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A certain half-reaction has a standard reduction potential E=-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that red must provide at least 0.80 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer reguires this half-reaction to happen at the cathode of the cell. dh Data Ag+ (aq) + e Ag (s) 0.7996 Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? O yes, there is a minimum. Ov %3D red Al3+ (aq) + 3e-- Al (s) -1.676 Au+ (aq) + e- Au (s) 1.692 If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. Au3+ (aq) + 3e Au (s) 1.498 O no minimum Ba2+ (aq) + 2e- Ba (s) -2.912 Brz (1) + 2e - - 2Br (aq) 1.066 Ca2+ (aq) + 2e - Ca (s) -2.868 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? = Ov Cl2 (9) + 2e → 2CI- (aq) O yes, there is a maximum. E red 1.35827 %3! Co2+ (aq) + 2e - Co (s) Co3+ (aq) + e Co2+ (aq) Cr2+ (aq) + 2e→ Cr (s) Cr3+ (aq) + 3e Cr (s) Cr+ (aq) + e Cr2+ (aq) Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 50H (aq) -0.28 If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. 1.92 O no maximum -0.913 -0.744 -0.407 -0.13 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Cu2+ (aq) + 2e- Cu (s) 0.3419 Cu2+ (ag) + e - Cu* (aq) 0.153 Note: write the half reaction as it would actually occur at the anode. Cu* (ag) + e - Cu (s) 0.521 F2 (9) + 2e 2F (aq) 2.866 Fe2+ (ag) + 2e- Fe (s) -0.447 Explanation Check Fe3+ (aq) + e - Fe2+ (aq) 0.771 1. Fe+ (aq) + 3e Fe (s) -0.037 MacBook Air 20 FI F12 D00 F4 F10 FS F6 F7 F8 F9 FI F2 F3

A certain half-reaction has a standard reduction potential E=-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that red must provide at least 0.80 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer reguires this half-reaction to happen at the cathode of the cell. dh Data Ag+ (aq) + e Ag (s) 0.7996 Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? O yes, there is a minimum. Ov %3D red Al3+ (aq) + 3e-- Al (s) -1.676 Au+ (aq) + e- Au (s) 1.692 If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. Au3+ (aq) + 3e Au (s) 1.498 O no minimum Ba2+ (aq) + 2e- Ba (s) -2.912 Brz (1) + 2e - - 2Br (aq) 1.066 Ca2+ (aq) + 2e - Ca (s) -2.868 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? = Ov Cl2 (9) + 2e → 2CI- (aq) O yes, there is a maximum. E red 1.35827 %3! Co2+ (aq) + 2e - Co (s) Co3+ (aq) + e Co2+ (aq) Cr2+ (aq) + 2e→ Cr (s) Cr3+ (aq) + 3e Cr (s) Cr+ (aq) + e Cr2+ (aq) Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 50H (aq) -0.28 If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. 1.92 O no maximum -0.913 -0.744 -0.407 -0.13 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Cu2+ (aq) + 2e- Cu (s) 0.3419 Cu2+ (ag) + e - Cu* (aq) 0.153 Note: write the half reaction as it would actually occur at the anode. Cu* (ag) + e - Cu (s) 0.521 F2 (9) + 2e 2F (aq) 2.866 Fe2+ (ag) + 2e- Fe (s) -0.447 Explanation Check Fe3+ (aq) + e - Fe2+ (aq) 0.771 1. Fe+ (aq) + 3e Fe (s) -0.037 MacBook Air 20 FI F12 D00 F4 F10 FS F6 F7 F8 F9 FI F2 F3

Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Steven S. Zumdahl, Susan A. Zumdahl
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 160MP
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A certain half-reaction has a standard reduction potential E =-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 0.80 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell. h Data Ag+ (aq) + e Ag (s) 0.7996 Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? yes, there is a minimum. Ered = v !! Al3+ (aq) + 3e-- Al (s) -1.676 Au+ (aq) + e Au (s) 1.692 If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. Au3+ (aq) + 3e - Au (s) 1.498 O no minimum Ba2+ (ag) + 2e - Ba (s) -2.912 Brz (1) + 2e- 2Br (aq) 1.066 Ca2+ (ag) + 2e - Ca (s) -2.868 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? E= Ov Cl2 (g) + 2e - 2CI- (aq) Co2+ (aq) + 2e- - Co (s) O yes, there is a maximum. 1.35827 red -0.28 If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. Co3+ (aq) + e- Co2+ (aq) 1.92 O no maximum Cr2+ (aq) + 2e→ Cr (s) Cr+ (aq) + 3e - Cr (s) Cr+ (aq) + e - Cr2+ (aq) -0.913 -0.744 -0.407 Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 5OH (aq) -0.13 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Cu2+ (aq) + 2e - Cu (s) 0.3419 Cu2+ (ag) + e - Cut (ag) 0.153 Note: write the half reaction as it would actually occur at the anode. Cu* (ag) + e→ Cu (s) 0.521 F2 (g) + 2e 2F (aq) 2.866 Fe2+ (aq) + 2e Fe (s) -0.447 Explanation Check Fe3+ (aq) + e-- Fe2+ (aq) 0.771 Fe+ (aq) + 3e Fe (s) -0.037 MacBook Air 20 esc 000 000 FA ►► F12 F8 F9 F10 FII F1 F2 F3 F5 F6 F7 24 % & *
