Page III-17-1 / Chapter Seventeen Lecture Notes

ELECTROCHEMISTRY Chapter 17 TRANSFER REACTIONS Atom / Group transfer

-1 + HOAc(aq) + H2O(l) → OAc (aq) + H3O (aq)

Electron transfer

Chemistry 223 Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) Professor Michael Russell

Last update: MAR 8/9/21 MAR

TRANSFER REACTIONS Electron Transfer Reactions

Electron Transfer Reactions: Electron transfer reactions are oxidation- reduction or redox reactions. Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s) Redox reactions can result in the Cu(s) → Cu2+(aq) + 2 e- generation of an electric current or be caused by imposing an electric current. and Therefore, this field of chemistry is often called ELECTROCHEMISTRY. 2 Ag+(aq) + 2 e- → 2 Ag(s)

MAR MAR

OXIDATION - loss of electron(s) by a species; increase in oxidation number. Blast from REDUCTION - gain of electron(s); the Past! decrease in oxidation number. OXIDIZING AGENT - electron acceptor; species is reduced. REDUCING AGENT - electron donor; species is oxidized.

Review of Terminology for Redox Reactions MAR MAR

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LEO says GER LEO Lose Gain says Electrons Electrons Oxidized Reduced GER Zn(s) → Zn2+ + 2e- Oxidized Cu2+ + 2e- → Cu(s) Reduced Can also use "OIL RIG": OIL = "Oxidation is Losing" (electrons) MAR MAR RIG = "Reduction is Gaining" (electrons)

Copper + Silver Ion OXIDATION-REDUCTION REACTIONS

Direct Redox Reaction Oxidizing and reducing agents in direct contact. Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

MAR MAR

OXIDATION-REDUCTION Why Study Electrochemistry? REACTIONS • Batteries • Corrosion Indirect Redox Reaction • Industrial A battery functions by transferring electrons production of through an external wire from the reducing chemicals such as agent to the oxidizing agent. Cl2, NaOH, F2 and Al

• Biological redox reactions A rusted car

MAR MAR The heme group

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Famous Electrochemists Electrochemical Cells An apparatus that allows a Alessandro Volta, redox reaction to occur by 1745-1827, Italian transferring electrons through an external scientist and connector. inventor. Product favored reaction ---> voltaic or galvanic cell ----> electric current created Reactant favored reaction ---> electrolytic cell ---> electric Luigi Galvani, current used to cause Batteries are voltaic 1737-1798, Italian cells chemical change scientist and MAR MAR inventor.

Balancing Equations for Redox Reactions Some redox reactions have equations that must be balanced by special techniques.

- 2+ + Frankenstein MnO4 + 5 Fe + 8 H Mn = +7 Fe = +2 2+ 3+ → Mn + 5 Fe + 4 H2O Mn = +2 Fe = +3

MAR MAR See: Redox Reactions Handout

Balancing Equations Balancing Equations Step 1: Divide the reaction into half-reactions, one for oxidation and the other for reduction. Consider the reduction of Ag+ ions with Ox Cu → Cu2+ copper metal. Red Ag+ → Ag Step 2: Balance each for mass. Already done in this case. Step 3: Balance each half-reaction for charge by adding electrons. Ox Cu → Cu2+ + 2e- Red Ag+ + e- Ag Cu + Ag+ --gives--> Cu2+ + Ag →

MAR MAR

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Balancing Equations + Reduction of VO2 with Zn Step 4: Multiply each half-reaction by a factor that means the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu → Cu2+ + 2e- Oxidizing agent 2 Ag+ + 2 e- → 2 Ag Step 5: Add half-reactions to give the overall equation. Cu + 2 Ag+ → Cu2+ + 2 Ag

The equation is now balanced for both charge and mass. MAR MAR

Balancing Equations Balancing Equations Balance the following in acidic solution- Balance the following in acidic solution- + 2+ 2+ + 2+ 2+ VO2 + Zn → VO + Zn VO2 + Zn → VO + Zn Step 1: Write the half-reactions Step 1: Write the half-reactions Ox Zn → Zn2+ Ox Zn → Zn2+ + 2+ 5+ 4+ + 2+ 5+ 4+ Red VO2 → VO (V → V ) Red VO2 → VO (V --> V ) Step 2: Balance each half-reaction for mass. Step 2: Balance each half-reaction for mass. Ox Zn → Zn2+ Ox Zn → Zn2+ + 2+ + + 2+ Red VO2 → VO + H2O Red 2 H + VO2 → VO + H2O

