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Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e¯ → Pb(s) E° = -0.126 V Fe3*(aq) + e¯ → Fe2*(s) E° = +0.771 V A Pb?*(aq) + 2 Fe²*(s) → Pb(s) + 2 Fe³*(aq); 'cell = +0.897 V B Pb2*(aq) + Fe²*(s) → Pb(s) + Fe³*(aq); cell = +0.645 V (©) Pb(s) + 2 Fe3*(aq) → Pb²*(aq) + 2 Fe²*(s); * cell = +1.416 V D Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); 'cell = +0.897 V Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); F. cell = +0.645 V

Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e¯ → Pb(s) E° = -0.126 V Fe3*(aq) + e¯ → Fe2*(s) E° = +0.771 V A Pb?*(aq) + 2 Fe²*(s) → Pb(s) + 2 Fe³*(aq); 'cell = +0.897 V B Pb2*(aq) + Fe²*(s) → Pb(s) + Fe³*(aq); cell = +0.645 V (©) Pb(s) + 2 Fe3*(aq) → Pb²*(aq) + 2 Fe²*(s); * cell = +1.416 V D Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); 'cell = +0.897 V Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); F. cell = +0.645 V

Chemistry
Chemistry
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...
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Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e- → Pb(s) E° = -0.126 V Fe3*(aq) + e- → Fe2*(s) E° = +0.771 V 000 Pb?*(aq) + 2 Fe²*(s) Pb(s) + 2 Fe*(aq); сеll = +0.897 V Pb2*(aq) + Fe2*(s) → Pb(s) + Fe*(aq); = +0.645 V cell Pb(s) + 2 Fe*(aq) → Pb²*(aq) + 2 Fe²*(s); = +1.416 V 'cell Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); p cell = +0.897 V E Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); = +0.645 V cell
Transcribed Image Text:Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e- → Pb(s) E° = -0.126 V Fe3*(aq) + e- → Fe2*(s) E° = +0.771 V 000 Pb?*(aq) + 2 Fe²*(s) Pb(s) + 2 Fe*(aq); сеll = +0.897 V Pb2*(aq) + Fe2*(s) → Pb(s) + Fe*(aq); = +0.645 V cell Pb(s) + 2 Fe*(aq) → Pb²*(aq) + 2 Fe²*(s); = +1.416 V 'cell Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); p cell = +0.897 V E Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); = +0.645 V cell
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