Chapter 17 Creative Commons License
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Chemistry: OpenStax 2
Electrochemistry Oxidation Numbers
Free elements: ox # = 0 (H (g), Hg(l), etc.) Electrochemistry is the study of batteries 2 and the conversion between chemical and Ions in binary ionic compounds: ox # = charge. Ex. For Al S , ox # for Al = +3, ox # for S = -2. electrical energy. 2 3 H: ox # = +1, except when H is with alkali Based on redox (oxidation-reduction) metals it is -1 (LiH, NaH, etc.) reactions in which one substance gains O: ox # = -2, except in peroxides it is -1 (H O , electrons and another loses electrons. 2 2 K2O2) These two processes MUST happen together. The sum of the oxidation numbers of all the Single replacement, combustion, atoms in a molecule = 0 combination and decomposition rxns are all For polyatomic ions, the sum of the oxidation examples of redox reactions. numbers must equal the charge on the ion.
Ox # examples Redox: LEO the lion goes GER
- Oxidation: loss of electrons; ox # , more + Find ox numbers for all the atoms in HCO3 H: ox # = +1 O: ox # = -2 Reduction: gain of electrons; ox # , more − 1H + 1C + 3O = -1 oxidizing agent: substance that is reduced; it caused oxidization of other substace. + 1 + C + 3x(-2) = -1 reducing agent: substance that is oxidized; it Ox # for C = +4 caused reduction of other substance. 2- What is the ox # for Cr in Cr2O7 ? Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s) 2Cr + 7O = -2 Identify atom oxidized, atom reduced, 2Cr + 7(-2) = -2 oxidizing agent and reducing agent. Cr = +6
1 1 Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s) Zn(s) + Cu(NO3)2(aq) Zn(NO3)2(aq) + Cu(s) 0 +2 +5 -2 +2 +5 -2 0
• Zn 0 to +2 ox # ; Zn becomes more + Zn(s) Zn2+(aq) + 2e- • Cu +2 to 0 ox # ; Cu2+ becomes more -
Zn(s) dissolves 2+ What is reduced: Cu in Cu(NO3)2
What is oxidizing agent: Cu(NO3)2 Cu2+(aq) + 2e- Cu(s) What is reducing agent: Zn(s)
What is oxidized: Zn(s) Cu plates on electrode
These are always reactants in an overall cell reaction!
Balancing Simple Redox Reactions Redox Reactions
What is oxidized/reduced, give the specific atom. E.g. How do we balance the following reaction? 3+ 2+ Agents: give the whole substance as an answer. Cr (aq) + Be(s) Cr(s) + Be (aq) - + 2+ 2+ MnO4 (aq) + 8H (aq) + 5 Fe (aq) Mn (aq) + 1) Break the redox rxn into two half reactions, the 3+ 5Fe (aq) + 4H2O(l) oxidation half reaction and reduction half reaction. For this reaction: 2) Balance charges by adding electrons (e-) to the - What is the ox # for Mn in MnO4 +7 more + side for each half reaction. What is oxidized? Fe2+ 3) Balance electrons by multiplying each half - What is reduced? Mn in MnO4 reaction by an integer so that the # electrons - gained = # electrons lost. What is the oxidizing agent? MnO4 What is the reducing agent? Fe2+
Example Balancing Redox Rxn Types of Cells
E.g. Cr3+(aq) + Be(s) Cr(s) + Be2+(aq) Galvanic cell (also called voltaic cell) Electrochemical cell in which a spontaneous Write the Half reactions reaction generates electricity. (Electric energy Oxidation: Be(s) Be2+(aq) + 2e- (e- = product) is produced.) Reduction: Cr3+(aq) + 3e- Cr(s) (e- = reactant) Balance the electrons Electrolytic cell 2+ - Oxidation: (Be(s) Be (aq) + 2e )x3 Electrochemical cell in which an electrical 3+ - Reduction: (Cr (aq) + 3e Cr(s))x2 current is used to drive a nonspontaneous Add reactions and cancel common terms: reaction. (Electric energy is consumed.) 2Cr3+(aq) + 3Be(s) 2Cr(s) + 3Be2+(aq)
2 2 Galvanic Cell Galvanic Cells Oxidation occurs at the anode - both vowels Figure 17.4: mass of anode decreases - it is dissolving as metal A Galvanic cell atoms lose electrons to form ions in solution provides energy = + V. Reduction occurs at the cathode - both consonants
mass of cathode increases as metal ions are reduced to form atoms that plate onto the cathode. External Circuit - electrons flow from the anode to the cathode via an external wire. Salt bridge - soluble salt solution in a bridge that connects the two half cells; ions flow through the bridge to complete the electrical circuit. Galvanic cell Youtube Videos: Voltaic Cell: movie http://www.youtube.com/watch?feature=player_detailpage&v=J1ljxodF9_g
Electron motion in a cell 13 http://www.youtube.com/watch?feature=player_detailpage&v=C26pH8kC_Wk
Dr. Lisa’s cell Mnemonic What happens in Salt Bridge?
