Reduction-Oxidation (REDOX) Reactions

Electrochemistry 1

Reduction-Oxidation (REDOX) Reactions In chemical reaction bonds, both covalent and ionic, are made and broken by moving electrons. So far all in reactions considered the number of electrons on each atom has been preserved 0 0 1+ 2-2 Na + ½ O 2 Na 2 O 2 Na + + O 2-2 Na 2 Na + + 2 e ½ O 2 + 2 e O 2- Oxidized Reduced Loss e s Oxidation = LEO Gain e s Reduction = GER In REDOX reaction electrons are transferred between atoms as well as bonds being broken and formed. The balance between ionic/covalency of the bonds in the reactants is different from that in the product. We therefore need to keep track of the number of electrons of each atom. The oxidation state of an atom is used for this purpose, which is related to the idea of the formal charge. 2

Oxidation State The oxidation state of an element is its charge assuming ionic bonding Electrons are not shared they are placed on the more electronegative element Rules for Assigning Oxidation States 1) Pure elements have oxidation states of 0 2) Ions have oxidation states that add up to the charge of the ion 3) Hydrogen has an oxidation state of +1 unless bonded to a less electronegative atom. When bound to metals or boron it has an oxidation state of -1. 4) Fluorine has an oxidation state of -1 5) Oxygen has an oxidation state of -2 unless bonded to fluorine or another oxygen. 6) Halogens other than fluorine have oxidation states of -1 unless bonded to oxygen or a more electronegative halogen 7) The rest are determined by the process of elimination, where the oxidation states must add up to the total charge of the molecule or ion. 3

Oxidation State 1) Determine the oxidation state of all the element in the following molecules: a) H 2 H = 0 b) CO 2 O = -2 0 = C + 2(-2) c) BF 3 F= -1 0 = B + 3(-1) C = 4 B = 3 d) H 2 SO 4 H = 1 O = -2 0 = S +(2(+1) + 4(-2)) = S - 6 S = 6 e) SO 2 ClF O = -2 F = -1 Cl = -1 0 = S +(2(-2) -1-1) = S - 6 S = 6 f) IO 2 F 2 _ O = -2 F = -1-1 = I + (2(-2) + 2(-1)) = I - 6 I = 5 g) HPO 4 2- O = -2 H = 1-2 = P + (4(-2) + 1) = P - 7 P = 5 4

Oxidation States of Poly Atomic Ions CO 3 2- NO 3-4 5 NO 2-3 PO 3- SO 2-4 4 5 6 ClO 3-5 PO 3 3- SO 3 2- ClO 2-3 4 3 5

Recognizing Redox Reactions Consider the following reaction: Cu (s) + 4HNO 3 (aq) Cu(NO ) +N O + 2H O 3 2 (aq) 2 4 (g) 2 (l) Reactants Products Cu N = 0 Cu 2+ N = 2 HNO 3 N = 5 Cu Cu 2+ + 2 e 4 HNO 3 + 2 e 2NO 3 - + N 2 O 4 + 2 H 2 O N 2 O 4 N = 4 Oxidation Reduction REDOX equation seem difficult to balance # of e s gained = # of e s lost 6

Recognizing Redox Reactions The species that is oxidized in a Redox reaction is called the reducing agent In the previous example Cu is the reducing agent The species that is reduced in a Redox reaction is called the oxidizing agent In the previous example HNO 3 is the oxidizing agent 7

Redox Reactions The term Oxidation originates from the fact that many of these reactions involve oxygen where the element loses electrons to it. For example, Burning magnesium: Mg Mg 2+ + 2 e Oxidation Half Reaction Oxygen gains electrons: ½ O 2 + 2 e O 2 Reduction Half Reaction and the net reaction is: 1 Mg + ½ O 2 1 MgO 8

Redox Reactions The term Reduction originates from the concept of reducing an ore, usually a metal oxide, to its elemental form, usually the pure metal. For example: Iron is made from iron ore, the iron reaction is: Fe 2+ + 2 Fe 3+ + 8 e 3 Fe Reduction Half Reaction and the carbon monoxide reaction is: 4 CO 4 CO 2+ + 8 e Oxidation Half Reaction Here the oxide ions are spectators, transferring from the iron to the oxidized carbon. The net reaction therefore becomes: 1 Fe 3 O 4 + 4 CO 3 Fe + 4 CO 2 9

Balancing Redox Reactions in Solution The method of half reactions Step 1. Step 2. Step 3. Recognize the reaction as an oxidation-reduction. Separate the overall process into half-reactions. Balance each half-reaction by mass. i) In acid solution, a) add H 2 O to the side requiring O atoms. b) add H + to balance any remaining unbalanced H atoms. ii) In basic solution, a) add 2 OH to the side requiring O atoms, and H 2 O to the other b) add H 2 O to the side requiring H atoms, and one OH to the other side. Step 4. Step 5. Step 6. Step 7. Step 8. Balance the half-reactions by charge. Multiply the balanced half-reactions by appropriate factors to achieve common whole number of electrons being transferred. Add the balanced half-reactions. Eliminate common reactants and products. Check the final result for mass and charge balance. 10

