Chem 1 B 118 Lecture Announcements Todays Lecture

Chem. 1 B – 11/8 Lecture

Announcements • Today’s Lecture – Electrochemistry • The Nernst Equation (for non-standard conditions, including determining Ksp and Kf values) • Batteries – applied voltaic cells • Electrolytic Cells

Chapter 18 Electrochemistry • Nernst Equation Application – Cont. • 2 nd Examples: – The following cell, Cd(s)|Cd. C 2 O 4(s)|C 2 O 42 -(aq)||Cu 2+(aq)1 M|Cu(s) is used to determine [C 2 O 42 -]. If Eº for the Cd reaction is -0. 522 V (reduction potential for oxidation reaction) and Eº for the Cu reaction is +0. 337 V, and the measured voltage is 0. 647 V, what is [C 2 O 42 -]?

Chapter 18 Electrochemistry • Nernst Equation Application – Cont. • 3 rd Type of Example: – While metal electrodes are commonly used, it is also possible to combine metal reactions with other aqueous phase metal ion reactions. For example, in the lab, we use electrochemical measurements to determine equilibrium constants (1/Kf for Cu(NH 3)42+ and Ksp for Ag. Cl)

Chapter 18 Electrochemistry • Nernst Equation Application – Cont. • 3 rd Type – Specific Example: – Using information given from the following cell and standard potentials, determine the Ksp for Hg 2 Cl 2(s) Hg(l)|Hg 2 Cl 2(s)|Hg 22+(X M), 0. 10 M Na. Cl||1. 0 M Cu 2+|Cu(s) Above cell has E = 0. 01 V , E°(Hg 22+ reduction) = 0. 80 V, and E°(Cu 2+ reduction) = 0. 34 V Calculate Ksp and E° for Hg 2 Cl 2(s) reduction.

Chapter 18 Electrochemistry • Voltaic Cells – Batteries • Requirements – General interest is in high energy density (large E° per unit mass or volume) – To get high E values, more highly reactive reagents are desired (provided package is stable) – Rechargeable batteries • When switching from voltaic cell (using battery) to electrolytic cell (charging battery), reverse reaction must occur (vs. producing other products)

Chapter 18 Electrochemistry • Voltaic Cells – Batteries • Examples: – Lead-Acid (standard Car battery) • Pb used in both oxidation and reduction to Pb 2+ (most stable form) • Oxidation: Pb(s) + HSO 4 -(aq) ↔ Pb. SO 4(s) + H+(aq) + 2 e • Reduction: Pb. O 2(s) + HSO 4 -(aq) + 3 H+(aq) + 2 e↔ Pb. SO 4(s) + H 2 O(l)

Chapter 18 Electrochemistry Batteries • Examples: – Lead-Acid (standard Car battery) • Net: Pb(s) + Pb. O 2(s) + 2 HSO 4 -(aq) + 2 H+(aq) ↔ Pb. SO 4(s) + 2 H 2 O(l) • So E = Eº – 0. 0592/2 log[1/([HSO 4 -]2[H+]2)] • As reaction proceed, voltage slowly drops (faster when nearly depleted) – due to loss of HSO 4 -, H+ • Rechargeable • Low Energy Density (Pb is heavy) • 12 V from 6 cells

Chapter 18 Electrochemistry Batteries • Examples: – Alkaline batteries • Net: Zn(s) + 2 Mn. O 2(s) + 2 H 2 O(l) ↔ Zn(OH)2(s) + 2 Mn. O(OH)(s) • Since all reactants and products are solids, E = Eº = constant • Better if V must stay constant, but harder to tell if battery is near end of lifetime • Non-rechargeable

Chapter 18 Electrochemistry Batteries - Questions 1. If 100 g. of Pb(s) is used in a lead acid battery, what mass of Pb. O 2(s) is required for best efficiency? 2. How many Amp–Hours will this provide in a 12 V battery? 3. Will the voltage generated by a lead acid battery be affected by how “depleted” it is? 4. What is the voltage when it is 90% depleted (vs. initial voltage)?

Chapter 18 Electrochemistry Other Voltaic Cells • Examples: – Fuel Cells • Voltaic cell where reactants (fuel plus oxygen) flow to electrodes to produce electricity • New Toyota Fuel Cell vehicle available (reported last year in Sacramento Bee) • Reactions: 2 H 2(g) + 4 OH-(aq) ↔ 4 H 2 O(l) + 4 e. And O 2(g) + 2 H 2 O(l) + 4 e- ↔ 4 OH-(aq) • If H 2 is produced from electrolysis using solar energy, 100% renewable • More commonly, H 2 is made from natural gas

Chapter 18 Electrochemistry Other Voltaic Cells • Examples: – Powering Medical Devices • Batteries have a limited lifetime, so pacemakers and defibrillators must be surgically removed to replace batteries • Another option is to run devices off of blood glucose oxidation (C 6 H 12 O 6(aq) + O 2(aq) ↔ C 6 H 10 O 6(aq) + H 2 O 2(aq) - requires enzyme) • Either attachment of enzyme to electrode or electrodes for H 2 O 2 oxidation or reduction can be used to generate electricity

Chapter 18 Electrochemistry Electrolytic Cells • Example Reactions 1. Electrolysis of water (opposite of fuel cell example) • Anode: H 2 O – oxygen is oxidized to O 2(g) • Cathode: H 2 O – hydrogen is reduced to H 2(g) 2. Industrial Use – Electroplating (Chrome, nickel, silver plating possible) – using external potential to deposit metal to electrode

Chapter 18 Electrochemistry Electrolytic Cells • Example Reactions 3. Electrolysis of Mixtures – e. g. analysis • External potential will work on easiest to oxidize/reduce pair • For example, if we have a mixture of Na. I and Na. Cl in water, electrolysis will cause the following reactions: – – – Na+(aq) + e- ↔ Na(s) Eº = -2. 71 V H 2 O(l) + 2 e- ↔ H 2(g) + 2 OH-(aq) Eº = -0. 83 V 2 Cl-(aq) ↔ Cl 2(g) + 2 e- Eº = +1. 36 V 2 I-(aq) ↔ I 2(aq) + 2 e- Eº = +0. 54 V 2 H 2 O(l) ↔ O 2(g) + 4 H+(aq) + 4 e- Eº = 1. 23 V

Chapter 18 Electrochemistry Electrolytic Cells - Questions 1. Which of the following changes in switching from a voltaic to an electrolytic cell? a) Charge on anode/cathode b) Which electrode (e. g. anode) does oxidation/reduction c) Ion migration to electrode 2. An anode in an electrolytic cell is used to measure oxalate (Eº = -0. 49 V) in the presence of pyruvate (Eº = -0. 70 V). Which will oxidize first in a mixture?

Chapter 18 Electrochemistry Electrolytic Cells – Questions – Cont. 3. When a battery is being recharged, which of the following happens? a) Electrode charge changes b) Electrode reaction changes (anode – oxidation becomes cathode) + + Pb(IV) → Pb(II) → Pb(IV) - - Used as voltaic cell – battery powers light bulb External Power Generator – to charge battery Reactions reverse (products to reactants to replenish charge)

Chapter 18 Electrochemistry Electrolytic Cells – Questions – cont. 4. A steel rod is being chrome plated. If the rod has a diameter of 10 cm and a length of 150 cm, how long does it take to chrome plate the rod to a thickness of 1 mm, if a current of 20 A is applied during the chrome plating process? V(rod) = pd 2 L/4 and AW(Cr) = 52. 00 g/mol