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Redox reactions involve transfer of e-s Mg0(s) + Cu2+(aq) Cu0(s) + Mg2+(aq) electrons transferred from Mg to Cu2+ Mg loses e-s and Cu2+ gains them Mg is oxidized, loses e-s Cu2+ is reduced, gains e-s Redox reactions involve transfer of e Key terms oxidation = loss of e-s (oxidation # goes up) reduction = gain of e-s (oxidation # goes down) oxidizing agent = e- acceptor (subst red) reducing agent = e- donor (subst oxid) Types of Electrochem Rx Direct – oxidizing and reducing agents come into contact with each other Produce heat, not electrical energy Indirect – oxidizing and reducing agents in separate compartments, connected by external wire that e-s pass thru Each compartment called ½ cell oxidation ½ cell reduction ½ cell ½ cells are connected by salt bridge Types Electrochemical cells Galvanic or Voltaic cell Apparatus that allows a redox rx to occur Product favored rxs push e-s thru wire thereby prod an electric potential (voltage) Electrolytic cell Reactant favored rx can be forced to occur by using an electrical current to force a nonspontaneous rx (electrolysis) Parts of Electrochemical Cell Oxidation ½ cell Anode (-) site of oxidation e-s flow from it Cathode (+) site of reduction e-s flow to it Voltmeter Reduction ½ cell Measures cell potential (pressure) Salt bridge Allows ions to migrate Parts of galvanic cell 1. Given the following data for Hg Hg(l) S˚ = 76.0 J/mol·K Hg(g) S˚ = 175.0 J/mol·K calculate the normal boiling point for Hg: 2. For the reaction: 2H2O(l) + 2Br-(aq) Hg(l) Hf˚ = 0 KJ/mol Hf˚ = 60.77 KJ/mol Hg(g) H2(g) + Br2(l) + 2OH-(aq) a. calc G˚ at 25˚C. b. calc G when pH2 = 0.100 atm, [Br-} = 0.750 M, and the pH of the solution is 9.73. Cell potential = E˚ amount of push with which the e-s are sent through the external circuit. e-s are pushed through a galvanic cell by electromotive force (emf) the diff in electrical potential energies (Ep) of the metals that make up the anode and cathode determine the potential (voltage), or electrical pressure, of the cell Units of Electricity Coulomb = unit of electrical charge Ampere = unit of electrical flow (current) 1 e- = 1.602 x 10-19 C Amp = 1 C/sec Joule = unit of energy 1J=1Cx1V Energy = charge x potential (push) Faraday = F = 96,480 C/mol e- Determination of E˚ E = electrical potential under standard conditions (1 atm, 1 molar) E˚cell = E˚oxid + E˚red E˚oxid = -(E˚red) (+) value for E˚cell = rx is NOT spontaneous (-) value for E˚cell = rx is spontaneous Units = volts 1 volt = 1 J/1 C Calculation of cell potential Zn0(s) + Cu2+(aq) Zn2+(aq) + Cu0(s) E˚= E˚oxid + E˚red = -(-0.762V) + (0.339V) = 1.10V (+) value for E˚ means rx is spontaneous (-) value for E˚ means rx is NOT spont. Cu0(s) + 2Ag1+(aq) Cu2+(aq) + 2Ag0(s) Relate E˚ and G˚ G˚ = -nFE˚ E˚ = standard cell potential n = # e-s transferred From balanced redox equation F = Faraday’s constant F = 9.648 x 104 C/mol e- Relate E˚ and K E˚ = RT (lnK) nF R = 8.31 J/mol K T = Kelvin temp n = # mol e-s transferred F = Faraday’s constant K = equilibrium constsnt [conc] or (pp) prod/ [conc] or (pp) reactant each raised to it’s coefficient as an exponent E˚ = RT (lnK) nF At 25˚ C equation becomes E˚ = (0.0257 V) ln K n If E˚ is (+) ln K is positive and K > 1, therefore the forward reaction is favored If E˚ is (-) ln K is negative and K < 1, therefore the reverse reaction is favored Nernst Equation Voltage will increase with increases in concentration Voltage will decrease with decreases in concentration E = E˚- RT (ln Q) n At 25˚C the equation becomes E = E˚- 0.0257 V (ln Q) n Electrolytic Cells Process of electrolysis A nonspontaneous reaction is made to occur by pumping electrical energy into the system Electrons are pushed from the anode to the cathode by a direct current Electroplating Reduction of metal ions to metal atoms occurs at the cathode Can calculate the mass of metal deposited through the pathway: Coul mol e-s F mol metal stoic factor mm of metal (F = 96,480 C/mol e-) g metal Amperes x seconds = coulombs Electroplating