Gibb's Free energy and Electric cells Supa Chemistry Mrs. Paparella Spring 2016

DG = -n F Eo (need to memorize) 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? 2 mol e- 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

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 E = 1.99 V

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.