Balancing Redox Reactions Sample Problems 2. Balance equations for the half-reactions. Assume these reactions occur in acidic solution, meaning that H + or H + and H 2 O may be used to balance the equation. (a) Br 2 (l) Br (aq) Ox. St. 0-1 reduction Mass balance: Br 2 (l) 2 Br (aq) Charge balance: Br 2 (l) + 2e - 2 Br (aq) (b) VO 2+ (aq) V 3+ (aq) Ox. St. +4 +3 reduction O balance: VO 2+ (aq) V 3+ (aq) + H 2 O H balance: VO 2+ (aq) + 2 H + V 3+ (aq) + H 2 O Charge balance: VO 2+ (aq) + 2 H + + e - V 3+ (aq) + H 2 O 11

Balancing Redox Reactions Sample Problems 3. The half-reactions here are in basic solution. You may need to use OH or the OH / H 2 O pair to balance the equation. (a) CrO 2 (aq) CrO 4 2 (aq) Ox. St. 3+ 6+ oxidation O balance: CrO 2 (aq) + 4 OH - CrO 2 4 (aq) + 2 H 2 O Charge balance: CrO 2 (aq) + 4 OH - CrO 2 4 (aq) + 2 H 2 O + 3 e - (b) Ni(OH) 2 (s) NiO 2 (s) Ox. St. 2+ 4+ oxidation H balance: Ni(OH) 2 (s) + 2 OH - NiO 2 (s) + 2 H 2 O Charge balance: Ni(OH) 2 (s) + 2 OH - NiO 2 (s) + 2 H 2 O + 2 e - 12

Balancing Redox Reactions Sample Problems 4. The reactions here are in acid solution, meaning that H + or H + and H 2 O may be used to balance the equation. MnO4 (aq) + HSO 3 (aq) Mn 2+ (aq) + SO 4 2 (aq) Ox. St. +7 +4 2+ 6+ Reduction half reaction: MnO 4 / Mn 2+ Mass Balance MnO 4 (aq) + Mn 2+ (aq) O balance: MnO 4 (aq) + 8H + Mn 2+ (aq) + 4 H 2 O Charge balance: MnO 4 (aq) + 8H + + 5 e Mn 2+ (aq) + 4 H 2 O Oxidation half reaction: HSO 3 / SO 4 2 Mass balance: HSO 3 (aq) SO 4 2 (aq) O balance: HSO 3 (aq) + H 2 O SO 4 2 (aq) + 3 H + Charge balance: HSO 3 (aq) + H 2 O SO 4 2 (aq) + 3 H + + 2 e x 5 x 2 X 2 MnO 4 (aq) + 5 HSO 3 (aq) + 16H + + 5 H 2 O 2 Mn 2+ (aq) + 5 SO 4 2 (aq) + 8 H 2 O + 15 H + 2 MnO 4 (aq) + 5 HSO 3 (aq) + 1 H + 2 Mn 2+ (aq) + 5 SO 4 2 (aq) + 3 H 2 O X X X 13

Balancing Redox Reactions Sample Problems 4. The reactions here are in acid solution, meaning that H + or H + and H 2 O may be used to balance the equation. Cr 2 O 7 2 (aq) + Fe 2+ (aq) Cr 3+ (aq) + Fe 3+ (aq) Ox. St. +6 +2 3+ 3+ Reduction half-reaction: Cr 2 O 7 2 / Cr 3+ Mass balance: Cr 2 O 7 2 (aq) 2Cr 3+ (aq) O balance: Cr 2 O 7 2 (aq) + 14H + 2 Cr 3+ (aq) + 7 H 2 O Charge balance: Cr 2 O 7 2 (aq) + 14H + + 6 e 2 Cr 3+ (aq) + 7 H 2 O Oxidation half-reaction: Fe 2+ / Fe 3+ Mass balance: Fe 2+ (aq) Fe 3+ (aq) Charge balance: Fe 2+ (aq) Fe 3+ (aq) + 1 e x 6 1 Cr 2 O 7 2 (aq) + 6 Fe 2+ (aq) + 14H + 2 Cr 3+ (aq) + 6 Fe 3+ (aq) + 7 H 2 O 14