Fat red cat eats electons! Migration of ions maintains charge e- neutrality in both compartments: Anions move into the anode where excess + charge builds up as metal Anorexic ox spits them out! cations are formed by oxidation. Cations move into the cathode where e- excess (-) charge builds up as the metal cations are reduced to form neutral metal atoms. Cathode is reduced (gains e-) & mass (plating) Anode is oxidized (loses e-) & mass (dissolves)
Shorthand Notation for Galvanic Cells Electromotive Force Anode half-reaction: Zn(s) Zn2+(aq) + 2e- emf = E = cell voltage: Cathode half-reaction: Cu2+(aq) + 2e- Cu(s) cell Electromotive force is the cell potential Overall cell reaction: Zn(s) + Cu2+ (aq) Zn2+(aq) + Cu(s) measured in volts. This is the driving force that Anode half-cell cathode half-cell pushes electrons away from the anode and Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) towards the cathode.
| = Phase boundary between components in same cell Joules = Coulombs x Volts || = Salt bridge between two half cells E is measured in volts: V = J/C • electrodes always on 2 ends: anode on left; cathode on right cell • Inert electrodes (Pt and graphite) are used for gas phase and coulomb - the quantity of charge that passes a aqueous reactions. They allow e- transfer to occur but don’t react with cell components. point in 1 sec when a current of 1 ampere flows.
3 3 Standard Cell Potential, Eo Galvanic Cells – SHE
Gases at 1 atm, Solutions at 1 M, An inert platinum o Temperature at 298 K (25 C) electrode allows Standard potential for any galvanic cell is electron transfer the sum of the half-cell potentials. to take place but o o o don’t take part in E cell = E ox + E red the reaction. All cell potentials are compared to hydrogen (SHE: standard hydrogen electrode) SHE consists of Pt electrode in contact with 1 M Pt metal + H solution and H2 gas at 1 atm pressure. + - o 2H + 2e H2(g) E red = 0 V
Reduction Potentials Calculating Cell potentials
We can use the table of Standard Reduction Ecell Ered (cathode)Eox(anode) Potentials for reduction half reactions to determine the cell potential of a galvanic cell. . For the oxidation reaction at the anode, Table 17.1: All potentials are listed as make sure you reverse the reaction and
reduction potentials. change the sign for Eox, then add the reduction and oxidation potentials. Oxidation potentials: reverse the reaction . E° is intensive; it does not depend on # of moles involved. Don't multiply E° by a factor and Eox (anode) Ered (cathode) if coefficients of reaction are changed.
Standard Reduction Potentials Strength of Oxidizing Agents Standard Reduction Potentials at 25oC (See Table 17.2 – partial
table). Full list is in Appendix L. - - o
F2(g) + 2e 2F (aq) E red = 2.866 V Half-Reaction Eo (V) o red F2 has the most positive E red value. – – F2 (g) + 2e → 2F (aq) 2.866 – – F is the easiest to reduce. (It wants e- Br2 (l) + 2e → 2Br (aq) 1.0873 2 Ag+ (aq) + e– → Ag(s) 0.7996 the most!) Cu2+ (aq) + 2e– → Cu(s) 0.337 Thus F2 is strongest oxidizing agent. + – 2H (aq) + 2e H2(g) 0.00 Pb2+ (aq) + 2e– → Pb(s) –0.126 Active nonmetals are good oxidizing Cd2+ (aq) + 2e– → Cd(s) –0.4030 agents. (ox agent = reduced = gain e-) Zn2+ (aq) + 2e– → Zn(s) –0.7618 o As E red, strength ox agent
Li+ (aq) + e– → Li(s) –3.04 Increasing strength as reducing agents reducing as strength Increasing Increasing strength as oxidizing agents oxidizing as strength Increasing 23
4 4 Strength of Reducing Agents Cell Potentials
+ - o o Li (aq) + e Li(s) E red = -3.04 V Based on their E red values, determine the best o Li has the most negative E red value, so it oxidizing agent, best reducing agent, worst is the easiest to oxidize. oxidizing agent, and worst reducing agent. o Li(s) is strongest reducing agent. Half-reaction E red 2+ - o Li(s) wants to lose electrons, so reverse Co (aq) + 2e Co(s) -0.28 V (most + E red) 2+ - o reaction occurs: Mn (aq) + 2e Mn(s) -1.18 V (most - E red) + - o 2+ - Li(s) Li (aq) + e E ox = +3.04 V Cd (aq) + 2e Cd(s) -0.40 V Active metals are good reducing agents. (red agent = oxidation = lose e-).
