Flipping the reactant and product in the reaction Zn → Zn²+ + 2 e will generate a non-spontaneous, reduction half-reaction. The aforementioned half-reaction is non-spontaneous when coupled with the following reaction, 2 H+2 e→ H₂ which is called the "standard hydrogen electrode". Yet, coupling with which of the following half-reactions will make Zn²+ + 2 e → Zn spontaneous without having to flip it back (while its E* remains at -0.76 V)? Cu-Cu²+2 e (E = -0.34 V) Fe-Fe²+2 e (E = +0.45 V) © Cd-Cd²+2 e" (E +0.40 V) O Mg Mg² +2e (E = +2.4V) PbPb²+ + 2 e (E = +0.13 V) 1
Flipping the reactant and product in the reaction Zn → Zn²+ + 2 e will generate a non-spontaneous, reduction half-reaction. The aforementioned half-reaction is non-spontaneous when coupled with the following reaction, 2 H+2 e→ H₂ which is called the "standard hydrogen electrode". Yet, coupling with which of the following half-reactions will make Zn²+ + 2 e → Zn spontaneous without having to flip it back (while its E* remains at -0.76 V)? Cu-Cu²+2 e (E = -0.34 V) Fe-Fe²+2 e (E = +0.45 V) © Cd-Cd²+2 e" (E +0.40 V) O Mg Mg² +2e (E = +2.4V) PbPb²+ + 2 e (E = +0.13 V) 1
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter17: Electrochemistry
Section: Chapter Questions
Problem 59P
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![Flipping the reactant and product in the reaction Zn → Zn²+ + 2 e will generate a non-spontaneous,
reduction half-reaction. The aforementioned half-reaction is non-spontaneous when coupled with
the following reaction, 2 H+ + 2 e→ H₂ which is called the "standard hydrogen electrode". Yet,
coupling with which of the following half-reactions will make Zn²+ + 2 e → Zn spontaneous without
having to flip it back (while its E* remains at -0.76 V)?
Cu Cu²+ +2 e (E* = -0.34 V)
Fe-Fe2+2 e (E = +0.45 V)
Cd Cd²+2 e (E = +0.40 V)
-
O Mg Mg2+ + 2e (E* = +2.4V)
PbPb²+ + 2 e (E = +0.13V)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f4e6b6d-fbfa-48c7-a58f-531bc05ede9a%2F9d7f8e29-9774-4c43-99b1-aa963dfa35a8%2F8i10z98w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Flipping the reactant and product in the reaction Zn → Zn²+ + 2 e will generate a non-spontaneous,
reduction half-reaction. The aforementioned half-reaction is non-spontaneous when coupled with
the following reaction, 2 H+ + 2 e→ H₂ which is called the "standard hydrogen electrode". Yet,
coupling with which of the following half-reactions will make Zn²+ + 2 e → Zn spontaneous without
having to flip it back (while its E* remains at -0.76 V)?
Cu Cu²+ +2 e (E* = -0.34 V)
Fe-Fe2+2 e (E = +0.45 V)
Cd Cd²+2 e (E = +0.40 V)
-
O Mg Mg2+ + 2e (E* = +2.4V)
PbPb²+ + 2 e (E = +0.13V)
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