of cells Overall Leclanché cell equation:

Zn+MnO2+NH4ClZnCl2+Mn2O3+NH3+H2O

Balance it!Zn Zn+2 + 2e-

2H+ + 2MnO2 + 2e- Mn2O3 + H2O

2H++2MnO2+ZnZn+2+Mn2O3+H2O

Add spectators (NH3 and Cl-):

2NH4Cl+2MnO2+ZnZnCl2+Mn2O3+H2O+2NH3

Types of cells Alkaline

manganese cell – used in alkaline batteries – uses KOH as electrolyte

Zn+MnO2+H2OZn(OH)2+

Mn2O3

Types of cells

Electrolytic cells – nonspontaneous redox reaction is forced to proceed by application of an electric current

Electrolysis of water – Hoffman apparatus

2H2O 2H2+ O2

Electrolysis of aluminum oxide

Alumina (Al2O3) from bauxite is dissolved in molten cryolite (Na3AlF6).

Electrolysis of aluminum oxide The steel container is coated with

carbon (graphite) and serves as the negative electrode (cathode).

Electrolysis of the alumina/cryolite solution gives aluminum at the cathode and oxygen at the anode.

Aluminum is more dense than the alumina/cryolite solution, and so falls to the bottom of the cell.

Electrolysis of aluminum oxide Aluminum can be tapped off the bottom as

pure liquid metal. The overall reaction is:

2Al2O3(l) 4Al(l) + 3O2(g)

Oxygen is discharged at the positive carbon (graphite) anode.

Oxygen reacts with the carbon anode to form carbon dioxide gas. The carbon anode slowly disappears as carbon dioxide and needs to be replaced regularly.

Anodic protection