Lab Report 6 - RT E E ln = 0 0 Q nF RT E E ln ln 2 2 Cu Cu...

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Results, Discussion and Questions: Part 1: Anode Cathode Exp. Voltage Theor. Voltage (V) 1 Mg (s) Mg +2 + 2e - Cu +2 + 2e - Cu 1.55 2.7119 2 Mg (s) Mg +2 + 2e - Zn +2 + 2e - Zn 0.56 1.6082 3 Mg (s) Mg +2 + 2e - Fe +2 + 2e - Fe 0.95 1.923 4 Zn (s) Zn +2 + 2e - Cu +2 + 2e - Cu 0.90 1.1037 5 Fe (s) Fe +2 + 2e - Cu +2 + 2e - Cu 0.55 0.7889 6 Zn (s) Zn +2 + 2e - Fe +2 + 2e - Fe 0.29 0.3148 Part 2: Voltage before NH 3 is added = 0 V Voltage after NH 3 is added = 0.3V Questions: 1. The reason why no voltage was recorded before the porous cup was immersed in the glass is because there was no barrier between the two electrodes which means the electrons could flow freely between them. 2. If both the electrodes where copper and had the same concentration, the resulting voltage would be zero, this is because, the ln(1) is zero and the electric potential is zero, so zero minus zero equals zero. Q nF
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Unformatted text preview: RT E E ln = 0 - 0 Q nF RT E E ln ] [ ] [ ln 2 2 Cu Cu nF RT E E L mol L mol mol C mol J E / 2 . / 002 . ln ) / 485 , 96 ( 2 15 . 298 / 314 . 8 030294 . = 0.089 V If the Cu2+ concentration in the anode half cell were 0.0020 mol/L and Cu2+ concentration in the cathode half cell were 0.20 mol/L, the observed voltage be 0.089V. The reason why it wouldn’t be practical to perform this experiment with the voltmeters used in this experiment is because they aren’t sensitive enough to record such a small voltage. 3. [Cu(NH 3 ) 4 +2 ] = [Cu +2 ]e [(E-E)nF)/RT] = 0.2e [(0-0.3)(2)(96485)/(8.314)(298.15)] = 1.44 x 10-11 M [Cu +2 ] = 0.2 M [NH 3 ] = 6.0 M K f = [Cu(NH 3 ) 4 +2 ]/ [Cu +2 ] [NH 3 ] 4 = (1.44 x 10-11 )/(0.2)(6) 4 = 5.56 x 10-14...
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This note was uploaded on 11/11/2010 for the course CHEMISTRY CHEM1800 taught by Professor Krista during the Winter '10 term at UOIT.

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Lab Report 6 - RT E E ln = 0 0 Q nF RT E E ln ln 2 2 Cu Cu...

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