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Solved: Calculating E°red from E°cellFor the Zn–Cu2+

Chemistry: The Central Science | 13th Edition | ISBN: 9780321910417 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward; Matthew E. Stoltzfus ISBN: 9780321910417 77

Solution for problem 1PE Chapter 20.5SE

Chemistry: The Central Science | 13th Edition

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Chemistry: The Central Science | 13th Edition | ISBN: 9780321910417 | Authors: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward; Matthew E. Stoltzfus

Chemistry: The Central Science | 13th Edition

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Problem 1PE

Calculating E°red from E°cell

For the Zn–Cu2+ voltaic cell shown in Figure, we have

How is electrical balance maintained in the left beaker as Zn2+ are formed at the anode?

Figure A voltaic cell that uses a salt bridge to complete the electrical circuit.

Zn(s) + Cu2+(aq, 1 M) → Zn2+(aq, 1 M) + Cu(s) E°cell = 1.10 V

Given that the standard reduction potential of Zn2+ to Zn(s) is –0.76 V, calculate the E°red for the reduction of Cu2+ to Cu:

Cu2+(aq, 1 M) + 2 e– → Cu(s)

A voltaic cell based on the reaction

2 Eu2+(aq) + Ni2+(aq) → 2 Eu3+(aq) + Ni(s) generates E°cell = 0.07 V. Given the standard reduction potential of Ni2+ given in Table what is the standard reduction potential for the reaction Eu3+(aq) + e–→ Eu2+(aq)?

Table Standard Reduction Potentials in Water at 25°c

Eºred(V)

Reduction Half–Reaction

+2.87

F2(g) + 2 e–→.2 F–(aq)

+1.51

MnO4–(aq) + 8 H+(aq) + 5 e–→.Mn2+(aq) + 4 H2O(l)

+1.36

Cl2(g) + 2 e–→.2 Cl–(aq)

+1.33

Cr2O72–(aq) + 14 H+(aq) + 6 e–→.2 Cr3+(aq) + 7 H2O(l)

+1.23

O2(g) + 4 H+(aq) + 4 e–→.2 H2O(l)

+1.06

Br2(l) + 2 e–→.2 Br–(aq)

+0.96

NO3–(aq) + 4 H+(aq) + 3 e–→.NO(g) + 2 H2O(l)

+0.80

Ag+(aq) + e–→.Ag(s)

+0.77

Fe3+(aq) + e–→.Fe2+(aq)

+0.68

O2(g) + 2 H+(aq) + 2 e–→.H2O2(aq)

+0.59

MnO4–(aq) + 2 H2O(l) + 3 e–→.MnO2(s) + 4 OH–(aq)

+0.54

l2(s) + 2 e–→.2 l–(aq)

+0.40

O2(g) + 2 H2O(l) + 4 e–→.4 OH–(aq)

+0.34

Cu2+(aq) + 2 e–→.Cu

0 [defined]

2 H+(aq) + 2 e–→.H2(g)

–0.28

Ni2+(aq) + 2 e–→.Ni(s)

–0.44

Fe2+ (aq) + 2 e–→.Fe(s)

–0.76

Zn2+(aq) + 2 e–→.Zn(s)

–0.83

2 H2O(l) + 2 e–→.H2(g) + 2 OH–(aq)

–1.66

Al3+(aq) + 3 e–→.Al(s)

–2.71

Na+(aq) + e–→.Na(s)

–3.05

Li+(aq) + e–→.Li(s)

(a) –0.35 V, (b) 0.35 V, (c) –0.21 V, (d) 0.21 V, (e) 0.07 V.

Step-by-Step Solution:
Step 1 of 3

Thermodynamics & Microstates • thermodynamics - describes the energy of a system —1st law: energy is conserved and cannot be created or destroyed (∆Euniv=∆Esys + ∆Esurr) —2nd law: entropy of an isolated system always increases during spontaneous processes • thermodynamic process—>heat (q) and work (w) • spontaneous - once started a reaction proceeds as long...

