Standard cell potentials are determined with 1.0 M solutions of ions. In this ex

Question

Standard cell potentials are determined with 1.0 M solutions of ions. In this experiment, we will use 0.10 M solutions of Zn2+, Cu2+, Pb2+, and Ag1+ to minimize the amount of hazardous waste generated. In this series of questions, you will figure out what effect this will have on the voltages you observe. Consider a cell consisting of a Cu2+/Cu couple and an Ag1+/Ag couple. (a) Starting with 1.0 M solutions of ions (standard conditions), evaluate Q in the Nernst Equation. (b) What is log Q for this cell? (c) The cell potential in this case should be which of the following? (d) Starting with 0.1 M solutions of ions (at room conditions), evaluate Q in the Nernst Equation. (e) What is log Q for this cell? (f) The cell potential in this case should be which of the following?..

Explanation / Answer

Cu(2+)(aq) + e? ? Cu+(aq) Eo = +0.16 V

Cu(2+)(aq) + 2e? ? Cu(s) Eo = +0.34 V
Cu+(aq) + e? ? Cu(s) Eo = +0.52 V

2Ag+aq + 2e- ? 2Ags and E0red = +0.80 V

reduction: Ag+ + e- ==> Ag

oxidation:       Cu ==> Cu2+ + 2e-

Ecell = E0cell - (RT/nF) ln Q or Ecell = E0cell - (2.303*RT/nF )logQ

Calculating for the Cu half cell

E cell = +0.34

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