+ Add H2O on O-deficient side and add H on MAR MAR other side for balancing hydrogen

Balancing Equations Balancing Equations Step 3: Balance half-reactions for charge. Balance the following in basic solution- 2+ 1- - 2- Ox Zn → Zn + 2e- MnO4 + HO2 → MnO4 + O2 + + 2+ Red e- + 2 H + VO2 → VO + H2O Step 1: Write the half-reactions - -1 0 Step 4: Multiply by an appropriate factor. Ox HO2 → O2 (peroxide: O → O ) 2+ 1- 2- 7+ 6+ Ox Zn → Zn + 2e- Red MnO4 → MnO4 (Mn → Mn ) + + Red 2 e- + 4 H + 2 VO2 Step 2: Balance each half-reaction for mass. 2 VO2+ + 2 H O 1- 2- → 2 Red MnO4 → MnO4 Step 5: Add half-reactions - Ox HO2 → O2 + H2O + + 2+ 2+ Zn + 4 H + 2 VO2 → Zn + 2 VO + 2 H2O MnO4-1 = permanganate MnO42- = manganate MAR MAR

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Balancing Equations Balancing Equations Balance the following in basic solution- Step 3: Balance half-reactions for charge. 1- - 2- - - MnO4 + HO2 → MnO4 + O2 Ox OH + HO2 → O2 + H2O + 2e- 1- 2- Step 1: Write the half-reactions Red e- + MnO4 → MnO4 - -1 0 Ox HO2 → O2 (peroxide: O → O ) Step 4: Multiply by an appropriate factor. 1- 2- 7+ 6+ - - Red MnO4 → MnO4 (Mn → Mn ) Ox OH + HO2 → O2 + H2O + 2e- 1- 2- Step 2: Balance each half-reaction for mass. Red 2 e- + 2 MnO4 → 2 MnO4 1- 2- Red MnO4 → MnO4 Step 5: Add half-reactions - - - - 1- Ox OH + HO2 → O2 + H2O OH + HO2 + 2 MnO4 2- → O2 + H2O + 2 MnO4 Add H O on H-deficient side and add OH- on - 2 Add H2O on H-deficient side and add OH on other MAR other side for balancing oxygen MAR side for balancing oxygen in basic solution

Alternate Basic Balancing Method Alternate Basic Balancing Method Balance basic reactions first with acid, then Step 3: Balance half-reactions for charge. "neutralize" with OH-. Previous example: - + Ox HO2 → O2 + H + 2e- 1- - 2- MnO4 + HO2 → MnO4 + O2 1- 2- Red e- + MnO4 → MnO4 Step 1: Write the half-reactions Step 4: Multiply by an appropriate factor. - Ox HO2 → O2 - + Ox HO2 → O2 + H + 2e- 1- 2- Red MnO4 → MnO4 1- 2- Red 2 e- + 2 MnO4 → 2 MnO4 Step 2: Balance each half-reaction for mass - + Step 5: Add half-reactions use H and/or H2O. - 1- + 2- 1- 2- HO2 + 2 MnO4 → O2 + H + 2 MnO4 Red MnO4 → MnO4 - + Ox HO2 → O2 + H This equation is balanced for pH < 7 but not base Add H+ to H-deficient side and H O to balance oxygen MAR 2 MAR

Tips on Balancing Equations Alternate Basic Balancing Method • Determine the pH of the Step 6: Neutralize H+ by adding OH-. reaction. H+ and OH- make H O. 2- 2 • Never add O2, O atoms, or O Add OH- to both sides of equation to balance oxygen.

