left right Exercise 7 Determine and based on the following half-reactions: Eocell Ecell VO2+ + 2H+ + e- VO2+ + H2O 1.00 V E°= → Zn2+ + 2e- Zn = -0.76V E° → Where = 25°C T [VO2+] = 2.0 M [H+] = 0.50 M [VO2+] = 1.0 x 10-2 M [Zn2+] = 1.0 x 10-1 M = 1.76 V Eocell = 1.89 V SUMMARY OF GIBB’S FREE ENERGY AND CELLS implies NONspontaneous. −Eo +Eo implies spontaneous (would be a good battery!) = 0, equilibrium reached (dead battery) E larger the voltage, more spontaneous the reaction will be negative in spontaneous reactions G > 1 are favored K Two important equations: [“minus nunfe”] G = −nFE [“ratlink”] G = = Gibbs free energy [Reaction is spontaneous if is G negative] ΔG = number of moles of electrons. n = Faraday constant 9.6485309 × 104 J/V (1 mol of electrons carries 96,500C ) F = cell potential E˚ = 8.31 J/molK R = Kelvin temperature T = equilibrium constant [products]coeff. /[reactants]coeff K **Favored conditions: > 0 < 0 > 1** G K Exercise 8 For the oxidation-reduction reaction
left right Exercise 7 Determine Eocell and Ecell based on the following half-reactions: VO2+ + 2H+ + e- → VO2+ + H2O E°= 1.00 V Zn2+ + 2e- → Zn E° = -0.76V Where T = 25°C [VO2+] = 2.0 M [H+] = 0.50 M [VO2+] = 1.0 x 10-2 M [Zn2+] = 1.0 x 10-1 M Eocell = 1.76 V = 1.89 V SUMMARY OF GIBB’S FREE ENERGY AND CELLS −Eo implies NONspontaneous. +Eo implies spontaneous (would be a good battery!) E = 0, equilibrium reached (dead battery) larger the voltage, more spontaneous the reaction G will be negative in spontaneous reactions K > 1 are favored Two important equations: G = −nFE [“minus nunfe”] G = [“ratlink”] G = Gibbs free energy [Reaction is spontaneous if ΔG is negative] n = number of moles of electrons. F = Faraday constant 9.6485309 × 104 J/V (1 mol of electrons carries 96,500C ) E˚ = cell potential R = 8.31 J/molK T = Kelvin temperature K = equilibrium constant [products]coeff. /[reactants]coeff **Favored conditions: > 0 G < 0 K > 1** Exercise 8 For the oxidation-reduction reaction
17%20Electrochemistry%20modified.docx