Balancing Redox Reactions Sample Problems 5. The reactions here are in basic solution. You may need to add OH or the OH / H 2 O pair to balance the equation. Fe(OH) 2 (s) + CrO 4 2 (aq) Fe 2 O 3 (s) + Cr(OH) 4 (aq) Ox. St. +2 +6 3+ 3+ Oxidation half-reaction: Fe(OH) 2 /Fe 2 O 3 Mass balance : 2 Fe(OH) 2 (s) + Fe 2 O 3 (s) O balance: 2 Fe(OH) 2 (s) + 2 OH Fe 2 O 3 (s) + 3 H 2 O Charge balance: 2 Fe(OH) 2 (s) + 2 OH Fe 2 O 3 (s) + 3 H 2 O + 2 e x 3 Reduction half-reaction: CrO 4 2 / Cr(OH) 4 Mass balance: CrO 4 2 (aq) Cr(OH) 4 (aq) O balance: CrO 4 2 (aq) + 4 H 2 O Cr(OH) 4 (aq) + 4 OH Charge balance: CrO 4 2 (aq) + 4 H 2 O + 3 e Cr(OH) 4 (aq) + 4 OH 6 Fe(OH) 2 (s) + 2 CrO 4 2 (aq) + 6 OH + 8 H 2 O 3 Fe 2 O 3 (s) + 2 Cr(OH) 4 (aq) + 9 H 2 O + 8 OH - 6 Fe(OH) 2 (s) + 2 CrO 4 2 (aq) 3 Fe 2 O 3 (s) + 2 Cr(OH) 4 (aq) + 1 H 2 O + 2 OH - x 2 X X X X 15

Balancing Redox Reactions Sample Problems 5. The reactions here are in basic solution. You may need to add OH or the OH / H 2 O pair to balance the equation. (b) PbO 2 (s) + Cl (aq) ClO (aq) + Pb(OH) 3 (aq) Ox. St. 4+ 1-1+ 3+ Reduction half-reaction: PbO 2 / Pb(OH) 3 Mass balance: PbO 2 (s) + Pb(OH) 3 (aq) O balance: PbO 2 (s) + 2 H 2 O Pb(OH) 3 + OH Charge balance: PbO 2 (s) + 2 H 2 O + 2 e Pb(OH) 3 + OH Oxidation half-reaction: Cl / ClO Mass balance: Cl (aq) ClO (aq) O balance: Cl (aq) + 2 OH ClO (aq) + H 2 O Charge balance: Cl (aq) + 2 OH ClO (aq) + H 2 O + 2 e X X X X PbO 2 (s) + Cl (aq) + 2 OH + 2 H 2 O Pb(OH) 3 (aq) + ClO (aq) + OH + H 2 O 1 PbO 2 (s) + 1 Cl (aq) + 1 OH + 1 H 2 O 1 Pb(OH) 3 (aq) + 1 ClO (aq) 16

Gibbs Free Energy and Cell Potential The Gibbs Free Energy of reaction tells us whether a certain reaction is product favored or reactantfavoured in the forward direction The electrochemical analogue is the cell potential given by the symbol E E is expressed in the common electrical unit of volts. The exact relationship between cell potential and Gibbs energy is given by: ΔG = -nfe in general, and for standard conditions where F = 96485 C/mol Faradays Constant The cell voltage E has the opposite sign convention to that of G: 0 0 ΔG = -nfe i) Product Favoured Reaction: ve G or +ve E ii) Reactant Favoured Reaction: +ve G or ve E A reaction which is product favoured produces a voltage, and is therefore called a Voltaic cell, in honor of Alessandro Volta. A reaction which is reactant favoured can be driven forward by the application of a greater opposite voltage; such reactions are called electrolytic cells, and the process is named electrolysis. 17

Voltaic Cells Consider the reaction: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) G f (kj mol 1 ) 0 65.52 147.0 0 G rxn = 212.5 kj mol 1 < 0 & n = 2 Anode: Ox Zn(s) Zn 2+ (aq) + 2 e Cathode: Red. Cu 2+ (aq) + 2 e Cu(s) ΔG = -nfe 0 0 -ΔG -212.5kJ E = = nf 2 96.485kC = +1.10V > 0 under zero load conditions 18

Reversible cell reactions and cell notation Although the reaction only requires a copper(ii) salt and metallic zinc, the cell is built in such a way that both metallic zinc and copper are present, and copper and zinc sulfate. This ensures that the reactions are reversible as accurate potentials can only be measured for reversible electrochemical cells. A short-hand notation indicates the complete composition of a given electrochemical cell. The notation utilizes the balanced redox reaction, including both the reactants and the products. The notation for the previous example would be: Zn(s) Zn 2+ (1 M) Cu 2+ (1 M) Cu(s) Left Hand Side Anode reaction Oxidation reaction. Right Hand Side Cathode reaction Reduction reaction The vertical lines,, indicate a phase boundary, such as between a solid and a liquid. Double lines,, indicate double boundaries, such as commonly occur when a salt bridge is placed between the two half cells. 19

