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A certain half-reaction has a standard reduction potential E=+0.40 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.10 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happ nn Data Half-Reaction E° (V) 2.866 F2 (g) + 2e- 2F (aq) Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? Co3+ (aq) + e-- Co2+ (aq) 1.92 O yes, there is a minimum. %3D red 1.776 H202 (aq) + 2H+ (aq) + 2e - 2H20 (1) 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. Au+ (aq) + e - Au (s) 1.692 MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (1) Au3+ (aq) + 3e- - Au (s) O no minimum 1.507 1.498 1.35827 Clz (g) + 2e 2CI (ag) 1.229 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? 02 (9) + 4H* (aq) + 4e + 2H20 (1) O yes, there is a maximum. E red 1.224 Mno2 (s) + 4H+ (aq) + 2e+ Mn2+ (aq) + 2H2O (1) 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. 1.195 2103 (aq) + 12H+ (aq) + 10e - 12 (s) + 6H20 (1) O no maximum Brz (1) + 2e 2Br (aq) 1.066 0.991 Vo2+ (aq) + 2H (aq) + e- Vo2+ (aq) + H20 (1) 0.983 HNO2 (aq) + H+ (aq) + e - NO (g) + H2O (1) 0.957 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. NO3 (aq) + 4H+ (ag) + 3e - NO (g) + 2H20 (1) 0.7996 Ag+ (aq) + e- Ag (s) 0.771 Fe3+ (ag) + e - Fe2+ (aq) Note: write the half reaction as it would actually occur at the anode. 0.695 O2 (9) + 2H* (aq) + 2e - H202 (aq) 0.595 Mno4" (aq) + 2H20 (1) + 3e - MnO2 (s) + 40H (aq) 0.5355 I2 (s) + 2e - 21 (aq) Explanation Check 0.521 Cut (aq) + e → Cu (s)

A certain half-reaction has a standard reduction potential E=+0.40 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.10 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happ nn Data Half-Reaction E° (V) 2.866 F2 (g) + 2e- 2F (aq) Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? Co3+ (aq) + e-- Co2+ (aq) 1.92 O yes, there is a minimum. %3D red 1.776 H202 (aq) + 2H+ (aq) + 2e - 2H20 (1) 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. Au+ (aq) + e - Au (s) 1.692 MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (1) Au3+ (aq) + 3e- - Au (s) O no minimum 1.507 1.498 1.35827 Clz (g) + 2e 2CI (ag) 1.229 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? 02 (9) + 4H* (aq) + 4e + 2H20 (1) O yes, there is a maximum. E red 1.224 Mno2 (s) + 4H+ (aq) + 2e+ Mn2+ (aq) + 2H2O (1) 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. 1.195 2103 (aq) + 12H+ (aq) + 10e - 12 (s) + 6H20 (1) O no maximum Brz (1) + 2e 2Br (aq) 1.066 0.991 Vo2+ (aq) + 2H (aq) + e- Vo2+ (aq) + H20 (1) 0.983 HNO2 (aq) + H+ (aq) + e - NO (g) + H2O (1) 0.957 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. NO3 (aq) + 4H+ (ag) + 3e - NO (g) + 2H20 (1) 0.7996 Ag+ (aq) + e- Ag (s) 0.771 Fe3+ (ag) + e - Fe2+ (aq) Note: write the half reaction as it would actually occur at the anode. 0.695 O2 (9) + 2H* (aq) + 2e - H202 (aq) 0.595 Mno4" (aq) + 2H20 (1) + 3e - MnO2 (s) + 40H (aq) 0.5355 I2 (s) + 2e - 21 (aq) Explanation Check 0.521 Cut (aq) + e → Cu (s)