Best/worst agents Cell Potentials For reduction half rxns, oxidizing agents are Positive Eo means the reaction is product- reactants and reducing agents are products. cell favored and spontaneous. 2+ Co is the easiest to reduce = best o A negative E cell means the reaction won’t o oxidizing agent (most + E red value). happen in the forward direction. Mn2+ is the hardest to reduce = worst Therefore, we want two half reactions that o yield the most positive Eo value. oxidizing agent. (most - E red value). cell o Mn(s) is the easiest to oxidize = best Assign rxn with more + E red as cathode! o reducing agent. (most + E ox value). o o o Co(s) is the hardest to oxidize = worst E cell = E red + E ox o reducing agent. (most - E red value).
Galvanic Cell Potentials Example: Fe(s) + Ni(s) 2+ - o If we were to make a Galvanic cell from the Fe (aq) + 2e Fe(s) E red = -0.44 V 2+ - o following metals, which would act as the Ni (aq) + 2e Ni(s) E red = -0.25 V anode and which as the cathode? Refer to the standard reduction potential table. Assign more + one (Ni) as cathode: 2+ - o Write the half-cell reactions and the Cathode: Ni (aq) + 2e Ni(s) E red = -0.25 V overall balanced reaction for each cell. For anode: reverse reaction and change sign o 2+ - o Calculate E cell for the Fe(s) - Ni(s) cell Anode: Fe(s) Fe (aq) + 2e E ox = +0.44 V Overall: Ni2+(aq) + Fe(s) Ni(s) + Fe2+(aq) o E cell = -0.25 + 0.44 = 0.19 V
5 5 Cell Potentials Cell Potentials in Reactions
o 2+ - Use these values for problem 5 on worksheet. 5. b) What will E cell be if Pb reacts with Br ? o o - 3+ - o E red Pt(s) = -0.13 V E ox Br (aq) = -1.07 V Au + 3e Au(s) E red = 1.50 V - - o Sign changed for bromide since it is oxidized! Br2(l) + 2e 2Br (aq) E red = 1.07 V o 2+ - o E cell = -1.07 V + -0.13 V = -1.20 V Pb + 2e Pb(s) E red = -0.13 V o 2+ - o c) Calculate E cell for the following cell Ni + 2e Ni(s) E red = -0.25 V 2+ - Ni(s) | Ni (aq) || Br2(l), Br (aq) | Pt(s) Ni = ox bromine = red o E cell = 0.25 V + 1.07 V = 1.32 V Reaction for part c is spontaneous (+ E)
Free Energy and Emf Calculate DG from E DG = -nFE Calculate DGo for a reaction between Ni(s) with Au3+. 3+ - o n = number of moles of e- transferred Cathode: 2(Au + 3e Au(s)) E red = 1.50 V 2+ - o E = Emf of cell Anode: 3(Ni(s) Ni + 2e ) E ox = -(-0.25 V) 3+ 2+ o F = Faraday's constant 2Au + 3Ni(s) 2Au(s) + 3Ni E cell = 1.75 V 96500 Coulombs 96500 J n = 6 mol e- transferred from anode to cathode 1F 1mol e V mol e 96500J DGo = 6mol e 1.75V V mol e spontaneous reaction: DG < 0 and E > 0 1kJ nonspontaneous reaction: DG > 0 and E < 0 = -1.01 × 106 J = -1.01 × 103 kJ 1000J At equilibrium: DG = 0 and E = 0 *note: the units for n may be omitted, and DGo may be expressed in kJ/mol
Standard emf & K Calculating K from E
At equilibrium, E = 0 and Q = K (plug these Calculate K at 25 C for Ni(s) + Br2 cell (#4c): 2+ - conditions into Nernst Equation) Ni(s) | Ni (aq) || Br2(l), Br (aq) | Pt(s) Anode Ni(s) Ni2+ + 2e- Eo = +0.25 V RT RT ox Cathode Br (l) + 2e- 2Br - Eo = 1.07 V 0 E lnK E lnK 2 red nF nF 2+ - Ni(s) + Br2(l) Ni + 2Br Ecell = 1.32 V, n = 2 J J R 8.314 , F 96500 From rearranging the equations relating E to K K mol V mol e nE 21.32V 0.0592V K 10 K 10 0.0592V 3.931044 At 298 K, 0.0592V E logK 96500J n 2 1.32V nFE Vmol e 8.314J RT 298K or K e K e Kmol 4.541044
