The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined. Concept introduction: According to the first law of thermodynamics , the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system. The equation is as follows. ΔU = Q - W ΔU = Change in internal energy Q = Heat added to the system W=Work done by the system In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell. ΔG 0 = -nFE 0 n = Number of moles transferred per mole of reactant and products F = Faradayconstant=96485C/mol E 0 = Volts = Work(J)/Charge(C) The relation between standard cell potential and equilibrium constant is as follows. lnK = nE 0 0 .0257 at 298K The relation between solubility product K sp and equilibrium constant is as follows. K sp = e +lnK

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Answer 1

Question

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

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Answer 2

Question

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

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Answer 3

Question

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

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Answer 4

Question

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

Expert Solution & Answer

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Answer 5

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

Answer 6

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

Answer 7

Chapter 19, Problem 76GQ

Interpretation Introduction

Interpretation:

The mass of Zinc consumed when 35 amp – hours are drawn from the cell has to be determined.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

$ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system$

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

$ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol E0= Volts = Work(J)/Charge(C)$

The relation between standard cell potential and equilibrium constant is as follows.

$lnK = nE00.0257 at 298K$

The relation between solubility product $Ksp$ and equilibrium constant is as follows.

$Ksp= e+lnK$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 19.1 - A common laboratory analysis for iron is to...Ch. 19.1 - Prob. 19.2CYU Ch. 19.2 - Describe how to set up a voltaic cell using the...Ch. 19.2 - The following overall chemical reaction occurs in...Ch. 19.4 - (a) Rank the following metals in their ability to...Ch. 19.5 - A voltaic cell is set up with an aluminum...Ch. 19.5 - The half-cells Ag+(aq. 1.0 M)|Ag(s) and H+(aq, ?...Ch. 19.6 - Prob. 19.8CYU Ch. 19.6 - Calculate the equilibrium constant at 25 C for the...Ch. 19.7 - Predict the chemical reactions that will occur at...

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