+ - = + - DSuniv Thermodynamics -DHsys T DSsys DSsurr Chapter 16 – A.P. Chemistry Thermodynamics Because of: Occurs only with outside intervention The 2nd Law Occurs w/o outside intervention The entropy of the univ. is increasing spontaneous nonspontaneous + DSuniv - -DHsys Which is equal to: 1. 2. 3. phases T # of gaseous particles A measure of disorder The change in the entropy of the universe = + DSsys DSsurr rankings complexity of particles measurable because of or 3rd Law The little superscript zeros mean: standard DSuniv = DSsys – DH/T Changed to DG° = DH° – TDS° Change in free energy Which can be changed to: Which says: -TDSuniv = DH – TDS = DG Standard ________ in ______________ Standard ________ in ___________ Standard ________ in __________ The entropy of a perfect crystal is zero change free energy change enthalpy change entropy - form that is usable Which must be ___ for spontaneous processes and for which info is easily found. But what if we are not at standard conditions? ←Q is the reaction quotient…the mass action expression w/ init. conditions DG = DG° + RTlnQ DG° = -RTlnK DG = 0 0 = DG° + RTlnK or And if we are at equilibrium?

E = E° -RT lnQ E = E° -0.0592 logQ Ch. 17 E °= 0 Electrochem. anode ion(s) ion(s) cathode spontaneous nonspontaneous ( Forcing a current through a cell or changing electrical energy to chemical energy) line notation Galvanic Cells Electrolytic Cells shorthand Device to: ___________ ____________________ change chem. energy to electrical energy 2H+(aq) + 2e-→ H2(g) standard Parts The test will have a blend of chapter 16 and 17 concepts focusing on using thermodynamic concepts in a electrochemical system E °= 0 (half-reactions) Plating reactions cathode anode (reduction) (oxidation) DG° = -nFE° Faraday 96500 C Info for these under ___________ conditions for____________ reactions standard reduction E = E° -RT lnQ ______ half reactions to get: ____________________ add nF Key: standard cell pot. (EMF), E° nonstandard conditions Nernst Eq. E = E° -0.0592 logQ n (At 298 K)

Cathode ► DG°= - RTln K K = 2.41 x 1061 ◄electron flow (because, for the electrode to gain mass, the Sn2+ must be reduced) +0.60 V = -0.14 V + Eº anode Eºanode = 0.74 V Eºreduction = -0.74 V Chromium, Cr 3Sn2+ (aq) + 2Cr (s) → 3Sn (s) + 2Cr3+ (aq) Find the cell potential under these conditions Ecell = 0.60 V – [(0.0592/6) log [(0.10)2/(0.50)3] = 0.61 V Using the standard cell potential calculate DGº for this reaction . Describe two other ways to find the standard change in free energy for this process. Determine the equilibrium constant for this reaction. DGº = - (6 mol) (96,500 C/mol) (0.60 J/V) = -350 kJ DG° = DH° – TDS° DG°= SnDG°f(prod.) - SnDG°f(react.) DG°= - RTln K K = 2.41 x 1061