Using the Table of Standard Reduction Potentials table shown below, calculate the standard cell potential for an electrolytic cell that has iron (Fe) and magnesium (Mg) electrodes and identify the cathode. a. +3.14 * V with Fe as the cathode d.-3.14 V with Mg as the cathode b. +3.14 V with Mg as the cathode e.+ 1.6V with Fe as the cathode c. -3.14 V with Fe as the cathode Standard Reduction Potentials (volts) in Aqueous Solution Pb^ 2+ 2e Pb^ 2- -1.80 Au^ 2- /3e^ - Au +1.50 F * e ^ (xx) + 3e -> Fe +0.771 I_{2} + 2c -> 21 +0.535 Pb^ 2^ - /2c^ - Pb -0.124 A * I ^ (3 + 3c) -> AI -1.66 Mg^ 2- + 2c -> Mg -2.37 K' + c -> K -2.93
Using the Table of Standard Reduction Potentials table shown below, calculate the standard cell potential for an electrolytic cell that has iron (Fe) and magnesium (Mg) electrodes and identify the cathode. a. +3.14 * V with Fe as the cathode d.-3.14 V with Mg as the cathode b. +3.14 V with Mg as the cathode e.+ 1.6V with Fe as the cathode c. -3.14 V with Fe as the cathode Standard Reduction Potentials (volts) in Aqueous Solution Pb^ 2+ 2e Pb^ 2- -1.80 Au^ 2- /3e^ - Au +1.50 F * e ^ (xx) + 3e -> Fe +0.771 I_{2} + 2c -> 21 +0.535 Pb^ 2^ - /2c^ - Pb -0.124 A * I ^ (3 + 3c) -> AI -1.66 Mg^ 2- + 2c -> Mg -2.37 K' + c -> K -2.93
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 77AP
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Question
Using the Table of Standard Reduction Potentials table shown below, calculate the standard cell potential for an electrolytic cell that has iron (Fe) and magnesium (Mg) electrodes and identify the cathode.
a. +3.14 * V with Fe as the cathode
d.-3.14 V with Mg as the cathode
b. +3.14 V with Mg as the cathode
e.+ 1.6V with Fe as the cathode
c. -3.14 V with Fe as the cathode
Standard Reduction Potentials (volts) in Aqueous Solution
Pb^ 2+ 2e Pb^ 2-
-1.80
Au^ 2- /3e^ - Au
+1.50
F * e ^ (xx) + 3e -> Fe
+0.771
I_{2} + 2c -> 21
+0.535
Pb^ 2^ - /2c^ - Pb
-0.124
A * I ^ (3 + 3c) -> AI
-1.66
Mg^ 2- + 2c -> Mg
-2.37
K' + c -> K
-2.93
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