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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
Chemistry
9th Edition
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Steven S. Zumdahl
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 3RQ: Table 17-1 lists common half-reactions along with the standard reduction potential associated with...
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17 25 30 « < 22 of Half-Reaction Standard Reductlon Potential, E (V) Ag+ (ag) + e- - Ag(s) +0.80 Cu?+ (ag) + 2 e- → Cu(s) +0.34 Pb2+ (aq) + 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 onll, and why? Assume all 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. B 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. 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
Transcribed Image Text:17 25 30 « < 22 of Half-Reaction Standard Reductlon Potential, E (V) Ag+ (ag) + e- - Ag(s) +0.80 Cu?+ (ag) + 2 e- → Cu(s) +0.34 Pb2+ (aq) + 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 onll, and why? Assume all 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. B 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. 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
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