2H+ (aq) + 2e + H2 (g)0.0002H20 (1) + 2e - H2 (g) + 20H (aq)-0.8277H2O2 (aq) + 2H+ (aq) + 2e¯ → 2H20 (1)1.77612 (s) + 2e 21- (aq)0.53552103 (aq) + 12H (aq) + 10e I2 (s) + 6H20 (I)1.195A Mg2+ (aq) + 2e Mg (s)-2.372Mn2+ (aq) + 2e Mn (s)-1.1851.224MnO2 (s) + 4H+ (aq) + 2e- Mn2+ (aq) + 2H20 (I)1.507MnO4 (aq) + 8H+ (aq) + 5e Mn2+ (aq) + 4H20 (I)0.595Mn04 (aq) + 2H20 (1) + 3e MnO2 (s) + 40H (aq)0.983HNO2 (aq) + H+ (aq) + e- → NO (g) + H20 (1)-1.16N2 (g) + 4H2O (1) + 4e¯ → 40H (aq) + N2H4 (aq)0.957NO3 (aq) + 4H+ (aq) + 3e → NO (g) + 2H20 (I)-2.71Na+ (aq) + e Na (s)-0.257Ni2+ (ag) + 2e Ni (s)1.229O2 (g) + 4H+ (aq) + 4e → 2H2O (I)0.401O2 (g) + 2H20 (I) + 4e¯ -→ 40H¯ (aq)0.695O2 (g) + 2H+ (aq) + 2e¯ → H2O2 (aq)-0.1262Pb2+ (aq) + 2e → Pb (s)-0.3588PbSO4 (s) + Ht (aq) + 2e → Pb (s) + HSO4 (aq)0.172HSO4 (aq) + 3H+ (aq) + 2e H2SO3 (aq) + H2O (1)2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Privacy Accessibilityok AirF11F12F9F10F7F8delete A certain half-reaction has a standard reduction potential E=+0.21 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that mustprovide at least 1.30 V of electrical power. The cell will operate under standard conditions. l DataNote for advanced students: assume the engineer requires this half-reaction to happen at tHalf-ReactionE° (V)Ag+ (aq) + e + Ag (s)A13+ (aq) + 3e - Al (s)Au+ (aq) + e - Au (s)Au3+ (aq) + 3e - Au (s)0.7996-1.676Is there a minimum standard reductionpotential that the half-reaction used atthe cathode of this cell can have?O yes, there is a minimum.E = Ov1.6921.498If so, check the "yes" box and calculatethe minimum. Round your answer to 2decimal places. If there is no lowerlimit, check the "no" box.Ba2+ (aq) + 2e- - Ba (s)-2.912O no minimumBr2 (1) + 2e - 2Br (aq)1.066Ca2+ (aq) + 2e-- Ca (s)-2.868Cl2 (g) + 2e - 2CI (aq)Co2+ (aq) + 2e-- Co (s)Co3+ (aq) + e - Co2+ (aq)Cr2+ (aq) + 2e → Cr (s)Cr3+ (aq) + 3e + Cr (s)1.35827Is there a maximum standard reductionpotential that the half-reaction used atthe cathode of this cell can have?O yes, there is a maximum.= OvE-0.28red1.92If so, check the "yes" box and calculatethe maximum. Round your answer to 2decimal places. If there is no upperlimit, check the "no" box.-0.913O no maximum-0.744Cr3+ (aq) + e - Cr2+ (aq)Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 50H (aq)-0.407-0.13Cu2+ (aq) + 2e - Cu (s)0.3419By using the information in the ALEKSData tab, write a balanced equationdescribing a half reaction that could beused at the cathode of this cell.Cu2+ (aq) + e- - Cut (aq)0.153Cu+ (aq) + e - Cu (s)0.521Note: write the half reaction as it wouldactually occur at the cathode.F2 (9) + 2e - 2F (aq)2.866Fe2+ (aq) + 2e Fe (s)Fe+ (ag) + e - Fe2+ (aq)Fe3+ (aq) + 3e- Fe (s)-0.4470.771-0.037ExplanationChecko 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Privacy AccessibilityMacBook AirFIV20 F3888F4escF10F7FIF2@232$%&*delete2345789.

Given: Introduction: The potential of the cell at standard conditions should be at least 1.30 V.…

A certain half-reaction has a standard reduction potential E=+0.21 V. An engineer proposes using this half-reaction at the anode of a galvanic cell that must provide at least 1.30 V of electrical power. The cell will operate under standard conditions. nla Data Note for advanced students: assume the engineer requires this half-reaction to happen at t Half-Reaction E° (V) Ag+ (aq) + e → Ag (s) Al3+ (aq) + 3e-- Al (s) 0.7996 -1.676 Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have? yes, there is a minimum. E = Ov red Au+ (aq) + e - Au (s) 1.692 Au3+ (aq) + 3e - Au (s) Ba2+ (aq) + 2e - Ba (s) 1.498 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. -2.912 O no minimum Br2 (1) + 2e- + 2Br (aq) 1.066 Ca2+ (aq) + 2e- - Ca (s) -2.868 Cl2 (9) + 2e- 2CI- (aq) 1.35827 Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have? = Ov red O yes, there is a maximum. Co2+ (aq) + 2e-- Co (s) -0.28 !! Co3+ (aq) + e- - Co2+ (aq) 1.92 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. Cr2+ (aq) + 2e- Cr (s) Cr3+ (ag) + 3e Cr (s) -0.913 O no maximum -0.744 Cr+ (aq) + e- - Cr2+ (aq) --0.407 Cro42- (aq) + 4H20 (1) + 3e - Cr(OH)3 (s) + 50H (aq) -0.13 Cu2+ (aq) + 2e - Cu (s) 0.3419 By using the information in the ALEKS Data tab, write a balanced equation describing a half reaction that could be used at the cathode of this cell. Cu2+ (aq) + e Cut (aq) Cut (aq) + e - Cu (s) 0.153 0.521 Note: write the half reaction as it would F2 (9) + 2e - 2F (aq) 2.866 actually occur at the cathode. Fe2+ (aq) + 2e - Fe (s) -0.447 Fe+ (aq) + e - Fe2+ (aq) 0.771

10th Edition

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...