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GIBB'S FREE ENERGY AND ELECTRIC CELLS SUPA CHEMISTRY MRS. PAPARELLA SPRING 2016 DG = -n F o (need to memorize) E • DG = Gibb’s Free Energy • n = number of moles of electrons exchanged in reaction • F = Faraday’s Constant = 96,500 Coulombs/mole of e- Given on Test • Eo = Cell Potential measured in Volts; Also referred to as emf or Electromotive Force. • Voltage = Work/Charge • 1 V= 1Joule/Coulomb of Charge • A Volt describes how much energy it takes to move 1 coulomb of charge • These problems will use KJ for DG that need to be changed back to Joules ( x by 1000) It’s never easy! HOW DOES E CELL RELATE TO GIBB'S FREE ENERGY? • https://www.youtube.com/watch?v= Xrzhby7bsJw • Spontaneous reactions have • a + E value ( + Volts) and • a – DG value. • Nonspontaneous Reactions have a + DG value and a – E value (- Volts) CALCULATE THE GIBB'S FREE ENERGY IN A ZINC /COPPER +2 ELECTROCHEMICAL CELL WHICH HAS A VOLTAGE OF 1.1 V • DG = -n F E How many e- are transferred? • 1.1 V = 1.1 J/C • Plug and Chug! • DG = -2 mol e- x 96,500 C/mol e-x 1.1J/C • DG = - 212,300 J or -212.3 KJ 2 mol e- YOU CAN ALSO FIND THE CELL POTENTIAL IF YOU ARE GIVEN THE EQUATION AND THE FREE ENERGY INFORMATION. • 2Al(s) + 3Cu2+(aq)-->2Al3+(aq) + 3Cu(s) • D G = -1152.2 KJ • Question: How many moles of e- are involved? • Answer: 6 moles of e• Now you can proceed! REMEMBER TO CHANGE KJ TO JOULES FOR DELTA G • DG = -n F E • -1,152,200 J = - 6moles of e- x 96,500 C/ mole of e- x E Solve for E (Cell Potential) -1,152,200 J/ -579,000 C= 1.99 J/C or 1.99 V ALTERNATIVE METHOD • 1 Faraday is equal to 96.5KJ / V mol e- This is also given on the test • Compare with 96,500 C /mol e• So, if you know the D G and the # of mol of e- you can find the cell voltage by DG = -n F E • • • • Using the previous problem data: 1,152.2KJ = -6 mol e- x 96.5KJ/Vmol e- x E and solve for E. 1,152.2KJ / 579 KJ/V = E = 1.99 V E BOOK PRACTICE: P 764 20.41 • You will first have to calculate the Cell Potential using the Eo Red half-reaction information given. Remember to change the sign if the substance is oxidized! For this set of problems, remember if something is oxidized its charge goes up! So Fe2+ is going to be oxidized to Fe 3+. You will be doing 3 separate problems.