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Data Table Metal Metal Half reaction that goes Half reaction that goes as at the at the as a reduction (this is (+) Termi Termi terminal)* Voltaic Measured Volt age (V) Cell (metals used) an oxidation (this is for the (-) for the metal at the (+) metal at the (-) terminal)* nal nal # 1 # 2 # 3 Cu-Zn 0.881 0.864 0.850 Cu - Pb 0.486 0.489 0.495 Cu-Ag -0.420 -0.415 -0.407 Cu - Fe 0.544 0.555 0.571

Data Table Metal Metal Half reaction that goes Half reaction that goes as at the at the as a reduction (this is (+) Termi Termi terminal)* Voltaic Measured Volt age (V) Cell (metals used) an oxidation (this is for the (-) for the metal at the (+) metal at the (-) terminal)* nal nal # 1 # 2 # 3 Cu-Zn 0.881 0.864 0.850 Cu - Pb 0.486 0.489 0.495 Cu-Ag -0.420 -0.415 -0.407 Cu - Fe 0.544 0.555 0.571

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Data Table Metal Metal Half reaction that goes Half reaction that goes as at the at the as a reduction (this is (+) Termi Termi terminal)* Voltaic Measured Volt age (V) an oxidation (this is for the for the metal at the (+) metal at the (-) terminal)* Cell (metals used) (-) nal nal #1 #2 #3 Cu - Zn 0.881 0.864 0.850 Cu – Pb 0.486 0.489 0.495 Cu - Ag -0.420 -0.415 -0.407 Cu – Fe 0.544 0.555 0.571 TO HELP YOU DECIDE THE CHARGE OF EACH TERMINAL OF A HALF CELL. An oxidation gives up electrons. Electrons are negative. So the half cell an oxidation is taking place is going to be the negative terminal negative electrons are coming out of this half cell. where because the A reduction takes electrons. Taking of electrons is essentially making the half cell positive because the negative electrons are half cell that goes as a reduction is going to be the positive going into the half cell. So the electrode. Results table 1 Metal at the Metal at (+) Terminal Voltaic Cell Reduction Adjusted reduction potential of metal reacting with Cu (assumes that Cu has a reduction Average the (-) Terminal (metals used) potential of metal reacting with Cu (assumes that Cu has a reduction Measured Voltage from the Data Table (V) voltage of 0.00 V)! voltage of 0.34 V)2 Cu – Zn 0.865 Cu Zn -0.865 -0.525 Cu – Pb 0.49 Cu Pb -0.49 -0.150 Cu - Ag -0.414 Ag Cu 0.414 0.754 Cu - Fe 0.556 Cu Fe -0.556 -0.216
Transcribed Image Text:Data Table Metal Metal Half reaction that goes Half reaction that goes as at the at the as a reduction (this is (+) Termi Termi terminal)* Voltaic Measured Volt age (V) an oxidation (this is for the for the metal at the (+) metal at the (-) terminal)* Cell (metals used) (-) nal nal #1 #2 #3 Cu - Zn 0.881 0.864 0.850 Cu – Pb 0.486 0.489 0.495 Cu - Ag -0.420 -0.415 -0.407 Cu – Fe 0.544 0.555 0.571 TO HELP YOU DECIDE THE CHARGE OF EACH TERMINAL OF A HALF CELL. An oxidation gives up electrons. Electrons are negative. So the half cell an oxidation is taking place is going to be the negative terminal negative electrons are coming out of this half cell. where because the A reduction takes electrons. Taking of electrons is essentially making the half cell positive because the negative electrons are half cell that goes as a reduction is going to be the positive going into the half cell. So the electrode. Results table 1 Metal at the Metal at (+) Terminal Voltaic Cell Reduction Adjusted reduction potential of metal reacting with Cu (assumes that Cu has a reduction Average the (-) Terminal (metals used) potential of metal reacting with Cu (assumes that Cu has a reduction Measured Voltage from the Data Table (V) voltage of 0.00 V)! voltage of 0.34 V)2 Cu – Zn 0.865 Cu Zn -0.865 -0.525 Cu – Pb 0.49 Cu Pb -0.49 -0.150 Cu - Ag -0.414 Ag Cu 0.414 0.754 Cu - Fe 0.556 Cu Fe -0.556 -0.216
