22 172530« < 22 ofHalf-ReactionStandard Reductlon Potential, E (V)Ag+ (ag) + e- - Ag(s)+0.80Cu?+ (ag) + 2 e- → Cu(s)+0.34Pb2+ (aq) + 2 e- → Pb(s)-0.13Al+ (ag) + 3 e- → Al(s)-1.66A standard galvanic cell is made using Pb-Pb(NO,), and Cu-Cu(NO,), half-cells. Which of the following modifications to the cell will cause the greatest increase in E onll, and why? Assumeall solutions are 1 M.= +0.67 V.Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-celI, because the reduction of 1 mol of Ag+ requires only1 mol of e-, resulting in E,Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-celI, because less current flows through the cell, resulting in E¨cell = +0.93 V.BReplacing the Pb-Pb(NO,), half-cell with an Al-Al(NO,), half-cell, because the oxidation of Al(s) is more thermodynamically favorable than the oxidation of Pb(s), resultingin E" cell = +2.00 V.Replacing the Pb-Pb(NO), half-cell with an Al-Al(NO,), half-cell, because 6 mol of e- flow through the cell, resulting in E cel = +5.66 V.US O

Given that: The cell is made of Pb.Pb(NO3)2 and Cu.Cu(NO3)2 half cells. To find, which among the…

10 16 18 19 20 26 (27 (28 29 « < Half-Reaction Standard Reduction Potentlal, E" (V) Ag+ (ag) + e- Ag(s) +0.80 Cu?+ (ag) + 2 e- + Cu(s) +0.34 Pb2+ (ag) + 2 e- Pb(s) -0.13 Al+ (ag) + 3 e- - Al(s) -1.66 A standard galvanic cell is made using Pb-Pb(NO,), and Cu-Cu(NO,), half-cells. Which of the following modifications to the cell will cause the greatest increase in E cell, and why? Assume all solutions are 1 M. Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO, half-cell, because the reduction of 1 mol of Ag+ requires only 1 mol of e, resulting in E celt = +0.67 V. B Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO, half-cell, because less current flows through the cell, resulting in E cell = +0.93 V. Replacing the Pb-Pb(NO), half-cell with an Al-Al(NO,), half-cell, because the oxidation of Al(s) is more thermodynamically favorable than the oxidation of Pb(s), resulting in E" cell = +2.00 V. cell = +5.66 V D Replacing the Pb-Pb(NO,), half-cell with an Al-Al(NO,), half-cell, because 6 mol of e flow through the cell, resulting in E™ US

10 16 18 19 20 26 (27 (28 29 « < Half-Reaction Standard Reduction Potentlal, E" (V) Ag+ (ag) + e- Ag(s) +0.80 Cu?+ (ag) + 2 e- + Cu(s) +0.34 Pb2+ (ag) + 2 e- Pb(s) -0.13 Al+ (ag) + 3 e- - Al(s) -1.66 A standard galvanic cell is made using Pb-Pb(NO,), and Cu-Cu(NO,), half-cells. Which of the following modifications to the cell will cause the greatest increase in E cell, and why? Assume all solutions are 1 M. Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO, half-cell, because the reduction of 1 mol of Ag+ requires only 1 mol of e, resulting in E celt = +0.67 V. B Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO, half-cell, because less current flows through the cell, resulting in E cell = +0.93 V. Replacing the Pb-Pb(NO), half-cell with an Al-Al(NO,), half-cell, because the oxidation of Al(s) is more thermodynamically favorable than the oxidation of Pb(s), resulting in E" cell = +2.00 V. cell = +5.66 V D Replacing the Pb-Pb(NO,), half-cell with an Al-Al(NO,), half-cell, because 6 mol of e flow through the cell, resulting in E™ US

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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