• Never add H2 or H atoms to - 1- - HO2 + 2 MnO4 + OH → balance hydrogen. + - 2- O2 + H + OH + 2 MnO4 • Be sure to write the correct charges on all the ions. - 1- - HO2 + 2 MnO4 + OH • Check your work at the end 2- → O2 + H2O + 2 MnO4 to make sure mass and charge are balanced. Use either method to balance basic redox reactions • See: Redox Reactions Also see the Redox Reactions Handout Handout MAR MAR

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CHEMICAL CHANGE --> CHEMICAL CHANGE --> ELECTRIC CURRENT ELECTRIC CURRENT Zn metal

With time, Cu Electrons are plates out onto transferred from Zn to Zn metal strip, 2+ 2+ Cu , but there is no and Zn strip Cu ions useful electric current. "disappears." Oxidation: Zn(s) → Zn2+(aq) + 2e- Zn is oxidized and is the reducing agent Reduction: Cu2+(aq) + 2e- → Cu(s) 2+ Zn(s) → Zn (aq) + 2e------2+ Cu is reduced and is the oxidizing agent Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) Cu2+(aq) + 2e- Cu(s) MAR → MAR

CHEMICAL CHANGE --> Cu2+ + 2e- → Cu Zn → Zn2+ + 2e- ELECTRIC CURRENT To obtain a useful current, we separate the oxidizing Reduction Oxidation and reducing agents so that Cathode Anode electron transfer occurs Positive Negative through an external wire. Anions--> This is accomplished in a <--Cations GALVANIC or VOLTAIC cell. Electrons travel through external wire. Salt bridge allows anions and cations to move A group of such cells is between electrode compartments, maintaining called a battery. electrical neutrality.

MAR MAR Salt Bridge

Red Cat = REDuction A Red Cat and An Ox SHORTHAND NOTATION for at the CAThode GALVANIC CELLS Cu2+ + 2e- → Cu Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) can also be written as:

anode half cell cathode half cell 2+ 2+ Zn(s) | Zn (aq) || Cu (aq) | Cu(s) An Ox = OXidation at the ANode phase boundary phase boundary Zn → Zn2+ + 2e- salt bridge Electrons flow this way also remember: oxidation = reducing agent oxidation reduction MAR reduction = oxidizing agent MAR FAT CAT = electrons flow From Anode To CAThode

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SHORTHAND NOTATION for GALVANIC CELLS Electrons move from anode to Example: Describe the following galvanic cell: cathode in the wire. Anions (mostly) Cu(s) | Cu2+(1.0 M) || Cl (1.0 atm) | Cl-(1.0 M) | Pt(s) 2 move through the salt bridge. Solution: 2+ - The anode (oxidation) is: Cu(s) → Cu (aq) + 2 e and [Cu2+] = 1.0 M - - The cathode (reduction) is: Cl2(g) + 2 e → 2 Cl (aq) - and [Cl ] = 1.0 M and P(Cl2) = 1.0 atm

The cathode uses a Platinum electrode to transfer electrons to the Cl2(g). The Pt does not react chemically with the electrons Cu | Cu2+ || Ag+ | Ag MAR MAR

Sign of Battery Terminals (Galvanic Cells Only) Electromotive Force (emf)

wire Anode, Water only electrons + Cathode, site of site of spontaneously oxidation, Zn salt Cu bridge reduction, flows one way in a negative positive waterfall. Likewise, electrons Zn2+ ions Cu2+ ions only spontaneously flow one way in a redox reaction— Electrons flow away from the "negative" terminal from higher to (anode) and to the "positive" terminal (cathode) in lower potential Galvanic cells energy. Electrolytic cells use opposite signs MAR MAR

1.10 V wire CELL POTENTIAL, E electrons salt Zn bridge Cu 1.10wire V CELL POTENTIAL, electrons E salt Cu Zn2+ ions Cu2+ ions Zn bridge 1.0 M 1.0 M For Zn/Cu cell, voltage is 1.10 V at 25 ˚C and 2+ 2+ Zn2+ ions Cu2+ ions when [Zn ] and [Cu ] = 1.0 M. 1.0 M 1.0 M This is the STANDARD CELL POTENTIAL, E˚ Electrons are "driven" from anode to cathode by an electromotive force or emf. E˚ (measured in Volts, V) is a quantitative For Zn/Cu cell, this is indicated by a voltage of measure of the tendency for reactants to 1.10 V at 25 ˚C and when [Zn2+] and [Cu2+] = 1.0 proceed to products when all are in their M. standard states at 1.0 M and 25 ˚C. MAR Note that 1 V = 1 J/C, more on this later MAR

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Calculating Cell Voltage CELL POTENTIALS, Eo Balanced half-reactions can be added Cannot measure 1/2 reaction Eo directly. together to get overall, balanced Therefore, measure it relative to a equation. STANDARD HYDROGEN CELL, SHE. Zn(s) → Zn2+(aq) + 2e- (SHE = Standard Hydrogen Electrode) Cu2+(aq) + 2e- → Cu(s) ------+ Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) 2 H (aq, 1 M) + 2e- ⇄ H2(g, 1 atm) If we know Eo for each half-reaction, we Eo = 0.00 V can calculate Eo for net reaction.