Transcribed Image Text:A certain half-reaction has a standard reduction potential E =-1.25 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 0.80 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the cathode of the cell. h Data Ag+ (aq) + e Ag (s) 0.7996 Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? yes, there is a minimum. Ered = v !! Al3+ (aq) + 3e-- Al (s) -1.676 Au+ (aq) + e Au (s) 1.692 If so, check the "yes" box and calculate the minimum. Round your answer to 2 decimal places. If there is no lower limit, check the "no" box. Au3+ (aq) + 3e - Au (s) 1.498 O no minimum Ba2+ (ag) + 2e - Ba (s) -2.912 Brz (1) + 2e- 2Br (aq) 1.066 Ca2+ (ag) + 2e - Ca (s) -2.868 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? E= Ov Cl2 (g) + 2e - 2CI- (aq) Co2+ (aq) + 2e- - Co (s) O yes, there is a maximum. 1.35827 red -0.28 If so, check the "yes" box and calculate the maximum. Round your answer to 2 decimal places. If there is no upper limit, check the "no" box. Co3+ (aq) + e- Co2+ (aq) 1.92 O no maximum Cr2+ (aq) + 2e→ Cr (s) Cr+ (aq) + 3e - Cr (s) Cr+ (aq) + e - Cr2+ (aq) -0.913 -0.744 -0.407 Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 5OH (aq) -0.13 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the anode of this cell. Cu2+ (aq) + 2e - Cu (s) 0.3419 Cu2+ (ag) + e - Cut (ag) 0.153 Note: write the half reaction as it would actually occur at the anode. Cu* (ag) + e→ Cu (s) 0.521 F2 (g) + 2e 2F (aq) 2.866 Fe2+ (aq) + 2e Fe (s) -0.447 Explanation Check Fe3+ (aq) + e-- Fe2+ (aq) 0.771 Fe+ (aq) + 3e Fe (s) -0.037 MacBook Air 20 esc 000 000 FA ►► F12 F8 F9 F10 FII F1 F2 F3 F5 F6 F7 24 % & *
2H (aq) + 2e H2 (g) 0.000 2H20 (1) + 2e → H2 (g) + 20H (aq) -0.8277 H202 (aq) + 2H+ (aq) + 2e → 2H20 (1) 1.776 I2 (s) + 2e 21 (aq) 0.5355 2103 (aq) + 12H+ (aq) + 10e I2 (s) + 6H20 (1) 1.195 Mg2+ (aq) + 2e Mg (s) -2.372 Mn2+ (aq) + 2e Mn (s) -1.185 MnO2 () + 4H+ (aq) + 2e- - Mn2+ (aq) + 2H20 (1) 1.224 MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (1) 1.507 MnO4 (aq) + 2H20 (1) + 3e MnO2 (s) + 40H (aq) 0.595 HNO2 (aq) + H+ (aq) + e- NO (g) + H20 (1) 0.983 N2 (g) + 4H2O (1) + 4e- → 40H (aq) + N2H4 (aq) -1.16 NO3 (aq) + 4H+ (aq) + 3e NO (g) + 2H20 (I) 0.957 Na+ (aq) + e Na (s) -2.71 Ni2+ (ag) + 2e Ni (s) -0.257 1.229 02 (g) + 4H+ (aq) + 4e- → 2H20 (1) 0.401 02 (g) + 2H20 (1) + 4e¬ → 40H (aq) 0.695 02 (g) + 2H+ (aq) + 2e → H2O2 (aq) -0.1262 Pb2+ (ag) + 2e → Pb (s) -0.3588 PBSO4 (s) + H (aq) + 2e Pb (s) + HSO4 (aq) k Air 4) F12 F11 II F8 F9 F10 F7 %3D delete + II
Transcribed Image Text:2H (aq) + 2e H2 (g) 0.000 2H20 (1) + 2e → H2 (g) + 20H (aq) -0.8277 H202 (aq) + 2H+ (aq) + 2e → 2H20 (1) 1.776 I2 (s) + 2e 21 (aq) 0.5355 2103 (aq) + 12H+ (aq) + 10e I2 (s) + 6H20 (1) 1.195 Mg2+ (aq) + 2e Mg (s) -2.372 Mn2+ (aq) + 2e Mn (s) -1.185 MnO2 () + 4H+ (aq) + 2e- - Mn2+ (aq) + 2H20 (1) 1.224 MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (1) 1.507 MnO4 (aq) + 2H20 (1) + 3e MnO2 (s) + 40H (aq) 0.595 HNO2 (aq) + H+ (aq) + e- NO (g) + H20 (1) 0.983 N2 (g) + 4H2O (1) + 4e- → 40H (aq) + N2H4 (aq) -1.16 NO3 (aq) + 4H+ (aq) + 3e NO (g) + 2H20 (I) 0.957 Na+ (aq) + e Na (s) -2.71 Ni2+ (ag) + 2e Ni (s) -0.257 1.229 02 (g) + 4H+ (aq) + 4e- → 2H20 (1) 0.401 02 (g) + 2H20 (1) + 4e¬ → 40H (aq) 0.695 02 (g) + 2H+ (aq) + 2e → H2O2 (aq) -0.1262 Pb2+ (ag) + 2e → Pb (s) -0.3588 PBSO4 (s) + H (aq) + 2e Pb (s) + HSO4 (aq) k Air 4) F12 F11 II F8 F9 F10 F7 %3D delete + II
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