Chem 04-22 - Any substance on the right will reduce any...

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Chemistry 102 Notes Final Exam 04/22/08 A voltaic cell is set up with an aluminum electrode in a .025 M Al^3+ (aq) solution and an iron electrode in a .50 M Fe^2+(aq) solution. Calculate the voltage produced by this cell at 25 ° C. Al^3+(aq) + 3e- Al(s) E ° = -1.66V Oxidation, Anode Fe^2+(aq) + 2e- Fe(s) E ° = -.44 V Reduction, Cathode A: Overall Reaction: 2 Al(s) + 3 Fe^2+ 2 Al^3+(aq) + 3 Fe(s) Ecell = Ecathode – Eanode -.44 V – (-1.66V) = 1.22 V E = E - .0257V/ln * ln Q; n = 6 Q= [Al^3+]^2/ [Fe^2+]^3 Ecell = 1.22 V - .0257 V/ 6 * ln [.025]^2/[.50]^3 1.22V = (-.023 V) = 1.24 V
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Unformatted text preview: Any substance on the right will reduce any substance higher than it on the left. ^ oxidizing ability & v reducing ability. E at Nonstandard Conditions E = E = .0257V/n * ln ([Products]/[Reactants]) Q E= potential under nonstandard conditions The Nernst equation N= number of electrons exchanged Ln= natural log If [P] and [R] = 1 mol/L, then E= E If [R] > [P], then E is ____________________ than E (A: more, higher, greater) If [R] < [P], then E is ____________________ than E (A: less, lower, smaller)...
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