MAR MAR

Volts + Zn/SHE half cell - Zn/SHE half cell Zn Salt Bridge H2

Negative Zn2+ H+ Positive Volts 2+ + electrode Zn Zn + 2e- 2 H + 2e- H2 electrode OXIDATION + REDUCTION - ANODE CATHODE Zn Salt Bridge o Zn and SHE How to find the zinc half cell potential, EZn : H2 2+ o Supplier of Acceptor of Zn(s) → Zn (aq) + 2e- EZn = ? + o electrons electrons 2 H (aq) + 2e- → H2(g) EH+ = 0.00 V Zn2+ H+ ------Zn → Zn2+ + 2e- 2 H+ + 2e-→ H Zn(s) + 2 H+ (aq) Zn2+ + H (g) E o = +0.76 V Zn Zn2+ + 2e- 2 H+ + 2e- H2 → 2 net Oxidation Reduction 2 o o o OXIDATION REDUCTION so Ezn = Enet - EH+ = 0.76 - 0.00 = +0.76 V Anode Cathode ANODE CATHODE Therefore, Eo for Zn → Zn2+ (aq) + 2e- is +0.76 V Eo for the cell = +0.76 V (measured) MAR MAR Zn is a better reducing agent than H2.

2+ + 2+ + Cu/Cu and H2/H Cell Cu/Cu and H2/H Cell

Volts - o + E = +0.34 V Cu Salt Bridge H Volts 2

Positive - Negative 2+ H+ + Cu Cu Salt Bridge 2+ + Cu + 2e- Cu H2 2 H + 2e- H REDUCTION OXIDATION 2 CATHODE Acceptor Supplier of ANODE 2+ of electrons Overall reaction is reduction of Cu by H2 gas. + electrons Cu2+ H 2+ + Cu (aq) + H2(g) → Cu(s) + 2 H (aq) 2+ + Cu + 2e- → Cu H2 → 2 H + 2e- Measured Eo = +0.34 V Cu2+ + 2e- Cu + Reduction H2 Oxidation 2 H + 2e- REDUCTION Therefore, Eo for Cu2+ + 2e- → Cu is Cathode OXIDATIONAnode CATHODE ANODE +0.34 V MAR MAR

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Zn/Cu Electrochemical Cell Uses of E° Values wire electrons + Half-reactions salt organized by Zn bridge Cu wire Anode, Cathode, relative ability to act electrons negative, as oxidizing agents salt positive, Zn bridge Cu source of sink for Zn2+ ions Cu2+ ions Use tables of electrons electrons reduction potentials in your textbook or Zn2+ ions Cu2+ ions Zn(s) → Zn2+(aq) + 2e- Eo = +0.76 V problem set to Cu2+(aq) + 2e- → Cu(s) Eo = +0.34 V predict the direction of redox reactions. ------Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) o MAR E (calc'd) = +1.10 V MAR

Reversing Half Reactions reducing ability Table of Standard Reversing half reactions changes the sign but not the magnitude of E˚ values increases Reduction Potentials or Example: oxidizing Zn(s) Zn2+(aq) + 2e- Eo = +0.76 V → agent would become capacity Zn2+(aq) + 2e- → Zn(s) Eo = -0.76 V increases oxidizing Because E˚ tables listed as reductions, many ability negative E˚ values will appear increases Negative E˚ values imply great oxidizers / or reducing reducing agents agent Positive E˚ values imply great reducers / capacity oxidizing agents increases MAR MAR

Standard Redox Potentials, E° Calculating Eo (Michael's Method) Cu2+ + 2 e- Cu +0.34 Flip one half-reaction from table (and E˚ value), then add oxidation and reduction values + 2 H + 2 e- H2 0.00 Example: Determine E˚ for the following: 2 Al + 3 Ni2+ → 2 Al3+ + 3 Ni Zn2+ + 2e- Zn -0.76 Solution: From redox tables we find: Any substance on the right will reduce any Al3+ + 3 e- → Al E˚ = -1.66 V substance higher than it on the left. Ni2+ + 2 e- → Ni E˚ = -0.25 V 2+ 3+ Northwest-southeast rule: product-favored Ni ok as written; need to flip Al reaction to: reactions occur between reducing agent at Al → Al3+ + 3 e- E˚ = +1.66 V southeast corner (anode) and oxidizing agent at So: E˚ = 1.66 - 0.25 = 1.41 V northwest corner (cathode).