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Chapter 20.5SE, Problem 1PE is Solved
Step 3 of 3

Textbook: Chemistry: The Central Science
Edition: 13
Author: Theodore E. Brown; H. Eugene LeMay; Bruce E. Bursten; Catherine Murphy; Patrick Woodward; Matthew E. Stoltzfus
ISBN: 9780321910417

Chemistry: The Central Science was written by and is associated to the ISBN: 9780321910417. The full step-by-step solution to problem: 1PE from chapter: 20.5SE was answered by , our top Chemistry solution expert on 09/04/17, 09:30PM. Since the solution to 1PE from 20.5SE chapter was answered, more than 285 students have viewed the full step-by-step answer. The answer to “Calculating E°red from E°cellFor the Zn–Cu2+ voltaic cell shown in Figure, we haveHow is electrical balance maintained in the left beaker as Zn2+ are formed at the anode?Figure A voltaic cell that uses a salt bridge to complete the electrical circuit. Zn(s) + Cu2+(aq, 1 M) ? Zn2+(aq, 1 M) + Cu(s) E°cell = 1.10 VGiven that the standard reduction potential of Zn2+ to Zn(s) is –0.76 V, calculate the E°red for the reduction of Cu2+ to Cu:Cu2+(aq, 1 M) + 2 e– ? Cu(s)A voltaic cell based on the reaction2 Eu2+(aq) + Ni2+(aq) ? 2 Eu3+(aq) + Ni(s) generates E°cell = 0.07 V. Given the standard reduction potential of Ni2+ given in Table what is the standard reduction potential for the reaction Eu3+(aq) + e–? Eu2+(aq)?Table Standard Reduction Potentials in Water at 25°cEºred(V)Reduction Half–Reaction+2.87F2(g) + 2 e–?.2 F–(aq)+1.51MnO4–(aq) + 8 H+(aq) + 5 e–?.Mn2+(aq) + 4 H2O(l)+1.36Cl2(g) + 2 e–?.2 Cl–(aq)+1.33Cr2O72–(aq) + 14 H+(aq) + 6 e–?.2 Cr3+(aq) + 7 H2O(l)+1.23O2(g) + 4 H+(aq) + 4 e–?.2 H2O(l)+1.06Br2(l) + 2 e–?.2 Br–(aq)+0.96NO3–(aq) + 4 H+(aq) + 3 e–?.NO(g) + 2 H2O(l)+0.80Ag+(aq) + e–?.Ag(s)+0.77Fe3+(aq) + e–?.Fe2+(aq)+0.68O2(g) + 2 H+(aq) + 2 e–?.H2O2(aq)+0.59MnO4–(aq) + 2 H2O(l) + 3 e–?.MnO2(s) + 4 OH–(aq)+0.54l2(s) + 2 e–?.2 l–(aq)+0.40O2(g) + 2 H2O(l) + 4 e–?.4 OH–(aq)+0.34Cu2+(aq) + 2 e–?.Cu0 [defined]2 H+(aq) + 2 e–?.H2(g)–0.28Ni2+(aq) + 2 e–?.Ni(s)–0.44Fe2+ (aq) + 2 e–?.Fe(s)–0.76Zn2+(aq) + 2 e–?.Zn(s)–0.832 H2O(l) + 2 e–?.H2(g) + 2 OH–(aq)–1.66Al3+(aq) + 3 e–?.Al(s)–2.71Na+(aq) + e–?.Na(s)–3.05Li+(aq) + e–?.Li(s)(a) –0.35 V, (b) 0.35 V, (c) –0.21 V, (d) 0.21 V, (e) 0.07 V.” is broken down into a number of easy to follow steps, and 258 words. This textbook survival guide was created for the textbook: Chemistry: The Central Science, edition: 13. This full solution covers the following key subjects: . This expansive textbook survival guide covers 305 chapters, and 6351 solutions.

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