Half-Reaction Standard Reduction Potential, E (V) Ag* (aq) +e- -→ Ag(s) +0.80 Cu?+ (aq) + 2 e- → Cu(s) +0.34 Pb²+(ag) + 2 e- → Pb(s) -0.13 A+ (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 are1 M. cell = +0.67 V A Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-cell, because the reduction of 1 mol of Agt requires only 1 mol of e, resulting in E'. B Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-cell, because less current flows through the cell, resulting in E = +0.93 V. Replacing the Pb-Pb(NO,), half-cell with an Al-Al(NO,), half-cell, because the oxidation of A1(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 Ei= +5.66 V.

Chemistry: An Atoms First Approach
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ISBN:9781305079243
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Chapter17: Electrochemistry
Section: Chapter Questions
Problem 135CWP: Consider a galvanic cell based on the following half-reactions: a. What is the expected cell...
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Half-Reaction
Standard Reduction Potential, E (V)
Ag* (aq) +e- -→ Ag(s)
+0.80
Cu?+ (aq) + 2 e- → Cu(s)
+0.34
Pb²+(ag) + 2 e- → Pb(s)
-0.13
A+ (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.
cell = +0.67 V
A
Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-cell, because the reduction of 1 mol of Agt requires only 1 mol of e, resulting in E' .
B
Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-cell, because less current flows through the cell, resulting in E = +0.93 V.
Replacing the Pb-Pb(NO,), half-cell with an Al-Al(NO,), half-cell, because the oxidation of A1(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 = +5.66 V.
Transcribed Image Text:Half-Reaction Standard Reduction Potential, E (V) Ag* (aq) +e- -→ Ag(s) +0.80 Cu?+ (aq) + 2 e- → Cu(s) +0.34 Pb²+(ag) + 2 e- → Pb(s) -0.13 A+ (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. cell = +0.67 V A Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-cell, because the reduction of 1 mol of Agt requires only 1 mol of e, resulting in E' . B Replacing the Cu-Cu(NO,), half-cell with a Ag-AGNO, half-cell, because less current flows through the cell, resulting in E = +0.93 V. Replacing the Pb-Pb(NO,), half-cell with an Al-Al(NO,), half-cell, because the oxidation of A1(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 = +5.66 V.
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