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uestion 9 D Half-Reaction Standard Reduction Potential, E (V) Ag+ (aq) +e -→ Ag(s) +0.80 Cu2+ (aq) + 2 e Cu(s) +0.34 Pb2+ (aq) + 2 e- Pb(s) -0.13 Al+ (aq) + 3 e- Al(s) -1.66 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 solutio e1 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 cell = +0.67 V. 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. 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" cell = +5.66 V D

uestion 9 D Half-Reaction Standard Reduction Potential, E (V) Ag+ (aq) +e -→ Ag(s) +0.80 Cu2+ (aq) + 2 e Cu(s) +0.34 Pb2+ (aq) + 2 e- Pb(s) -0.13 Al+ (aq) + 3 e- Al(s) -1.66 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 solutio e1 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 cell = +0.67 V. 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. 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" cell = +5.66 V D

Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Chapter24: Coulometry
Section: Chapter Questions
Problem 24.4QAP: Halide ions can he deposited at a silver anode, the reaction being Ag(s) + X- AgX(s) +e- Suppose...
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(12 Question 9 h Half-Reaction Standard Reduction Potential, E (V) Ag* (aq) + e- → Ag(s) +0.80 Cu2+ (aq) + 2 e- → Cu(s) +0.34 Pb2+ (ag) + 2 e- → Pb(s) -0.13 Alš+ (aq) + 3 e → Al(s) -1.66 A standard galvanic cell is made using Pb-Pb(NO3), 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. A Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO3 half-cell, because the reduction of 1 mol of Ag requires only 1 mol of e, resulting in E cell = +0.67 V. B Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO3 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. 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" cell = +5.66 V. Q Search or enter website name
Transcribed Image Text:(12 Question 9 h Half-Reaction Standard Reduction Potential, E (V) Ag* (aq) + e- → Ag(s) +0.80 Cu2+ (aq) + 2 e- → Cu(s) +0.34 Pb2+ (ag) + 2 e- → Pb(s) -0.13 Alš+ (aq) + 3 e → Al(s) -1.66 A standard galvanic cell is made using Pb-Pb(NO3), 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. A Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO3 half-cell, because the reduction of 1 mol of Ag requires only 1 mol of e, resulting in E cell = +0.67 V. B Replacing the Cu-Cu(NO,), half-cell with a Ag-AgNO3 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. 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" cell = +5.66 V. Q Search or enter website name
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