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Electrochemistry Chem 35.5

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Electrochemistry Chem 35.5 Announcements Problems Chapter 21: 2,9,10,11,13,17,22,29,31,38,40,44,46,50,53,58,62,64,65,70, 72,73,82,85,87 Wonderful job with the presentations I was very impressed with all of them. Exam 3 March 17 Comprehensive Final Exam: March 24 7:30AM - 9:30 C114 This is an electrochemical cell, also called a voltaic or galvanic cell. Chemists name parts electrochemical cell names. Salt bridge with non-reacting Electron but conductive salt to replace - -------> lost electrons and reduced Anode (-): Flow e cations Oxidation always occurs here Cathode (+): Reduction always - + occurs here. Anode Cathode Electrolyte Electrolyte Example Problem: Making a Voltaic Cell Diagram Diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr solid electrode in a Cr(NO3)3 solution, another half-cell with an Ag solid electrode in an AgNO3 solution, connected by a KNO3 salt bridge. What would the cell potential be under standard state conditions? 3+ 0.74 = Cr + 2 e− Cr(s) − −−→ Example Problem: Making a Voltaic Cell Diagram Cr(s) | Cr3+(aq) || Ag+(aq) | Ag(s) oxidation half-reaction: Cr(s) Cr3+(aq) + 3e- reduction half-reaction: Ag+(aq) + e- Ag(s) Cr(s) + Ag+(aq) Cr3+(aq) + Ag(s) Galvanic cells can also be made using “non-active electrode”, or electrodes that don’t undergo reaction directly in the cell. The non-active electrodes are typically carbon or platinum. reduction half-reaction: - + - MnO4 (aq) + 8H (aq) + 5e oxidation half-reaction: 2+ Mn (aq) + 4H2O(l) - - 2I (aq) I2(s) + 2e overall (cell) reaction: - + - 2+ 2MnO4 (aq) + 16H (aq) + 10I (aq) 2Mn (aq) + 5I2(s) + 8H2O(l) We have and use standard cell notation for non- active electrodes too as shown below. - + - 2+ graphite | I (aq) | I2(s) || H (aq), MnO4 (aq) | Mn (aq) | graphite inert electrode inert electrode anode--oxidation occurs cathode--reduction occurs Salt Bridge The electric potential difference (electromotive force (emf)) is caused by differences in electron donating tendencies of metals and ions. This tendency is tabulated in a table of reduction potentials. • In order to tell what predict what is oxidized and what species is reduced you must refer to a table of reduction potentials or be given the reaction that occurs. The purpose of a galvanic cell is to convert electrochemical potential energy to work! We need to relate electric potential E in volts to energy as they have different units. 1. The SI unit for electric potential is the Volt and for charge it is the Coulomb. By definition a 1 V potential across two electrodes can provide 1 J of energy for each Coulomb of charge that is 1 Joule the moves between the 1 Volt = electrodes. Coulomb The cell potential that is generated between a redox pair (called E˚cell) depends on four main factors which include: (1) molarity of solutions in each half-cell (2) the metals and ions in each half-cell (redox potential) (3) temperature of the solution. We must adopt a “yardstick” or a single half-cell from which we measure every other half-cell. It’s called the standard hydrogen electrode or SHE. The electrochemical potential (or emf) can do work just like a gravitational potential can do work. We say the system moves from higher chemical potential to lower chemical potential. Strong oxidant + Strong reductant Chemical Potential Weak Oxidant + Weak Reductant Reactants ==> Products An analogy of “water pressure” to electrochemical potential difference in a spontaneous redox system. It is caused by the different abilities of metals to give up electrons (strong to weak). Zn(s) + Cu2+(aq) Chemical Potential Eoxidized E Zn2+(aq) + Cu(s) reduced 3+ 0.74 = Cr + 2 e− Cr(s) − −−→ The Standard Hydrogen Electrode consists of an inert Pt wire electrode placed in a tube containing 1M HCl solution. H2 gas is bubbled through the tube into the acid and connected to another arbitrary half-cell. The reduction half-reaction is defined as zero! - + 2e + 2H (1 M) H2 (1 atm) ˚ E cell = 0 V This electrode is our zero point or “sea-level” Standard cell potential with the SHE are all done under “standard conditions” which means 1 M for electrolyte concentrations, any gases are at 1 atm, temperature at 298K. Standard Conditions The other part of --1 Molar the galvanic cell is any metal and its Electrolytes electrolyte that we choose. --All gases (if any) 1 atmosphere --Temperature = 298K = 25˚C Suppose we assemble the following electrochemical cell and ask what the cell potential would be? 2+ + Zn (s) | Zn (1 M) || H (1 M) | H2 (1 atm) | Pt (s) 2+ 0 Zn(s) Zn (1 M) + 2e E oxidized = 0.76 V + - 0 2H (1 M) + 2e H2 E reduced = 0.0 V 0 E cell = 0.76 V The Standard cell potential for all half-cells combinations under standard conditions is given by a simple equation shown below: E˚cell = E˚red(cathode) - E˚red(anode) standard reduction standard reduction E˚cell = potential of the - potential of the substance reduced substance oxidized Notice that both numbers will be the magnitude and sign of given in the standard reduction table. What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? Cd2+ (aq) + 2e- Cd (s) E0 = -0.40 V Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V What will be oxidized and what will be reduced? How can we tell? What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? 2+ - 0 Cd (aq) + 2e Cd (s) E = -0.40 V Cd is the stronger oxidizer Cr3+ (aq) + 3e- Cr (s) E0 = -0.74 V Cd will oxidize Cr Anode (oxidation): Cr (s) Cr3+ (1 M) + 3e- x 2 Cathode (reduction): 2e- + Cd2+ (1 M) Cd (s) x 3 2Cr (s) + 3Cd2+ (1 M) 3Cd (s) + 2Cr3+ (1 M) 0 0 0 Ecell = Ecathode - Eanode 0 Ecell = -0.40 – (-0.74) 0 Ecell = 0.34 V .
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