Electrochemical Cell Conventions 20

Standard Half-Cell Reduction Potentials The Cu/Zn cell has a standard potential of +1.10 V Can we relate this to the voltage of the two half-cells? Choose an arbitrary reference point: Standard Hydrogen Electrode, or SHE Thus Zn/Zn 2+ has a voltage of +0.76 V against a SHE 21

Standard Half-Cell Reduction Potentials The Cu/Cu 2+ has half cell has a voltage of +0.34 V against a SHE Combining the two half-cell voltages gives +0.76 + +0.34 = +1.10 V This is in good agreement with the measured voltage These values have been compiled into a general table To account for the sign convention, the table is written only as reduction reactions 22

Standard Hydrogen Electrode The glass tube is a hydrogen electrode that can be used as real reference electrode The picture below is a diagram showing the function of this device. + (s) 2 (g) 3 (aq) Pt H H O 23

Strongest oxidizers Reduction Half-Reaction E (V) F 2 (g) + 2 e 2 F (aq) +2.87 H 2 O 2 (aq) + 2 H 3 O + (aq) + 2 e 4 H 2 O(l) +1.77 PbO 2 (s) + SO 2 4 (aq) + 4 H 3 O + (aq) + 2 e PbSO4(s) + 6 H2O(l) +1.685 MnO 4 (aq) + 8 H 3 O + (aq) + 5 e Mn 2+ (aq) + 12 H 2 O(l) +1.52 Au 3+ (aq) + 3 e Au(s) +1.50 Cl 2 (g) + 2 e 2 Cl (aq) +1.360 Cr 2 O 2 7 (aq) + 14 H 3 O + (aq) + 6 e 2 Cr 3+ (aq) + 21 H 2 O(l) +1.33 O 2 (g) + 4 H 3 O + (aq) + 4 e 6 H 2 O(l) +1.229 Br 2 (l) + 2 e 2 Br (aq) +1.08 NO 3 (aq) + 4 H 3 O + (aq) + 3 e NO(g) + 6 H 2 O(l) +0.96 OCl (aq) + H 2 O(l) + 2 e Cl (aq) + 2 OH (aq) +0.89 Hg 2+ (aq) + 2 e Hg(l) +0.855 Ag + (aq) + e Ag(s) +0.80 Hg 2+ 2 (aq) + 2 e 2 Hg(l) +0.789 Fe 3+ (aq) + e Fe 2+ (aq) +0.771 I 2 (s) + 2 e 2 I (aq) +0.535 O 2 (g) + 2 H 2 O(l) + 4 e 4 OH (aq) +0.40 Cu 2+ (aq) + 2 e Cu(s) +0.337 Sn 4+ (aq) + 2 e Sn 2+ (aq) +0.15 2 H 3 O + (aq) + 2 e H 2 (g) + 2 H 2 O(l) 0.00 24

Reduction Half-Reaction E (V) 2 H 3 O + (aq) + 2 e H 2 (g) + 2 H 2 O(l) 0.00 Sn 2+ (aq) + 2 e Sn(s) 0.14 Ni 2+ (aq) + 2 e Ni(s) 0.25 V 3+ (aq) + e V 2+ (aq) 0.255 PbSO 4 (s) + 2 e Pb(s) + SO 2 4 (aq) 0.356 Cd 2+ (aq) + 2 e Cd(s) 0.40 Fe 2+ (aq) + 2 e Fe(s) 0.44 Zn 2+ (aq) + 2 e Zn(s) 0.763 2 H 2 O(l) + 2 e H 2 (g) + 2 OH (aq) 0.8277 Al 3+ (aq) + 3 e Al(s) 1.66 Mg 2+ (aq) + 2 e Mg(s) 2.37 Na + (aq) + e Na(s) 2.714 K + (aq) + e K(s) 2.925 Li + (aq) + e Li(s) 3.045 In volts (V) versus the standard hydrogen electrode. Strongest reducing agents Such tables may have the strongest reducing agent at the top, or at the bottom, as is the case in this table. Reactions are always written as reductions, but this is correlated to the voltages. For negative potentials, the reverse reaction is predicted. 25

Copper/Lithium cell Examples From the table we get the standard reduction potentials of the half-reactions: Cu 2+ (aq) + 2 e Cu(s) E = +0.337 V Reduction Li + (aq) + e Li(s) E = 3.045 V We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the lithium equation, making it the anode (where the oxidation will take place.) Li(s) Li + (aq) + e E = +3.045 V Anode now we have the cell potential for the overall reaction Cu 2+ (aq) + 2 Li(s) Cu(s) + 2 Li + (aq) E cell = +3.382 V Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in the value of n. 26

