Electrochemistry

17.1/17.2 GALVANIC CELLS AND STANDARD REDUCTION POTENTIALS Day 1

Electrochemistry Terminology #1  Oxidation  Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na + + e -  Reduction  Reduction – A process in which an element attains a more negative oxidation state Cl 2 + 2e -  2Cl -

Electrochemistry Terminology #2 G ain E lectrons = R eduction An old memory device for oxidation and reduction goes like this… LEO says GER L ose E lectrons = O xidation

Electrochemistry Terminology #3 Oxidizing agent  Oxidizing agent The substance that is reduced is the oxidizing agent Reducing agent  Reducing agent The substance that is oxidized is the reducing agent

Electrochemistry Terminology #4 Anode  Anode The electrode where oxidation occurs Cathode  Cathode The electrode where reduction occurs Memory device: Reduction at the Cathode

Table of Reduction Potentials Measured against the Standard Hydrogen Electrode Always written as REDUCTION potentials

Measuring Standard Electrode Potential Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

Relative Strength of Oxidizing Reducing Agents The more + the reduction potential, the stronger the Oxidizing Agent, (easier to reduce) 1. The more + the reduction potential, the stronger the Oxidizing Agent, (easier to reduce) –Compare the reduction potential of each substance –Whichever has the more positive reduction potential is the better oxidizing agent

Galvanic (Electrochemical) Cells

Spontaneous redox processes have: A positive cell potential, E 0 cell A negative free energy change, (  G is negative)

Calculation of Eº cell Zn e -  Zn Eº = -0.76V Cu e -  Cu Eº = +0.34V From a table of reduction potentials:

Calculation of Eº cell Eº cell = Eº red + Eº oxid Both values are written as reduction potentials in the table Determine what is oxidized Flip the equation Flip the sign Never multiply the potential by and integer, it is an intensive propertyNever multiply the potential by and integer, it is an intensive property

Zn - Cu Galvanic Cell Cu e -  Cu Eº (red) = +0.34V The less positive, or more negative reduction potential becomes the oxidation. Zn  Zn e - Eº (oxid) = +0.76V Cu Zn +  Zn 2+ + Cu E º cell = V

Line Notation Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) An abbreviated representation of an electrochemical cell Anodesolution Anodematerial Cathodesolution Cathodematerial ||||

Complete description of a galvanic cell 1.The cell potential Eºcell 2.Drawing of the cell* 3.Direction of electron flow a.Designation of anode and cathode b.Composition of the electrode (Pt or Graphite if no metals are in the half reactions) c.All ions in the solution from the half reactions *line notation is used if indicated

17.3 CELL POTENTIAL, ELECTRICAL WORK AND FREE ENERGY Day 2

Cell Potential and Work Galvanic Cell: E cell >0 and *work <0 *The amount of work a cell can perform (by the system on the surroundings)

Charge and Chemical Reactions Charge (q): number of coulombs transferred during a redox reaction n=number of moles of electrons transferred F= Faraday Constant (96485 C/mole e -)

Charge and Chemical Reactions Combine Need balanced reaction and cell potential

Calculating  G 0 for a Cell  G 0 = -nFE 0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e - E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = V

17.4 DEPENDENCE OF CELL POTENTIAL ON CONCENTRATION Day 3

The Nernst Equation Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R = 8.31 J/(mol  K) T = Temperature in K n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e -

Nernst Equation Simplified At 25  C (298 K) the Nernst Equation is simplified this way:

Equilibrium Constants and Cell Potential equilibrium At equilibrium, forward and reverse reactions occur at equal rates, therefore: The battery is “dead” 2. The cell potential, E, is zero volts Modifying the Nernst Equation (at 25  C):

E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = V Calculating an Equilibrium Constant from a Cell Potential

Concentration Cell Step 1: Determine which side undergoes oxidation, and which side undergoes reduction. Both sides have the same components but at different concentrations. ???

Concentration Cell Both sides have the same components but at different concentrations. The 1.0 M Zn 2+ must decrease in concentration, and the 0.10 M Zn 2+ must increase in concentration Zn 2+ (1.0M) + 2e -  Zn (reduction) Zn  Zn 2+ (0.10M) + 2e - (oxidation) ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M)

Concentration Cell Step 2: Calculate cell potential using the Nernst Equation (assuming 25  C). Both sides have the same components but at different concentrations. ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M) Concentration Cell

Nernst Calculations Zn 2+ (1.0M)  Zn 2+ (0.10M)

17.7 ELECTROLYSIS Day 3

Electrolytic Processes A negative -E 0 cell potential, ( -E 0 ) +  G A positive free energy change, ( +  G ) NOT Electrolytic processes are NOT spontaneous. They have:

Electrolysis of Water In acidic solution Anode rxn: Cathode rxn: V V V

Electroplating of Silver Anode reaction: Ag  Ag + + e - Electroplating requirements: 1. Solution of the plating metal 3. Cathode with the object to be plated 2. Anode made of the plating metal 4. Source of current Cathode reaction: Ag + + e -  Ag

Solving an Electroplating Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO 3 using a 20.0 Ampere current? 5.0 g Ag + + e -  Ag 1 mol Ag g 1 mol e - 1 mol Ag C 1 mol e - 1 s 20.0 C = 2.2 x 10 2 s

Uses of electrolysis Electrolysis of water –Produces H 2 and O 2 gas –2H 2 O  H 2 + O 2 Production of aluminium, lithium, sodium, potassium, magnesium, calciumaluminiumlithiumsodium potassiummagnesiumcalcium Production of chlorine and sodium hydroxidechlorinesodium hydroxide

Other uses Seperating mixtures of ions. –More positive reduction potential means the reaction proceeds forward. –Ion with highest reduction potential will plate out first or at the lowest voltage.

Consider Ag, Cu, and Zn ions, which will plate out first? Ag + + e -  AgE=0.80 V Cu e -  CuE= 0.35 V Zn e -  ZnE = V Silver, then copper, then zinc