Chemistry
Chemistry
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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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A certain half-reaction has a standard reduction potential Ed=+0.40 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.10 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happ nn Data Half-Reaction E° (V) 2.866 F2 (g) + 2e-→ 2F (ag) Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? = Ov Co3+ (aq) + e-- Co2+ (aq) 1.92 O yes, there is a minimum. Ered 1.776 H202 (aq) + 2H+ (aq) + 2e - 2H20 (1) 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. 1.692 Au+ (aq) + e Au (s) O no minimum 1.507 MnO4 (aq) + 8H* (aq) + 5e → Mn2+ (aq) + 4H20 (1) Au3+ (aq) + 3e- Au (s) 1.498 1.35827 Cl2 (g) + 2e 2CI- (aq) 1.229 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? = Dv 02 (g) + 4H* (aq) + 4e - 2H20 (1) O yes, there is a maximum. "red 1.224 Mno2 (s) + 4H+ (aq) + 2e Mn2+ (aq) + 2H2O (1) 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. 1.195 2103 (aq) + 12H+ (aq) + 10e → I2 (s) + 6H2O (1) O no maximum Brz (1) + 2e 2Br (aq) 1.066 0.991 Vo2+ (aq) + 2H+ (aq) + e- Vo2+ (aq) + H20 (1) 0.983 HNO2 (aq) + H* (aq) + e - NO (g) + H2O (1) 0.957 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. NO3 (aq) + 4H+ (aq) + 3e - NO (g) + 2H20 (1) 0.7996 Ag+ (aq) + e - Ag (s) 0.771 Fe3+ (ag) + e - Fe2+ (aq) Note: write the half reaction as it would 0.695 actually occur at the anode. O2 (9) + 2H* (aq) + 2e - H202 (aq) 0.595 Mno4" (aq) + 2H20 (1) + 3e - MnO2 (s) + 40H (aq) I2 (s) + 2e - 21- (aq) - 21 0.5355 Explanation Check 0.521 Cut (aq) + e → Cu (s) MacBook Air F12 F10 F11 F8 F9 F6 F7 O00 F4 F5 F3 #3 $4 & @
Transcribed Image Text:A certain half-reaction has a standard reduction potential Ed=+0.40 V. An engineer proposes using this half-reaction at the cathode of a galvanic cell that must provide at least 1.10 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happ nn Data Half-Reaction E° (V) 2.866 F2 (g) + 2e-→ 2F (ag) Is there a minimum standard reduction potential that the half-reaction used at the anode of this cell can have? = Ov Co3+ (aq) + e-- Co2+ (aq) 1.92 O yes, there is a minimum. Ered 1.776 H202 (aq) + 2H+ (aq) + 2e - 2H20 (1) 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. 