Some examples of using the Tables Silver/zinc cell From the table we get the standard reduction potential for the half-reactions: Zn 2+ (aq) + 2 e Zn(s) E = 0.763 V Reduction Ag + (aq) + e Ag(s) E = +0.80 V We now combine them in such a way as to get a positive overall cell potential. This means we must reverse the zinc equation, making it the anode (where the oxidation will take place.) Zn(s) Zn 2+ (aq) + 2 e E = +0.763 V Oxidation now we have the cell potential for the overall reaction Zn (s) + 2 Ag + (aq) Zn 2+ (aq) + 2 Ag (s) E cell = +1.563 V Note here very carefully, that the voltage was not doubled when the coefficients are doubled. However, n for this reaction = 2. Cell voltages are therefore independent of stoichiometry. The stoichiometric information is stored in the value of n. 27

Nernst Equation - Cells Non-Standard Conditions Voltaic cells eventually get depleted and stop producing a voltage. In fact, as the reactants in the cell are consumed, the voltage produces decreases. Temperature effects cell potential the potential is reduced upon cooling. Walter Nernst developed an expression, the Nernst equation, defined for the general equilibrium equation: aa + bb cc + dd c C D A B d o RT o RT E actual = E - ln = E - lnq a b nf Frequently the equation is given with all the constants applied out at 298.15K 0.0257V n o E actual =E - lnq nf 28

ph meter: the glass electrode A ph meter is actually a sensitive voltmeter attached to a glass electrode The electrode response to changes in [H 3 O + ] by altering its voltage as described by the Nernst equation The key to its function as a meter is the very thin and fragile glass membrane made of special materials that allow [H 3 O + ] but not other cations or anions to be sensed on the outer surface of the membrane The meter is calibrated in ph units by rewriting the Nernst equation in base-10 logarithms + 0.0592 H E glass_electrode = 0 - log + 1 H Filling in the constants at 298.15K this becomes: ph= E glass_electrode 0.0592 - const. inside outside Note that calibration is required to determine the constant Silver wire AgCl (s) KCl (saturated) Porous membrane 0.1 M HCl Glass membrane The Glass Electrode 29

Electrochemical Cells and Equilibrium Constants We have already seen that ΔG and E are related: 0 0 ΔG = -nfe We also know that ΔG and K are related: 0 ΔG = -RTlnK So its stands to reason that E and K are related, and you already know the relationship: An electrochemical reaction is at equilibrium when the cell voltage is zero So using the Nernst equation: RT nf o E actual =E - lnq when E= 0 V RT nf o 0 V =E - lnk Gives the relationship between cell potential and equilibrium constant Many equilibria are actually determined in this way nf lnk = E RT o Note that nf is on top in this equation! 30

Electrochemistry Example Problems 1. Calculate the E cell of a cell formed from a Zn 2+ /Zn half-cell in which [Zn 2+ ] = 0.0500 M and a Cl 2 /Cl half cell in which the [Cl ] = 0.0500 M and P(Cl 2 ) = 1.25 atm. In addition, provide a cell notation. From the Standard Reduction Potentials Table: Cl 2 (g) + 2 e 2 Cl (aq) +1.360 V Zn 2+ (aq) + 2 e Zn(s) 0.763 V Rev.#2: Zn(s) Zn 2+ (aq) + 2 e +0.763 V Add: Cl 2 (g) + Zn(s) Zn 2+ (aq) + 2 Cl (aq) +2.123 V o E actual = E - ln RT [Zn ][Cl ] nf p Cl 2+ - 2 2 actual J 2 8.314 Kmol 298.15K (0.0500)(0.0500) 2 96485 C mol (1.25atm) Voltage higher E = +2.123V - ln = +2.203V - 2+ (s) 2 (g) (s) Pt Cl Cl(0.0500M)Zn (0.0500M) Zn than standard! 31

More oxidizing Electrochemistry Example Problems 2. Consider the following half-reactions: Half-Reaction E (V) Ce 4+ (aq) + e Ce 3+ (aq) + 1.61 Ag + (aq) + e Ag(s) + 0.80 Hg 2+ 2 (aq) + 2e 2 Hg(l) + 0.79 Sn 2+ (aq) + 2 e Sn(s) 0.14 Ni 2+ (aq) + 2 e Ni(s) 0.25 Al 3+ (aq) + 3 e Al(s) 1.66 (a) Which is the weakest oxidizing agent in the list? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Does Sn(s) reduce Ag + (aq) to Ag(s)? (f) Does Hg(l) reduce Sn 2+ (aq) to Sn(s)? (g) Name the ions that can be reduced by Sn(s). (h) What metals can be oxidized by Ag + (aq)? More reducing Al 3+ Ce 4+ Al Ce 3+ +0.80 V + 0.25 V = +0.94 V: YES -0.79 V + -0.14 V = -0.93 V: NO Hg 2 2+, Ag +, Ce 4+ Hg, Sn, Ni, Al 32