6 6 The Nernst Equation: Calculate emf for Cell Potentials Summary nonstandard conditions
o o Recall: DG = DG + RT ln Q Positive E cell value o o Provides energy Plugging in DG = - nFE and DG = - nFE o DGo is negative -nFE = -nFE + RT ln Q K is large RT Nernst Equation: E E lnQ Reaction is product-favored nF o Negative E cell value 2.303RT J J E E logQ R 8.314 , F 96500 Consumes energy nF K mol V mol e DGo is positive 0.0592V K is small E E logQ At 298 K: n Reaction is reactant-favored Both [ ]’s & P’s can be plugged into Q; V = volts (unit)
Pb and Ag cell (worksheet) Nernst Equation: Pb/Ag+ cell #8. A galvanic cell utilizes the reaction: 8c) Calculate the cell potential if [Pb2+] = 0.88 M and [Ag+] = 0.14 M. 2Ag+ + Pb(s) Pb2+ + 2Ag(s) 2Ag+(aq) + Pb(s) 2Ag(s) + Pb2+(aq) o a) Write the half reactions and calculate E cell. o E cell = 0.93 V, n = 2 b) Write the short-hand notation for this cell. Q = (0.88 M) / (0.14 M)2 = 44.9 a) Look up reduction potentials E = Eo – (0.0592 V/n) log Q + E = 0.93 V - (0.0592 V/2)log 44.9 Cat: Ag (aq) + e- Ag(s) E red = 0.80 V +2 E = 0.93 V- (0.0296 V)(1.65) = 0.93 - 0.049 = 0.88 V An: Pb(s) Pb (aq) + 2e- E ox = -(-0.13) V E = 0.88 V (E because Q > 1) E = 0.80 V + 0.13 V = 0.93 V cell Do Worksheet problem #9 b) Pb(s)|Pb+2(aq)||Ag+(aq)|Ag(s)
Electrolysis Electrolysis o Electrolytic Cell: Electrical energy from an external Figure 17.19 Electrolysis of molten sodium chloride; E cell = – 4 V source (outlet or a battery) is used to force a nonspontaneous redox reaction to proceed. The reactions are not spontaneous in Water doesn’t naturally decompose into H2 and O2. A battery the forward direction: must supply energy to drive this reaction forward. Na likes to lose e- 2H2O(l) 2H2(g) + O2(g) Ecell = -1.23 V and Cl2 likes to gain e-. + - Anode: 2H2O(l) → O2(g) + 4H (aq) + 4e
- - Cathode: 4H2O(l) + 4e → 2H2(g) + 4OH (aq) – – The electrodes are the still the same for electrolysis: Anode 2Cl (l) → Cl2(g) + 2e + – (oxidation at anode, reduction at cathode). Cathode 2Na (l) + 2e → 2Na(l) + – 42 Overall: 2Na (l) + 2Cl (l) → 2Na(l) + Cl2(g)
7 7 Electrolysis Calcs Electrolysis Calculations
Used to find mass or volume How many grams of Cu can be collected in 1.00 of product produced by hour by a current of 1.62 A from a CuSO4 solution? passing current through cell. Reduction reaction: Cu2+ + 2 e- Cu(s) Current: measured in Amps Calculate coulombs: Current (C/s) x time (s) = C A = C/s C mol e- mol solid g solid
ampere: unit of electric convert time to secs: 1 hr = 3600 secs current; rate of flow of e-. 1.62C 1mol e 1molCu 63.55g 3600 s = 1.92 g Cu charge = current x time s 96,500C 2mol e mol coulombs = amps x seconds
Batteries Car Batteries
Figure 17.4: Batteries are examples of Galvanic cells we use Lead Storage in everyday life. (Car) Batteries For a Galvanic cell the voltage keeps dropping . 6 cells as the spontaneous reaction proceeds right. . 2 V per cell . 12 V total Battery is dead when Ecell = 0 (at equilibrium . Rechargeable the rxn no longer proceeds forward; this is minimum pt on free energy curve). Bigger batteries only last longer, they don’t
