Electrochemistry Interchange of electrical and chemical energy.

Electron Transfer Reactions Electron transfer reactions are oxidation-reduction or redox reactions. the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

Electrochemistry Terminology #1 Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na+ + e- Reduction – A process in which an element attains a more negative oxidation state Cl2 + 2e-  2Cl-

Electrochemistry Terminology #2 An old memory device for oxidation and reduction goes like this… LEO says GER Lose Electrons = Oxidation Gain Electrons = Reduction

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

2Mg (s) + O2 (g) 2MgO (s) O2 + 4e- 2O2- 2+ 2- 2Mg (s) + O2 (g) 2MgO (s) 2Mg 2Mg2+ + 4e- Oxidation half-reaction (lose e-) O2 + 4e- 2O2- Reduction half-reaction (gain e-) 19.1

Electrochemistry Terminology #4 Anode The electrode where oxidation occurs Cathode The electrode where reduction occurs Memory device: An ox Red Cat

Cr2O72-(aq) + SO32-(aq)  Cr3+(aq) + SO42-(aq) How can we balance this equation? First Steps: Separate into half-reactions. Balance elements except H and O.

Method of Half Reactions Cr2O72-(aq)  Cr3+(aq)   SO32-(aq)  SO42-(aq) Now, balance all elements except O and H.

Method of Half Reactions (continued) Cr2O72-(aq)  2Cr3+(aq)  SO32-(aq)  + SO42-(aq) How can we balance the oxygen atoms? Copyright © Cengage Learning. All rights reserved

Method of Half Reactions (continued) Cr2O72-(aq)  Cr3+(aq) + 7H2O  H2O +SO32-(aq)  + SO42-(aq) How can we balance the hydrogen atoms?

Method of Half Reactions (continued) This reaction occurs in an acidic solution. 14H+ + Cr2O72-  2Cr3+ + 7H2O  H2O +SO32-  SO42- + 2H+ How can we write the electrons?

Method of Half Reactions (continued) This reaction occurs in an acidic solution. 6e- + 14H+ + Cr2O72-  2Cr3+ + 7H2O  H2O +SO32-  SO42- + 2H+ + 2e- How can we balance the electrons?

Method of Half Reactions (continued) 14H+ + 6e- + Cr2O72-  2Cr3+ + 7H2O 3[H2O +SO32-  SO42- + 2e- + 2H+] Final Balanced Equation: Cr2O72- + 3SO32- + 8H+  2Cr3+ + 3SO42- + 4H2O

Exercise Balance the following oxidation–reduction reaction that occurs in acidic solution. Br–(aq) + MnO4–(aq)  Br2(l)+ Mn2+(aq) 10Br–(aq) + 16H+(aq) + 2MnO4–(aq)  5Br2(l)+ 2Mn2+(aq) + 8H2O(l) 10Br-(aq) + 16H+(aq) + 2MnO4-(aq)  5Br2(l)+ 2Mn2+(aq) + 8H2O(l)

Half–Reactions The overall reaction is split into two half–reactions, one involving oxidation and one reduction. 8H+ + MnO4- + 5Fe2+  Mn2+ + 5Fe3+ + 4H2O Reduction: 8H+ + MnO4- + 5e-  Mn2+ + 4H2O Oxidation: 5Fe2+  5Fe3+ + 5e-

How do we harness the energy from the reaction? If they are in the same solution; energy is released as heat and can do no work. Separate the solution; MnO4- in one beaker and Fe2+ in another requiring the e- to be transferred through a wire producing a current.

Figure 17.1: Schematic of a method to separate the oxidizing and reducing agents of a redox reaction. (The solutions also contain counterions to balance the charge.) Current stop flowing due to separation of charge.

Keeps the net charge at zero; allows Figure 17.2: Galvanic cells can contain a salt bridge as in (a) or a porous-disk connection as in (b). Keeps the net charge at zero; allows the ions to flow w/o mixing the solutions.

Galvanic Cell A device in which chemical energy is changed to electrical energy.