Electrochemistry Example Problems 3. Calculate equilibrium constant for the following reaction: Zn 2+ (aq) + Ni(s) Zn(s) + Ni 2+ (aq) Locate in tables: Zn 2+ (aq) + 2 e - (aq) Zn(s) -0.763 V Ni(s) Ni 2+ (aq) + 2 e - (aq) +0.25 V (reversed, oxidation) So E cell = -0.763 + 0.25 = -0.51 V (two sig. fig. from the subtraction) We use the equation: nf lnk = E RT o which becomes: nf E o K = e RT 2 96485C mol -0.51V J 8.314 Kmol 298.15K -18 K = e = 5.7 10 33

Commercial Voltaic Cells and Storage Batteries Strictly speaking, a battery refers to an assembly, or battery, of 2 or more individual voltaic cells. In this sense, a typical flashlight battery is misnamed: it is really just a single cell. If cells are arranged in series their voltages add together. Thus for example a 12 V car battery is an assembly of six cells producing about 2 V each. A 9 V radio battery is also an assembly of 6 cells, each delivering about 1.5 V. Commercial voltaic cells are divided into two main categories: i) primary cells which are essentially the throw-away single use type, and ii) secondary cells which are capable of multiple recharges. In charging mode, these cells function as electrolysis cells. Nowadays there are many different cell technologies in the marketplace. Let us consider just a few examples of both categories of cells We will first look at several primary cells, then after consider a few secondary cells 34

Some typical Primary and Secondary Cells Battery Primary Battery and cell (throwaway) Battery Battery Secondary Battery and cell (rechargeable) Cell Cells Cells Battery Cell? Cells Battery 35

Zinc-carbon cell (Leclenché cell) CATEGORY: (Zn/MnO 2 in NH 4 Cl) Primary (Throwaway) Zinc Family CONSTRUCTION: The Leclanché dry cell or ordinary "flashlight battery." The anode is a zinc container covered with an insulating wrapper, the cathode is the carbon rod at the center. The electrolyte paste contains ZnCl 2, NH 4 Cl, MnO 2, starch, graphite, and water. A fresh dry cell delivers about 1.5 V. REDOX REACTIONS Positive terminal: 2MnO 2 (s) + 2NH 4+ (aq) + 2e Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) 0.74 V Negative terminal: Zn (s) > Zn 2+ (aq) + 2e Nominal cell voltage = +1.5 V 0.76 V COMMENTS The original "flashlight battery" was developed in 1865 by Georges Leclanché. It is often called a dry cell because the electrolyte is incorporated into a non-flowing paste. 36

Alkaline (Zinc/MnO 2 in KOH) CATEGORY: Primary (Throwaway) Zinc Family (+) CONSTRUCTION: The alkaline dry cell. The anode is a paste of zinc, KOH, and water, which donates electrons to the cell base via a brass collector. The cathode is a paste of MnO 2, graphite, and water, which takes electrons from the inner steel case. A plastic sleeve separates the inner steel case from the outer steel jacket. REDOX REACTIONS Nominal cell voltage = +1.5 V Positive terminal: 2MnO 2 (s) + H 2 O(l) + 2e Mn 2 O 3 (s) + 2OH - (aq) 0.15 V Negative terminal: Zn (s) + 2OH - (aq) Zn 2+ (aq) + H 2 O(l) + 2e Outer steel jacket Plastic sleeve Inner steel jacket 1.25 V (-) Cell base Cathode(+): paste containing MnO 2, graphite, and water Anode(-): Paste containing powdered zinc, KOH, and water Brass collector COMMENTS The alkaline dry cell is more expensive than the Leclanché cell, but it is also more efficient. Again, zinc is the anode and manganese dioxide the oxidizing agent. The electrolyte is 40% KOH saturated with zinc oxide (ZnO). 37

Alkaline Button Cell (Zinc/MnO 2 in KOH) CATEGORY: Primary (Throwaway) Zinc Family CONSTRUCTION: The alkaline dry cell. The anode is a paste of zinc, KOH, and water, which donates electrons to the cell base via a brass collector. The cathode is a paste of MnO 2, graphite, and water, which takes electrons from the inner steel case. A plastic sleeve separates the inner steel case from the outer steel jacket. REDOX REACTIONS Positive terminal: 2MnO 2 (s) + H 2 O(l) + 2e Mn 2 O 3 (s) + 2OH - (aq) Negative terminal: Nominal cell voltage = +1.5 V Zn (s) + 2OH - (aq) Zn 2+ (aq) + H 2 O(l) + 2e 0.15 V 1.25 V COMMENTS Apart from the different type of container, the chemistry of this cell is the same as that of the standard alkali dry cell. 38