have more volts. Volts are determined by the – + – Anode Pb(s) + 2HSO (aq) → PbSO (s) + H (aq) + 2e chemical reaction that occurs. 4 4 + – Cathode PbO2 (s) + 3H (aq) + HSO4 (aq) + 2e– → PbSO4 (s) + 2H2O(l)
Overall: Pb(s) + PbO2 (s) + 2H2SO4 (aq) → 2PbSO4 (s) + 2H2O(l)
Dry Cell Batteries Alkaline Batteries Figure 17.10: Dry Cell (or Laclanche) Batteries: Zinc Figure 17.11: Alkaline Batteries: Zinc container container (anode), Graphite rod (cathode), 1.5 V (anode), Graphite rod (cathode), 1.43 V D, C, A, AA, and AAA batteries all Alkaline batteries were have the same Voltage rating. designed as direct However, larger batteries can deliver replacements for zinc-carbon more moles of electrons. As the zinc (dry cell) batteries in the container oxidizes, its contents 1950’s. They can deliver eventually leak out, so this type of about 3 – 5x the energy of a battery should not be left in any zinc-carbon dry cell battery of electrical device for extended periods. similar size.
2+ – - – Anode Zn(s) → Zn (aq) + 2e Anode Zn(s) + 2OH (aq) → ZnO(s) + H2O(l) + 2e + – – - Cathode 2 NH4 (aq) + 2MnO2(s) + 2e → Mn2O3(s) + 2NH3(aq) + H2O(l) Cathode 2MnO2(s) +H2O(l) + 2e → Mn2O3(s) + 2OH (aq)
+ 2+ Overall: Zn(s) + 2NH4 (aq) + 2MnO2(s) → Zn (aq)+ Mn2O3(s) + 2NH3(aq) + H2O(l) 47 Overall: Zn(s) + 2MnO2(s) → ZnO(s)+ Mn2O3(s) 48
8 8 9 V Batteries NiCd Batteries
9 Volt Batteries – 6 x 1.5 V Dry Cell batteries connected in series NiCd Batteries: Nickel-plated cathode and cadmium-plated anode.
NiCd batteries use a “jelly-roll” design that significantly increases the amount of current the battery can deliver as compared to a similar- sized alkaline battery.
- – Anode Cd(s) + 2OH (aq) → Cd(OH)2(s) + 2e – - Cathode NiO2(s) + 2H2O(l) + 2e → Ni(OH02(s) + 2OH (aq) 49 Overall: Cd(s) + NiO (s)+ 2H O(l) → Cd(OH) (s)+ Ni(OH) (s) 50 https://upload.wikimedia.org/wikipedia/commons/c/ce/9V_innards_AAAA_removal-1.jpg 2 2 2 2
Lithium ion Batteries Fuel Cells How it works
Lithium ion (Rechargeable) batteries: charge flows between the Fuel Cells; Eo = 1.23 V electrodes as the lithium ions move between the anode and cathode cell . Cell potential: 3.7 V In this hydrogen fuel-cell schematic, oxygen from the air reacts with hydrogen, producing water and electricity.
+ – Li(s) → Li + e 2- – Anode Anode 2H2(g) + 2O (aq) → 2H2O(l) + 4e + – Li (aq) + CoO + e → LiCoO (s) – 2- Cathode 2 2 Cathode O2 (aq) + 4e → 2O (aq) Overall: Li+(s) + CoO → LiCoO (s) 51 52 2 2 Overall: 2H2(g) + O2(g) → 2H2O(l)
17.6 Corrosion The term corrosion generally refers to the deterioration of a metal by an electrochemical process (e.g., rusting of iron). A view from the top of the Statue of Liberty, showing the green patina coating the statue. The patina is formed by corrosion of the copper skin of the statue, which forms a thin layer of an insoluble compound that contains copper(II), sulfate, and hydroxide ions.
(a) The Statue of Liberty is covered with a copper skin, and was originally brown, as shown in this painting. (b) Exposure to the elements has resulted in the formation of the blue-green patina seen today. Image from Principles of General Chemistry http://2012books.lardbucket.org/books/principles-of-general-chemistry53- v1.0/s23-02-standard-potentials.html CC-BY-NC-SA 3.0 license
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