As we expected from earlier. Example :W h a ti s ε for a Daniel cell at 300 K when the concentration of Cu 2+ is 0.100 M and the concentration of Zn 2+ is 0.001 M? Zn(s) + Cu 2+ (aq) → Cu(s) + Zn 2+ (30.6) First we need ε ° which we found last lecture to be 1.10 V. Then we use the Nernst equation, ε =1 . 10 − 8 . 314 × 3002 × 96485ln 0 . 100 0 . 001 =1 . 04 V The n =2 because 2 electrons are transferred from the zinc to the copper during the reaction. Notice that even though the concentrations di f er by a factor of 100, the di f erence in voltage from the standard potential is minimal. Group Work1. Explicitly derive the Nernst equation from the 4 G equa-
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.