- + anode oxidation cathode reduction spontaneous redox reaction 19.2 http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf 19.2

Cell Potential A galvanic cell consists of an oxidizing agent in one compartment that pulls electrons through a wire from a reducing agent in the other compartment. The “pull”, or driving force, on the electrons is called the cell potential (º), or the electromotive force (emf) of the cell. Unit of electrical potential is the volt (V). 1 joule of work per coulomb of charge transferred. (volt= Joule/coulomb)

Table of Reduction Potentials Measured against the Standard Hydrogen Electrode

Figure 17.5: (a) A galvanic cell involving the reactions Zn → Zn2+ + 2e- (at the anode/oxidation) and 2H+ + 2e-→ H2 (at the cathode/reduction) has a potential of 0.76 V. (b) The standard hydrogen electrode where H2(g) at 1 atm is passed over a platinum electrode in contact with 1 M H+ ions. This electrode process (assuming ideal behavior) is arbitrarily assigned a value of exactly zero volts.

Standard Reduction Potentials Potential of half-reactions are summed to obtain cell potential. 2H+(aq) + Zn(s) → Zn2+(aq) + H2(g) The anode compartment? The cathode compartment? Cathode consists of a platinum electrode (inert conduction) in contact w/1M H+ and bathed by H gas at 1atm. Standard Hydrogen Electrode. Zn → Zn2+ + 2e- oxidation 2H+ + 2e- → H2 reduction

Total cell potential = 0.76V; must be divided to get the half-potentials Assign this reaction 2H+ + 2e- → H2 reduction Where: [H+]= 1M and PH2=1 atm A potential of exactly 0 V, then the reaction will have a potential of 0.76 V because Zn → Zn2+ + 2e- oxidation

Example: Fe3+(aq) + Cu(s) → Cu2+(aq) + Fe2+(aq) Half-Reactions: Fe3+ + e– → Fe2+ E ° = 0.77 V Cu2+ + 2e– → Cu E ° = 0.34 V To calculate the cell potential, we must reverse reaction 2. Cu → Cu2+ + 2e– – E ° = – 0.34 V

Overall Balanced Cell Reaction 2Fe3+ + 2e– → 2Fe2+ E ° = 0.77 V (cathode) Cu → Cu2+ + 2e– – E ° = – 0.34 V (anode) Cell Potential: E °cell = E °(cathode) – E °(anode) **not on formula sheet! *** the – in the equation automatically flips the anode E °cell = 0.77 V – 0.34 V = 0.43 V

Manipulations to obtain a balanced oxidation-reduction reaction One half-reaction must be reversed, which means the sign must be reversed. Since the # of e- gained must equal the # of e- lost, the half-reactions must be multiplied by integers. But the value of º is not changed since it is an intensive property. A galvanic cell runs spontaneously in the direction that gives a positive value for E °cell.

Line Notation Anode components are listed on the left and the cathode on the right. Mg(s) Mg2+(aq) Al3+(aq) Al(s) Mg → Mg2+ + 2e– (anode) Al3+ + 3e– → Al (cathode) Phase difference- single line Salt bridge- double line

The relationship between thermodynamics and electrochemistry

Cell Potential, Electrical Work, and Free Energy Work that can be done by a cell depends on the “push” driving force. (emf) Defined in terms of potential difference (V) between two points in a circuit. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Volt – J of work/C of charge transferred emf = potential difference(V) = work (J) charge(C) 1J work is produced or needed, when 1C of charge is transferred between two points in the circuit that differ by a potential of 1 volt.

Work is from the point of view of the system. Work flowing out is (-) Cell produces current, ξ is (+), current is used to do work (-). Thus, ξ and w have opposite signs ξ= -w/q (charge) -w= ξq Max work is obtained at max ξ (cell potential) -wmax=q ξmax wmax= -q ξmax

Problem? Wmax will never be reached Why?? *in any real spontaneous process some E is always lost . Actual work is less that max work. Entropy??? What is the only type of process in which you can reach wmax ?

Although we may never achieve wmax, we can calculate max potential Although we may never achieve wmax, we can calculate max potential.We use wmax to calculate efficiency. Suppose we have a galvanic cell wmax= 2.50V And 1.33 mol of e- were passed through the cell and the actual potential of 2.10V is registered. The actual work done is: w= - ξq ξ = actual potential difference at which the current flowed (2.10V or 2.10J/C) q = quantity of charge in coulombs transferred

Faraday (F) The charge on 1mol of e- = faraday(F) 96,485 C of charge/mol e- q=nF= (1.33mol)(96,485C/mol) w= -q ξ = -(1.33mol)(96,485C/mol)(2.10J/C) = -2.96x105J wmax= -(1.33mol)(96,485C)(2.50J/C) = -3.21x105J Efficiency = w/wmax x 100% = 83.8%

Relate potential of cell to free energy at constant T and P : wmax=ΔG wmax= -qξmax = ΔG q=nF ΔG=-nF ξ Assume at this point any potential is max.