CATEGORY: Lithium (Li/MnO 2 in KOH) Primary (Throwaway) Lithium Family CONSTRUCTION: Very similar to alkaline cell in design, except much more MnO 2 paste is used compared to Li due to the light weight of Lithium metal. REDOX REACTIONS Nominal cell voltage = +3.0 V Positive terminal: 2MnO 2 (s) + H 2 O(l) + 2e Mn 2 O 3 (s) + 2OH - (aq) 0.15 V Negative terminal: Li (s) Li + (aq) + 3.04 V Outer casing Positive terminal MnO2, graphite KoH, and water Absorbent fabric sat'd in KOH and moistened (+) COMMENTS The electrolyte consists of between 20 and 40% by mass of KOH, which is impregnated on the absorbent material between the two half-cells. This is only one of a whole family of Li based batters. Others are: Li/SO 2, Li/SOCl 2, Li/CuO, Li-poly(vinyl pyridine)/i 2 solid electrolyte. Thus the lithium batteries can replace the zinc family of batteries with a whole new range of high-power, low mass cells. The next 10-20 years should see substantial development in this area. Wire mesh cont. lithium metal Inner steel jacket (-) 39

Mercury (Zn/HgO in KOH) CATEGORY: Primary (Throwaway) Zinc Family CONSTRUCTION: A mercury battery. A zinc-mercury amalgam is the anode; the cathode is a paste of HgO, graphite, and water, Mercury batteries, some of which are smaller than a pencil eraser, deliver about 1.34 V. REDOX REACTIONS Positive terminal: HgO (s) + H 2 O(l) + 2e Hg(l) + 2OH - (aq) Negative terminal: Zn (s) + 2OH - (aq) ZnO(s) + H 2 O(l) + 2e COMMENTS 0.09 V 1.25 V The mercury cell, developed in 1942, is another zinc dry cell devised for use in small appliances such as watches, and us usually produced as a button battery. It delivers only 1.34 V, which is significantly less than the 1.5 V of a standard dry-cell. It has the advantage of maintaining a fairly constant voltage during its lifetime. Nominal cell voltage = +1.34 V 40

Silver oxide (Zn/Ag 2 O in KOH) CATEGORY: Primary (Throwaway) Zinc Family CONSTRUCTION: A silver oxide button battery, similar to the mercury and alkaline manganese cell. Also a reliable source of voltage. Used in medical devices, calculators, older cameras. REDOX REACTIONS Positive terminal: Ag 2 O (s) + H 2 O(l) + 2e 2 Ag(s) + 2OH - (aq) Negative terminal: Zn (s) + 2OH - (aq) ZnO(s) + H 2 O(l) + 2e COMMENTS 0.34 V 1.25 V Nominal cell voltage = +1.5 V Because no solution species is involved in the net reaction, the quantity of electrolyte is very small, and the electrodes can be maintained very close together. The storage capacity of a silverzinc cell is about six times as great as a lead-acid cell of the same size. They are expensive because of the use of silver oxide. 41

Zinc-Air (Zn/O 2 in KOH) CATEGORY: Primary (Throwaway) Zinc Family CONSTRUCTION: Constructions very similar to mercury or silver oxide button cells. But there are holes in the base to allow air in! REDOX REACTIONS Positive terminal: ½ O 2 (s) + H 2 O(l) + 2e 2OH - (aq) Negative terminal: Zn (s) + 2OH - (aq) ZnO(s) + H 2 O(l) + 2e COMMENTS 0.40 V 1.25 V Mercury is added to the zinc, forming a liquid mixture called an amalgam, which prevents the formation of an insulating layer of zinc oxide. The zinc-air cell is sold with a patch covering the air holes. This patch must be removed before the battery develops a voltage. When oxygen gas from the air percolates into the hydroxide solution in the positive half-cell, reaction commences and a voltage develops in the cell. Can you see why the voltage is less than predicted by E? Nominal cell voltage = +1.4 V p(o 2 ) 0.20 atm 42

Lead Acid (PbO 2 /Pb) CATEGORY: Secondary/Rechargeable CONSTRUCTION: Shown in the picture is a 6-V (i.e. motorcycle) battery, which contains 3 individual cells in series. The anode (negative terminal) consists of a lead grid filled with spongy lead, and the cathode (positive terminal) is a lead grid filled with lead dioxide. The cell also contains 38% (by mass) sulfuric acid. REDOX REACTIONS Nominal cell voltage = +2.0 V Positive terminal: PbO 2 (s) + HSO 4- (aq) + 3H + + 2e PbSO 4 (s) + 2H 2 O (l) 1.69 V Negative terminal: Pb (s) + HSO 4- (aq) PbSO 4 (s) + H + + 2e 0.34 V COMMENTS The cell equations shows that discharging depletes the sulfuric acid electrolyte and deposits solid lead sulfate on both electrodes. The electrolyte density can be measured by a hydrometer; it is charged when d = 1.28 (37% H 2 SO 4 ) and discharged when it is below 1.15 g/ml (21% H 2 SO 4 ). The lead acid battery is extremely reliable, but heavy. Electrical cars using such batteries are too inefficient. How many cells are there in a 12 V automotive battery? How can you tell? 43