Footnote #1: If Cu is the (+) terminal and the other metal (M) is the (-) terminal, the Cu is going as a reduction and M is going as an oxidation. By definition, the reduction potential of M is (-). If Cu is (-) and M is (+), then Cu is going as an oxidation and other metal is going as a reduction. By definition, the reduction potential of M is (+). Essentially, this means that the sign of each voltage in the first column is to be reversed when entered in this column. Footnote #2: In the previous column, Cu was arbitrarily assigned a voltage of 0.00 V. However, the actual standard reduction potential of Cu is 0.34 V. Thus, to convert the reduction potential in the previous column to actual reduction potentials, add 0.34 V to each entry. Results Table 2 Metal Actual reduction Percent error between your Reduction potential in descending order from most (+) to most (-) reordered from the last column in Results Table 1. (Don't forget to insert Cu at the +0.34 V position in this table) potential of each measured reduction metal from the textbook voltage (column #1) and the actual reduction (Zumdahl, 6th ed, voltage (column #3). (Cu p 833)* will, of course, have a 0% error)** * Be careful to chose the proper half reactions. These involve the metal and the ions listed in the materials section on page L–1. **Students in the past have found that iron gives the largest percent error. It turns out that the "iron" we are using is really mild steel and is alloyed with other metals, so a large error is to be expected. We have not found a source of pure iron at this time. Is the order of the metals in Results Table 2 the same as that in the table of Standard Reduction Potentials on p 833 in Zumdahl's text? Yes No (circle one) If you circled No, check all your calculations. The order should come out the same as in the standard table. The formula for percent error: % error=yourvalue-book valuex 100% book value The vertical lines stand for absolute value, meaning that the sign of the number between the vertical lines is always positive.
Transcribed Image Text:Footnote #1: If Cu is the (+) terminal and the other metal (M) is the (-) terminal, the Cu is going as a reduction and M is going as an oxidation. By definition, the reduction potential of M is (-). If Cu is (-) and M is (+), then Cu is going as an oxidation and other metal is going as a reduction. By definition, the reduction potential of M is (+). Essentially, this means that the sign of each voltage in the first column is to be reversed when entered in this column. Footnote #2: In the previous column, Cu was arbitrarily assigned a voltage of 0.00 V. However, the actual standard reduction potential of Cu is 0.34 V. Thus, to convert the reduction potential in the previous column to actual reduction potentials, add 0.34 V to each entry. Results Table 2 Metal Actual reduction Percent error between your Reduction potential in descending order from most (+) to most (-) reordered from the last column in Results Table 1. (Don't forget to insert Cu at the +0.34 V position in this table) potential of each measured reduction metal from the textbook voltage (column #1) and the actual reduction (Zumdahl, 6th ed, voltage (column #3). (Cu p 833)* will, of course, have a 0% error)** * Be careful to chose the proper half reactions. These involve the metal and the ions listed in the materials section on page L–1. **Students in the past have found that iron gives the largest percent error. It turns out that the "iron" we are using is really mild steel and is alloyed with other metals, so a large error is to be expected. We have not found a source of pure iron at this time. Is the order of the metals in Results Table 2 the same as that in the table of Standard Reduction Potentials on p 833 in Zumdahl's text? Yes No (circle one) If you circled No, check all your calculations. The order should come out the same as in the standard table. The formula for percent error: % error=yourvalue-book valuex 100% book value The vertical lines stand for absolute value, meaning that the sign of the number between the vertical lines is always positive.
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