For standard conditions: ΔG = nFE The max cell potential is directly related to the free energy difference between the reactants and the products in the cell. (experimental means!) A positive cell potential yeilds a -ΔG

Calculate ΔG° for the reaction Cu2+(aq) + Fe(s) → Cu(s) + Fe2+(aq) Is this reaction spontaneous? Half reaction: Cu2+ + 2e- → Cu ξ° = 0.34 V Fe(s) → Fe2+ + 2e- -ξ° = 0.44 V Cu2+ + Fe → Fe2+ + Cu ξ° = 0.78 V

Copyright©2000 by Houghton Mifflin Company. All rights reserved. Predict whether 1M HNO3 will dissolve gold metal to form a 1 M Au3+ solution. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Concentration Cell . . . a cell in which both compartments have the same components but at different concentrations.

Cell Potential and concentration Standard conditions (1M) Cu(s) + 2Ce4+(aq) → Cu2+ + 2Ce3+(aq) ξ= 1.36V What will the potential be if [Ce4+] is greater than 1M? LeChatelier? Increase or decrease cell potential? Increase driving force on the e-

For the reaction a. ξ°< 0.48V b. ξ°> 0.48V 2Al(s) + 3Mn2+(aq) → 2Al3+(aq) + 3Mn(s) ξ°= 0.48 V Will ξcell be larger or smaller than ξ° in the following cases? a. [Al3+] = 2.0M [Mn2+] = 1.0M b. [Al3+] = 1.0M [Mn2+] = 3.0M a. ξ°< 0.48V b. ξ°> 0.48V

Cell potential and concentration An increase in [reactant] caused an increase in cell potential, and vice versa. Therefore we can construct [cells] of the same components.

Which way will electron flow to equilibrium? Ag+ + e- → Ag ξ°=0.80V Figure 17.9: A concentration cell that contains a silver electrode and aqueous silver nitrate in both compartments Ag+ + e- → Ag ξ°=0.80V e-  Which way will electron flow to equilibrium? anode cathode

Nernst Equation Dependence of cell potential on concentration, results from the dependence ΔG° on concentration. ΔG= ΔG° + RTln(Q) Q= reaction quotient ΔG = -nF ξ -nFξ =-nF ξ° + RTln(Q) ξ = ξ ° -RT/nF ln(Q)

The Nernst Equation R = 8.31 J/(molK) T = Temperature in K 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:

Both sides have the same components but at different concentrations. ??? Concentration Cell Both sides have the same components but at different concentrations. 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. Anode Cathode The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration Zn2+ (1.0M) + 2e-  Zn (reduction) Zn  Zn2+ (0.10M) + 2e- (oxidation) Zn2+ (1.0M)  Zn2+ (0.10M)

Both sides have the same components but at different concentrations. Concentration Cell ??? Concentration Cell Both sides have the same components but at different concentrations. Anode Cathode Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C). Zn2+ (1.0M)  Zn2+ (0.10M)

Nernst Calculations Zn2+ (1.0M)  Zn2+ (0.10M)

ξ°cell =0.48V for a cell based on the reaction: 2Al(s) + 3Mn2+(aq)→2Al3+(aq) + 3Mn(s) [Mn2+] = 0.50M [Al3+] = 1.50M

2Al(s) + 3Mn2+(aq)→2Al3+(aq) + 3Mn(s) [Mn2+] = 0.50M [Al3+] = 1.50M Half reactions 2Al →2Al3+ + 6e- 3Mn3+ + 6e- →3Mn n= 6 Voltage reduced (Le Chatelier) since the [reactant] is lower than 1.0M and the [products] is higher than 1.0M, ξcell is less than ξ°cell

*The cell will spontaneously discharge until it reaches equilibrium. at that point… Q=K (the equilibrium constant) and ξcell=0.

Dead battery is at equilibrium. At equilibrium the components in the two cell compartments have the same free energy, and ΔG=0. The cell can no longer do work.

Calculation of Equilibrium Constants for Redox Reactions At equilibrium, Ecell = 0 and Q = K.

Spontaneity of Redox Reactions 19.4

Calculating an Equilibrium Constant from a Cell Potential Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

One of the Six Cells in a 12–V Lead Storage Battery Anode rxn: Pb + HSO4-  PbSO4 + H+ + 2e- Cathode rxn: PbO2 + HSO4- + 3H+ + 2e- + PbSO4 + 2H2O Cell rxn: Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq)  2PbSO4(s) + 2H2O(l) Discharge lowers the concentration of sulfuric acid, and the density of the solution. PbSO4(s) forms on the grids (electrodes) and can be recharged.