MAR MAR

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o Calculating Eo E for a Voltaic Cell

+ Calculate E˚net for reaction between Cu and Ag ions Volts

Cd Salt Bridge Fe

Cd2+ Fe2+ Ag+ + e- → Ag E˚ = 0.80 V Cu2+ + 2 e- → Cu E˚ = 0.337 V Cd → Cd2+ + 2e- and Fe2+ + 2e- → Fe Ag+ ok as written; need to flip Cu2+ reaction to: or or Cd2+ + 2e- → Cd and Fe → Fe2+ + 2e- Cu → Cu2+ + 2 e- E˚ = -0.337 V

o All ingredients are present. E net for Cu/Ag+ reaction = 0.80 - 0.337 = +0.46 V Which way does the reaction proceed? MAR balanced reaction: Cu + 2 Ag+ → 2 Ag + Cu2+ MAR

Eo for a Voltaic Cell More About Calculating Cell Voltage From tables, you see: Volts Can I- ion reduce water? From tables: Fe2+ + 2 e- → Fe E˚ = -.40 - Cd Salt Bridge Fe Cd2+ + 2 e- → Cd E˚ = -.44 2 H2O + 2e- → H2 + 2 OH E˚ = -.828 V - 2 I → I2 + 2e- E˚ = -.535 V - - 2 I + 2 H2O I2 + 2 OH + H2 Cd2+ Fe2+ → Assuming reaction occurs as written, Overall reaction: Reaction occurs E˚net = E˚red + E˚ox = -0.828 V - 0.535 V = -1.363 V Fe2+ + Cd → Cd2+ + Fe spontaneously o Minus E˚ means rxn occurs in opposite direction! E = E˚cathode - E˚anode when E° values - - = (-0.40 V) - (-0.44 V) I2 + 2OH + H2 → 2I + 2H2O E˚ = +1.363 V are positive = +0.04 V - MAR MAR I ion does not reduce water spontaneously!

More About E° and ∆G° Calculating Cell Voltage Can we make the I- ion reduce water? - - 2 I + 2 H2O → I2 + 2 OH + H2 E° = -1.363 V E° is related to ∆G°, the free energy Non-spontaneous (negative E°) reactions can be change for the reaction. "forced" to occur with external voltage ∆G° = - n F E° Voltage can be applied through battery, other where F = Faraday constant = 9.6485 x 104 C/mol e- voltaic cells, etc. and n is the number of moles of Applying 1.363 V to the above electrolytic cell will electrons transferred cause the I- reduce water - electrolysis Are we cheating the second law of thermodynamics? Memorize the value of F! Michael Faraday MAR MAR 1791-1867

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E° and ∆G° Quantitative Aspects of Electrochemistry ° ° ∆G = - n F E Consider electrolysis of aqueous silver ion. For a product-favored reaction Ag+ (aq) + e- → Ag(s) Reactants → Products 1 mol e- → 1 mol Ag ∆G° < 0 and so E° > 0 If we could measure the moles of e-, we could E° is positive know the quantity of Ag formed. For a reactant-favored reaction But how to measure moles of e-? Reactants ⃪ Products charge passing coulombs Current = I (amps) = ∆G° > 0 and so E° < 0 time seconds E° is negative MAR X MAR

Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry coulombs coulombs Amps = Amps = seconds seconds A current of 1.50 amps (1.50 C/s) flows through A current of 1.50 amps (1.50 C/s) flows through a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited? Solution Solution (a) Charge = 1350 C (a) Calculate charge in Coulombs (C) (b) Calculate moles of e- used (F = 96,485 C/mol e-) 1.50 amps = 1.50 C/s 1 mol e- = (1.50 C/s)(60 s/min)(15.0 min) = 1350 C 1350 C • = 0.0140 mol e- 96,485 C (c) Calc. quantity of Ag 1 mol Ag 0.0140 mol e- • = 0.0140 mol Ag or 1.51 g Ag 1 mol e- MAR MAR

