Chp9_PExp9pt5

Chp9_PExp9pt5 - Practice Example 9.5 2 2 1 The cell Zn Zn...

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Unformatted text preview: Practice Example 9.5 2 2 1 The cell: Zn Zn 0.066M Cu 0.188M , Cu 0.084 M Pt Pt is an inert electrode NO copper metal involved The Zn couple is the anode The Cu ion couple is the cathode 1. The Cell Reaction & Standard State Cell Potential Cu 2 1e1 Cu 1 E 0 .16 0 Zn Zn 2 2e1 E 0 0.76 Note that the E0 value is the reduction potential even though we note here that the Zn couple is undergoing oxidation. The net reaction is thus (note that we must balance the number of electrons): 2Cu 2 Zn 2Cu 1 Zn 2 n=2 0 0 E 0 E Cath E Anode .16 0.76 .92 0 0 2. The Reaction Quotient Zn Cu Q Cu 2 1 2 2 2 0.066 0.084 2 0.188 2 1.32 10 2 note carefully the distinction of charges and exponents. 3. The Cell Potential. We use the Nernst Equation since we require the cell potential at the specific concentrations specified above. 0.0592 log Q n 0.0592 .92 0 log 1.32 10 2 2 .98 0 E E 0 Note that E > E0 (compare below with Q vis-à-vis K). 4. The Equilibrium Constant K e G 0 RT e nE 0 RT 2 96,485 0.92 72 31 exp e 1.32 10 8.314 298 Note all SI units. Note that Q <<< K yet E is only 0.06V greater than E0. See comment at the end of text example 9.5, p. 231. 5. The Gibbs Energy G 0 nE 0 2 96,485 0.92 178kJ G nE 2 96,485 0.98 189kJ ...
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Chp9_PExp9pt5 - Practice Example 9.5 2 2 1 The cell Zn Zn...

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