Charging a Battery When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal. In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.

A Common Dry Cell Battery Acidic version 1.5V

A Mercury Battery Basic version: last longer, less corrosion your calculator

Schematic of the Hydrogen-Oxygen Fuel Cell galvanic cells for which the reactants are continuously supplied. 2H2(g) + O2(g)  2H2O(l) anode: 2H2 + 4OH  4H2O + 4e cathode: 4e + O2 + 2H2O  4OH

Chemistry In Action: Dental Filling Discomfort Sn /Ag3Sn -0.05 V 2+ Hg2 /Ag2Hg3 0.85 V 2+ Sn /Ag3Sn -0.05 V 2+

Corrosion: Process of returning metals to their natural state – the ores from which they were originally obtained. Involves oxidation of the metal. Some metals, such as copper, gold, silver and platinum, are relatively difficult to oxidize. These are often called noble metals.

The Electrochemical Corrosion of Iron Steel is not homogeneous and has nonuniformities where oxidation occurs easily (anodic regions) E- flow through the moisture on the surface of the steel (salt bridge) to the cathodic region and react with O2

Corrosion Prevention 1. Application of a coating (like paint or metal plating) Chromium and Tin form durable oxide coating Galvanizing with Zinc 2. Alloying: Stainless steel contains chromium and nickel (changes the reduction potential) 3. Cathodic Protection: Protects steel in buried fuel tanks and pipelines.

Cathodic Protection Mg: active metal furnish the e- instead of the iron, preventing the iron from oxidizing

Producing aluminum metal Chrome plating Electrolysis: Forcing a current through a cell to produce a chemical change for which the cell potential is negative. Electrolytic cell Charging a battery Producing aluminum metal Chrome plating

Figure 17.19: (a) A standard galvanic cell based on the spontaneous reaction (b) A standard electrolytic cell. A power source forces the opposite reaction

Electrolytic Processes Electrolytic processes are NOT spontaneous. They have: A negative cell potential, (-E0) A positive free energy change, (+G)

Stoichiometry of Electrolysis How much chemical change occurs with the flow of a given current for a specified time? current and time  quantity of charge  moles of electrons  moles of analyte  grams of analyte

Stoichiometry of Electrolysis current and time  quantity of charge 1 ampere = 1 coulomb/second 1 mole of e- = 96,485 coulombs Coulombs of charge = amps (C/s) × seconds(s) quantity of charge  moles of electrons

How much Ca will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: 2Cl- (l) Cl2 (g) + 2e- Cathode: 2 mole e- = 1 mole Ca Ca2+ (l) + 2e- Ca (s) Ca2+ (l) + 2Cl- (l) Ca (s) + Cl2 (g) mol Ca = 0.452 C s x 1.5 hr x 3600 s hr 96,500 C 1 mol e- x 2 mol e- 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca 19.8

Electrolysis of Water 19.8

Electrolysis of Water In acidic solution Anode rxn: Cathode rxn: -1.23 V Cathode rxn: -0.83 V -2.06 V

Electroplating a Spoon

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

Solving an Electroplating Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO3 using a 20.0 Ampere current? Ag+ + e-  Ag 5.0 g 1 mol Ag 1 mol e- 96 485 C 1 s 1 mol e- 20.0 C 107.87 g 1 mol Ag = 2.2 x 102 s

Ag+ + e- → Ag ξ°= 0.80V Cu2+ + 2e- → Cu ξ°= 0.34V Suppose a solution in an electrolytic cell contains the inos Cu2+, Ag+, and Zn2+. If the voltage is very low and is gradually turned up, in which order will the metals be plated out onto the cathode? Ag+ + e- → Ag ξ°= 0.80V Cu2+ + 2e- → Cu ξ°= 0.34V Zn2+ + 2e- → Zn ξ°= -0.76V

Zn - Cu Galvanic Cell Zn  Zn2+ + 2e- E = +0.76V The less positive, or more negative reduction potential becomes the oxidation… Cu2+ + 2e-  Cu E = +0.34V Zn  Zn2+ + 2e- E = +0.76V Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

Line Notation Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) | || | An abbreviated representation of an electrochemical cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Anode material Anode solution Cathode solution Cathode material | || |