1.692 Au+ (aq) + e Au (s) O no minimum 1.507 MnO4 (aq) + 8H* (aq) + 5e → Mn2+ (aq) + 4H20 (1) Au3+ (aq) + 3e- Au (s) 1.498 1.35827 Cl2 (g) + 2e 2CI- (aq) 1.229 Is there a maximum standard reduction potential that the half-reaction used at the anode of this cell can have? = Dv 02 (g) + 4H* (aq) + 4e - 2H20 (1) O yes, there is a maximum. "red 1.224 Mno2 (s) + 4H+ (aq) + 2e Mn2+ (aq) + 2H2O (1) 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. 1.195 2103 (aq) + 12H+ (aq) + 10e → I2 (s) + 6H2O (1) O no maximum Brz (1) + 2e 2Br (aq) 1.066 0.991 Vo2+ (aq) + 2H+ (aq) + e- Vo2+ (aq) + H20 (1) 0.983 HNO2 (aq) + H* (aq) + e - NO (g) + H2O (1) 0.957 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. NO3 (aq) + 4H+ (aq) + 3e - NO (g) + 2H20 (1) 0.7996 Ag+ (aq) + e - Ag (s) 0.771 Fe3+ (ag) + e - Fe2+ (aq) Note: write the half reaction as it would 0.695 actually occur at the anode. O2 (9) + 2H* (aq) + 2e - H202 (aq) 0.595 Mno4" (aq) + 2H20 (1) + 3e - MnO2 (s) + 40H (aq) I2 (s) + 2e - 21- (aq) - 21 0.5355 Explanation Check 0.521 Cut (aq) + e → Cu (s) MacBook Air F12 F10 F11 F8 F9 F6 F7 O00 F4 F5 F3 #3 $4 & @
d conditions. - olo Data 02 (g) + 2H20 (1) + 4e 40H- (aq) 0.401 Cu2+ (aq) + 2e Cu (s) 0.3419 HSO4 (aq) + 3H+ (aq) + 2e → H2SO3 (aq) + H20 (1) 0.172 Cu2+ (aq) + e Cut (aq) 0.153 Sn4+ (aq) + 2e Sn2+ (aq) 0.151 2H+ (aq) + 2e H2 (9) 0.000 Fe3+ (aq) + 3e- → Fe (s) -0.037 Pb2+ (aq) + 2e Pb (s) -0.1262 CrO42- (aq) + 4H20 (1) + 3e¯ → Cr(OH)3 (s) + 5OH (aq) -0.13 Sn2+ (aq) + 2e → Sn (s) -0.1375 Ni2+ (aq) + 2e Ni (s) -0.257 Co2+ (aq) + 2e Co (s) -0.28 PbSO4 (s) + H+ (aq) + 2e → Pb (s) + HSO4- (aq) -0.3588 Cr3+ (aq) + e-→ Cr2+ (aq) -0.407 Fe2+ (ag) + 2e Fe (s) -0.447 Cr3+ (aq) + 3e Cr (s) -0.744 Zn2+ (aq) + 2e Zn (s) -0.7618 2H20 (1) + 2e- → H2 (g) + 20H (aq) -0.8277 Cr2+ (aq) + 2e Cr (s) -0.913 N2 (g) + 4H20 (1) + 4e¯ → 40H (aq) + N2H4 (aq) -1.16 Mn2+ (aq) + 2e Mn (s) -1.185 Al3+ (aq) + 3e- → Al (s) -1.676 acBook Air F8 F9 F10 F11 F12 8 delete + || P
Transcribed Image Text:d conditions. - olo Data 02 (g) + 2H20 (1) + 4e 40H- (aq) 0.401 Cu2+ (aq) + 2e Cu (s) 0.3419 HSO4 (aq) + 3H+ (aq) + 2e → H2SO3 (aq) + H20 (1) 0.172 Cu2+ (aq) + e Cut (aq) 0.153 Sn4+ (aq) + 2e Sn2+ (aq) 0.151 2H+ (aq) + 2e H2 (9) 0.000 Fe3+ (aq) + 3e- → Fe (s) -0.037 Pb2+ (aq) + 2e Pb (s) -0.1262 CrO42- (aq) + 4H20 (1) + 3e¯ → Cr(OH)3 (s) + 5OH (aq) -0.13 Sn2+ (aq) + 2e → Sn (s) -0.1375 Ni2+ (aq) + 2e Ni (s) -0.257 Co2+ (aq) + 2e Co (s) -0.28 PbSO4 (s) + H+ (aq) + 2e → Pb (s) + HSO4- (aq) -0.3588 Cr3+ (aq) + e-→ Cr2+ (aq) -0.407 Fe2+ (ag) + 2e Fe (s) -0.447 Cr3+ (aq) + 3e Cr (s) -0.744 Zn2+ (aq) + 2e Zn (s) -0.7618 2H20 (1) + 2e- → H2 (g) + 20H (aq) -0.8277 Cr2+ (aq) + 2e Cr (s) -0.913 N2 (g) + 4H20 (1) + 4e¯ → 40H (aq) + N2H4 (aq) -1.16 Mn2+ (aq) + 2e Mn (s) -1.185 Al3+ (aq) + 3e- → Al (s) -1.676 acBook Air F8 F9 F10 F11 F12 8 delete + || P
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