Quantitative Aspects of Electrochemistry Quantitative Aspects of Electrochemistry The anode reaction in a lead storage battery is The anode reaction in a lead storage battery is - + - + Pb(s) + HSO4 (aq) → PbSO4(s) + H (aq) + 2e- Pb(s) + HSO4 (aq) → PbSO4(s) + H (aq) + 2e- If a battery delivers 1.50 amp, and you have 454 g of Pb, how long If a battery delivers 1.50 amp, and you have 454 g of Pb, will the battery last? how long will the battery last? Solution Solution a) 454 g Pb = 2.19 mol Pb a) 454 g Pb = 2.19 mol Pb b) Mol of e- = 4.38 mol b) Calculate moles of e- c) Charge = 423,000 C 2 mol e- 2.19 mol Pb • = 4.38 mol e- Charge (C) d) Calculate time Time (s) = 1 mol Pb I (amps) c) Calculate charge 4.38 mol e- • 96,485 C/mol e- = 423,000 C 423,000 C Time (s) = = 282,000 s About 78.3 hours 1.50 amp MAR MAR

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The Nernst Equation The Nernst Equation

Q, the reaction quotient, can be related to non Cu2+(aq) + Zn(s) → Zn2+(aq) + Cu(s) standard cell potentials, E For Zn/Cu cell, voltage is 1.10 V at 25 ˚C when [Zn2+] and [Cu2+] = 1.0 M E is related to Q, the reaction quotient, by: E = E° - (RT/nF) ln Q where R = Gas constant (8.3145 J/K mol) T = Temperature (K) F = Faraday constant E° = standard cell potential Adding Cu2+ shifts reaction right (Le and n = the number of moles of (aq) electrons transferred Chatelier's Principle), making reaction more MAR This is the Nernst Equation (Handout) MAR spontaneous (∆G) and E° more positive

The Nernst Equation The Nernst Equation

2+ 2+ Example: Find E when [Zn ] = 0.0010 M, P(H2) = 0.10 Example: Find E when [Zn ] = 0.0010 M, P(H2) = atm and pH = 0 at 290. K. [H+] = 1.0 M 0.10 atm and pH = 0 at 290. K. + 2+ 2 H (aq) + Zn(s) → Zn (aq) + H2(g) E°net = 0.76 V Solution: Find E° for reaction under standard conditions first Solution Use Nernst Equation: Zn(s) → Zn2+(aq) + 2e- E° = 0.76 V E = E° - (RT/nF) ln Q + E = 0.76 V - (8.3145 * 290. K / 2 * 96485) ln Q 2 H (aq) + 2e- → H2(g) E° = 0.00 V ------[H+] = 10-pH M = 10-0 M = 1.0 M + 2+ Q = [Zn2+] * P / [H+]2 = [0.0010 * 0.10 / (1.0)2] 2 H (aq) + Zn(s) → Zn (aq) + H2(g) H2 E°net = 0.76 V E = 0.76 V + 0.12 V = 0.88 V Note that n = 2 MAR MAR

E° and K Electrochemical Processes in Batteries A battery consists of self-contained At Equilibrium (K), combine ∆G expressions voltaic cells arranged in series, so their for E° and K to get: individual voltages are added. ! RT E = ln K A primary battery cannot be recharged. The battery is “dead” when the cell nF reaction has reached equilibrium. If at 298 K, can use: A secondary battery is rechargeable. ! 0.0257 Once it has run down, electrical energy E = ln K is supplied to reverse the cell reaction n and form more reactant. Only valid at 298 K! Find equilibrium constants from E˚ data! MAR MAR

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Dry Cell Battery Alkaline Battery A primary battery - Nearly same reactions as uses reactions that in common dry cell, but cannot be recharged under basic conditions.

