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 KNO 3 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 – amperes 1 amp = 1 coulomb/sec 1 coulomb = 1/96490 of a mole of electrons (6.24x10 18 e-)  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 H 2, 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 e -  Pb Eº = V  Lead is the anode, so it is oxidized (reduction equation is reversed)

Cell potential Hg +2 +2e -  Hg Eº = 0.851V Pb  Pb e - Eº = V  Cell potential is the sum of the half cell potentials Hg +2 + Pb  Pb +2 + Hg Eº = 0.851V V = 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. H 2,Pt|H + ║ Ag + |Ag +

Galvanic cell calculations  Calculate the voltage of this cell: Mg|Mg +2 ║ Au +3 |Au Mg  Mg e - E º = +2.37V cathode: Au e -  Au E º = +1.50V  Total cell voltage = 3.87V

Current calculations EExtent of oxidation/reduction and amount of material oxidized or reduced is directly related to the number of electrons transferred MMoles e - = current x time / Faraday’s # = It/ F F = 96485coul/mole

 A zinc anode with mass 2.30g is used in a copper/zinc cell. The cell has a current of amps. How long will the electrode last?  Solution: 2.30 g Zn is mole zinc.  Each mole zinc requires 2e-, so mole e- are required to completely oxidize the anode.

Current calculations  There are 96,485 coulombs/mole, so 6790 coulombs are needed mol e - x coul/mol = 6790c  At coulombs/sec, the electrode will last 4.85x10 6 sec, or 1350 hours.

Types of cells  Concentration cell – uses identical electrodes – potential difference due to concentration differences in the cell

Nonstandard cells  E cell = Eº cell – (RT/n F )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) MnO 2, water, NH 4 Cl in a paste around a graphite cathode, with a zinc anode

Types of cells OOverall Leclanché cell equation: Zn+MnO 2 +NH 4 Cl  ZnCl 2 +Mn 2 O 3 +NH 3 +H 2 O Balance it! Zn  Zn e - 2H + + 2MnO 2 + 2e -  Mn 2 O 3 + H 2 O 2H + +2MnO 2 +Zn  Zn +2 +Mn 2 O 3 +H 2 O Add spectators (NH 3 and Cl - ): 2NH 4 Cl+2MnO 2 +Zn  ZnCl 2 +Mn 2 O 3 +H 2 O+2NH 3

 Alkaline manganese cell – used in alkaline batteries – uses KOH as electrolyte  Zn+MnO 2 +H 2 O  Zn(OH) 2 + Mn 2 O 3

Types of cells  Electrolytic cells – nonspontaneous redox reaction is forced to proceed by application of an electric current  Electrolysis of water – Hoffman apparatus 2H 2 O  2H 2 + O 2

Electrolysis of aluminum oxide  Alumina (Al 2 O 3 ) from bauxite is dissolved in molten cryolite (Na 3 AlF 6 ).

Electrolysis of aluminum oxide  The steel container is coated with carbon (graphite) and serves as the negative electrode (anode).  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: 2Al 2 O 3(l)  4Al (l) + 3O 2(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