(a) Interpretation: To show the reaction is likely to go to completion or very nearly so. Concept introduction: Reaction which have K eq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate K eq . E cell 0 = R T n F ln K eq Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, K eq represents equilibrium constant, and E 0 cell represents standard electrode potential. Also, R = 8.314 J K − 1 mol − 1 F = 94580 T = 298 K Putting the values, E cell 0 = R T n F ln K eq E cell 0 = 8.314 J K − 1 mol − 1 × 298 K n × 96480 ln K eq Or, E cell 0 = 0.0257 n ln K eq ....... (1)

Question

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Answer 1

Question

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

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6. Consider the following exothermic reaction below. 2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq) a. If Cul is added, there will be a shift left/shift right/no shift (circle one). b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one). c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one). d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle one). Hint: one of the reaction species is more soluble in hexane than in water. e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one). f. Which of the changes above will change the equilibrium constant, K?

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Answer 2

Question

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 6

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 7

Chapter 22, Problem 39E

Interpretation Introduction

(a)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 8

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 9

Interpretation Introduction

(b)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 10

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 11

Interpretation Introduction

(c)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 12

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 13

Interpretation Introduction

(d)

Interpretation:

To show the reaction is likely to go to completion or very nearly so.

Concept introduction:

Reaction which have Keq value greater than 1000 then they are considered to be essentially quantitative. Therefore, below expression is used to calculate Keq.

Ecell0=RTnFlnKeq

Here, R represents gas constant, T represents temperature, n represents number of electron, F represents Faraday’s constant, Keq represents equilibrium constant, and E0cell represents standard electrode potential.

Also,

R=8.314 J K−1 mol−1F=94580T=298 K

Putting the values,

E cell0=RTnFlnKeqE cell0=8.314 J K −1 mol −1×298 Kn×96480 lnKeq

Or,

Ecell0=0.0257nlnKeq ....... (1)

Answer 14

Expert Solution & Answer

Answer 15

Expert Solution & Answer

Answer 16

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See solution

Answer 17

Ch. 22 - Give the formula of the stable fluride by Li, Be,...Ch. 22 - Fluorine is able to stabilize element’s in very...Ch. 22 - Prob. 3E Ch. 22 - Prob. 4E Ch. 22 - Prob. 5E Ch. 22 - Prob. 6E Ch. 22 - Prob. 7E Ch. 22 - Use VSEPR theory to predict the probable geometric...Ch. 22 - Prob. 9E Ch. 22 - Prob. 10E

Solution Summary: The author explains that the reaction is likely to go to completion or very nearly so.

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