- Anode (-): Zn + 2 OH ---> ZnO + H2O + 2e- - Anode (-) Cathode (+): 2 MnO2 + H2O + 2e- ---> Mn2O3 + 2 OH Zn → Zn2+ + 2e- Zn(s) + MnO2(s) + H2O(l) → ZnO(s) + Mn(OH)2(s) Ecell = 1.5 V

Cathode (+) + 2 NH4 + 2e- → 2 NH3 + H2 MAR MAR

Silver Button Battery Mercury Button Battery Anode: Zn is reducing agent under basic conditions Cathode: - HgO + H2O + 2e- → Hg + 2 OH

- - Anode (oxidation): Zn(s) + 2OH (aq) → ZnO(s) + H2O(l) + 2e - - Cathode (reduction): Ag2O(s) + H2O(l) + 2e → 2Ag(s) + 2OH (aq)

Overall (cell) reaction: Zn(s) + Ag2O(s) → ZnO(s) + 2Ag(s)

Ecell = 1.6 V

MAR MAR

Lead-Acid Storage Battery Nickel-Cadmium Battery

- + - Anode (0.36 V): Pb(s) + HSO4 (aq) → PbSO4(s) + H (aq) + 2e Cathode (1.68 V): + - - Anode (-): PbO2(s) + 3H (aq) + HSO4 (aq) + 2e → PbSO4(s) + 2H2O(l) Overall (cell) reaction (discharge): - Cd + 2 OH → Cd(OH)2 + 2e- PbO2(s) + Pb(s) + H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Ecell = 2.04 V Overall (cell) reaction (recharge): Cathode (+): 2PbSO4(s) + 2H2O(l) → PbO2(s) + Pb(s) + H2SO4(aq) NiO(OH) + H2O + e- → - Ni(OH)2 + OH Overall cell: A secondary 2 NiO(OH) + 2 H2O + Cd → battery - can be 2 Ni(OH) + Cd(OH) recharged, 2 2 reversible redox Ecell = 1.4 V reactions

MAR MAR

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Lithium Battery Lithium-Ion Battery

The primary lithium battery is widely used in watches, implanted medical devices, and remote-control devices. Anode (oxidation): + - LixC6(s) → xLi + xe + C6(s) Cathode (reduction): Anode (oxidation): + - Li1-xMn2O4(s) + xLi + xe → LiMn2O4(s) + - 3.5Li(s) → 3.5Li + 3.5e Overall (cell) reaction: Cathode (reduction): LixC6(s) + Li1-xMn2O4(s) → LiMn2O4(s) - - AgV2O5.5 + 3.5Li + 3.5e → Li3.5V2O5.5 Ecell = 3.7 V Overall (cell) reaction:

AgV2O5.5 + 3.5Li(s) → Li3.5V2O5.5

The secondary (rechargeable) lithium-ion battery is used to power laptop computers, cell phones, and camcorders. MAR MAR

Hydrogen Fuel Cell Electrolysis

In a fuel cell reactants enter the cell Using external electrical energy to produce chemical and products leave, generating change in a nonspontaneous reaction 2+ - electricity through controlled Sn (aq) + 2 Cl (aq) → Sn(s) + Cl2(g) combustion. Reaction rates are lower in fuel cells than in other batteries, so an SnCl2(aq) electrocatalyst is used to decrease the activation energy.

+ - Cl Sn Anode (oxidation): 2H2(g) → 4H (aq) + 4e 2 + - Cathode (reduction): O2(g) + 4H (aq) + 4e → 2H2O(g)

Overall (cell) reaction: 2H2(g) + O2(g) → 2H2O(g) Ecell = 1.2 V MAR MAR

A tin-copper reaction as a voltaic and electrolytic cell Electrolysis of Aqueous NaOH Electric Energy ----> Chemical Change

Anode (+) Eo = -0.40 V - 4 OH → O2(g) + 2 H2O + 2e-

o Sn(s) → Sn2+(aq) + 2e- Cu(s) → Cu2+(aq) + 2e- Cathode (-) E = -0.83 V Cu2+(aq) + 2e- → Cu(s) Sn2+(aq) + 2e- → Sn(s) - 4 H2O + 4e- → 2 H2 + 4 OH Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq) Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq) Eo for cell = -1.23 V voltaic cell, Ecell = 0.48 V electrolytic cell, Ecell = -0.48 V

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Page III-17-14 / Chapter Seventeen Lecture Notes Page III-17-15 / Chapter Seventeen Lecture Notes