Nickel-Cadmium (Nicad) CATEGORY: Secondary/Rechargeable CONSTRUCTION: The Nicad cell consists of many layers of alternating Ni and Cd electrodes. The actual electrodes consist of steel foil with lots of holes in them, onto which the active components are smeared. Each electrode is surrounded by an absorbent pad which contains the electrolyte. The nickel hydroxide electrodes are slightly longer at the top, and are soldered to a collector along the top of the cell; the cadmium electrodes are similarly joined along the bottom. REDOX REACTIONS Nominal cell voltage = +1.2 V Positive terminal: NiO(OH)(s) + H 2 O(l) + e NiO 2 (s) + OH - (aq) 0.82 V Negative terminal: Cd (s) + 2 OH - (aq Cd(OH) 2 (s) 0.49 V Positive terminal collector Concentric rings of mesh covered alternatively with NiO(OH) and Cd Negative terminal collector COMMENTS The nickel-cadium cell, for example, is used in hand calculators, electronic flash units, and other small battery-operated devices. The nickel-cadmium cell has electrodes of cadmium and NiO(OH), and the electrolyte is a 20% aqueous KOH solution. 44

Electrolysis Reactions So far, voltaic cells, which are product favoured reactions doing useful work Many electrochemical reactions are truly reversible, and can be driven against the spontaneous direction by applying a greater voltage of opposite polarity Driven electrochemical reactions are called electrolytic reactions, and the process is know as electrolysis. Important industrial electrolytic reactions: Aluminum Magnesium Sodium Many of the rarer metals Electro-refining of copper depends in part on electrolysis Elemental chlorine Sodium hydroxide Electroplating technologies, e.g. chrome plating The aluminum refining industry is of great significance in Canada 45

Electro-Refining of Copper 46

Electrolytic Production of Magnesium 47

Counting Electrons It is both desirable and straightforward to relate the electrical current in an electrochemical process to the quantity of materials produced and/or consumed during an electrochemical reaction. This is best understood as a branch of stoichiometry in which the number of moles of electrons is used as a stoichiometric factor. The basis of this is the Faraday constant which we have already dealt with and the equation relating current to charge from basic physics. This equation is: q I= t in which I is the current in Amperes, q is the charge in Coulombs and t is the time in seconds. Note that these precise units must be used. Rearrangment allows the calculation of the number of Coulombs of electrical charge that is consumed in any given process. Then we remember that Faraday s constant is 96 485 C/mol of electrons Once we have the number of moles of electrons, and knowing the value of n in the balanced redox reaction, the whole problem boils down to simple stoichiometry. 48

Counting Electron Example Problems 1. If you electrolyze a solution of Ni 2+ (aq) to form Ni(s), and use a current of 0.15 amp for 10 min, how many grams of Ni(s) is produced? Ni 2+ (aq) + 2 e - Ni (s) n = 2 q I= ; q =I t t q = 0.15amp 10min 60 90C #mol e = 96485 1 #mol Ni = 2 s min = 90C - -4 C mol = 9.327 10 mol - -4-4 #mol e = 0.5 9.327 10 mol = 4.66 10 mol -4 g #gni = 4.66 10 mol 58.69 mol = 0.027g 49

Counting Electron Example Problems 2. An electrolysis cell for aluminum production operates at 5.0 V and a current of 1.0 10 5 amp. Calculate the number of kilowatt-hours of energy required to produce 1 tonne (1.0 10 3 kg) of aluminum. 1 tonne = 1.0 x 10 6 g #mol Al = 6 1.0 10 mol 26.9815 g mol Al 3+ (aq) + 3 e - Al (s) n = 3 4 = 3.706 10 mol - 4 5 #mol e =3 #mol Al=3 3.706 10 mol=1.11 10 mol - 5 C 10 #C e =1.11 10 mol 96485 mol =1.07 10 C J=C V; 6 W=J s; 1kWh=3.60 10 J #kwh = 10 1.07 10 C 5.0V 3.60 10 6 JkWh 4 =1.5 10 kwh 50

Important Concepts In Electrochemistry Assigning oxidation states, redox reactions, etc. Using half-reactions to balance redox reactions Voltaic cells vs. electrolytic cells Cell potential (and how it relates to Gibbs free energy) Components of a voltaic cell Electrodes (cathode and anode) Salt bridge Cell notation Cells under nonstandard conditions (the Nernst equation) Relationship between E and equilibrium constant Practical considerations for batteries, ph meters, etc. Applications of electrolytic cells Determining efficiency of electrolytic cells 51