Electrolysis Electrolysis of Molten NaCl Electric Energy ---> Chemical Change Anode (+) Eo = -1.36 V electrons - 2 Cl → Cl2(g) + 2e- Electrolysis of BATTERY molten NaCl. electrons Here a battery + BATTERY Cathode (-) Eo = -2.71 V "pumps" electrons + Anode Cathode - + + from Cl to Na . Anode Cathode Na + e- → Na Polarity of Cl- Na+ Cl- Na+ electrodes is reversed from Eo for cell = -4.07 V batteries. External energy needed because Eo is (-). MAR MAR

Electrolysis of Aqueous NaCl Electrolysis of Aqueous NaCl o Anode (+) E = -1.36 V Cells like these are the source of NaOH and Cl2. - 2 Cl → Cl2(g) + 2e- In 2006 Cathode (-) Eo = -0.83 V 65 million tons Cl2 - 81 million tons NaOH 2 H2O + 2e- → H2 + 2 OH Eo for cell = -2.19 V electrons

BATTERY Note that H2O is more easily reduced than Na+. + Also the source Anode Cathode of NaOCl for use Also, Cl- is oxidized in in bleach. preference to H2O because - + of kinetics. Cl Na H2O MAR MAR

Producing Aluminum End of

2 Al2O3 + 3 C → 4 Al + 3 CO2 Chapter 17

See: • Chapter Seventeen Study Guide • Chapter Seventeen Concept Guide • Types of Equilibrium Constants • Important Equations (following this Charles Hall (1863-1914) developed electrolysis slide) process, founded Alcoa (alcoa.com) • End of Chapter Problems MAR MAR (following this slide)

Page III-17-15 / Chapter Seventeen Lecture Notes Page III-17-16 / Chapter Seventeen Lecture Notes

End of Chapter Problems: Test Yourself Important Equations, Constants, and Handouts from this Chapter:

Redox Reactions: oxidation, You will need the table of reduction potentials found in problem set #5 ∆G° = - n F E° reduction, LEO, GER, oxidizing coulombs -1 2+ agent, reducing agent, anode, Amps = 1. Write a balanced equation for the following half-reaction: VO3 (aq) → V seconds (aq) (in acid) cathode, galvanic/voltaic cells, E = E° - (RT/nF) ln Q 2. Write a balanced equation for the following half-reaction: Ag(s) → electrolysis (electrolytic cells), Ag2O(s) (in base) shorthand notation for galvanic 3. Balance the following redox equation in acid: Zn(s) + NO3-1(aq) → Zn2+ R = 8.3145 J mol-1 K-1 cells, SHE electrode (aq) + N2O(g) F = 9.6485 x 104 C/mol e- 4. Balance the following redox equation in base: CrO42-(aq) + SO32-(aq) → Cr(OH)3(s) + SO42-(aq) 5. Balance the following unbalanced equation in acid, then calculate the ! RT -1 2+ • know how to balance E = ln K standard redox potential, E°: Cu(s) + NO3 (aq) → Cu (aq) + NO(g) nF 6. Calculate E°, ∆G° and the equilibrium constant for the following reaction: redox reactions in acid or 2 Fe3+(aq) + 2 I-1(aq) → 2 Fe2+(aq) + I2(aq) base conditions • be able to calculate Eo andHandouts: E for cells • Thermodynamic Values and Electrochemical Cell Values (Problem Set #5) MAR MAR

End of Chapter Problems: Answers

1. VO3–(aq) + 6 H+(aq) + 3 e– → V2+(aq) + 3 H2O(l) 2. 2 Ag(s) + 2 OH–(aq) → Ag2O(s) + H2O(l) + 2 e– 3. 4 Zn(s) + 2 NO3–(aq) + 10 H+(aq) → 5 H2O(l) + 4 Zn2+(aq) + N2O(g) 4. 2 CrO42-(aq) + 5 H2O(l) + 3 SO32-(aq) → 2 Cr(OH)3(s) + 4 OH-(aq) + 3 SO42-(aq) 5. 3 Cu(s) + 2 NO3-1(aq) + 8 H+(aq) → 2 NO(g) + 3 Cu2+(aq) + 4 H2O(l), E° = 0.62 V 6. E° = 0.236 V, ∆G° = -45.5 kJ, K = 9 x 107

You will need the table of reduction potentials found in problem set #5

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Page III-17-16 / Chapter Seventeen Lecture Notes