Chemistry
Chemistry
9th Edition
ISBN: 9781133611097
Author: Steven S. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 18, Problem 85E

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed?

a. Crystals of I2 are added to a solution of NaCl.

b. Cl2 gas is bubbled into a solution of NaI.

c. A silver wire is placed in a solution of CuCl2

d. An acidic solution of FeSO4 is exposed to air.

For the reactions that occur, write a balanced equation and calculate Chapter 18, Problem 85E, Under standard conditions, what reaction occurs, if any, when each of the following operations is ,ΔG°, and K at 25°C.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the E° , ΔG° and K for the possible reaction are to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

To determine: The reaction, if any, when crystals of I2 are added to solution of NaCl .

Answer to Problem 85E

  1. a. On the addition of crystals of iodine to NaCl no reaction occurs.

Explanation of Solution

The reduction potential value for I2 is,

I2+2e2IE°=0.52V

The reduction potential value for Cl2 is,

Cl2+2e2ClE°=1.36V

As the value of reduction potential for iodine is lower than that of chlorine. So, it won’t be possible for iodine to replace ions from its solution.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the E° , ΔG° and K for the possible reaction are to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

To determine: The reaction that takes place when Cl2 gas is bubbled into a solution of NaI .

Answer to Problem 85E

  1. b. The value of E° is 0.82V_ , the value of ΔG° is -158.23kJ_ and the value of K is when Cl2 is bubbled into a solution of NaI 1096.47_ .

Explanation of Solution

Explanation

The reduction potential value for I2 is,

I2+2e2IE°=0.52V

The reduction potential value for Cl2 is,

Cl2+2e2ClE°=1.36V

As the reduction potential of chlorine is greater than iodine. Therefore, it is possible for chlorine to replace iodide ion from its solution and this leads to the liberation of iodine gas.

The reaction taking place at cathode is,

Cl2+2e2ClEred°=1.36V

The reaction taking place at anode is,

2II2+2eEox°=0.54V

Add both the oxidation and reduction half-reaction,

Cl2+2e2Cl2II2+2e

The final equation is,

Cl2+2I2Cl+I2

The value of E° is calculated as,

E°=Eox+Ered=0.54V+1.36V=0.82V_

The value of E° is calculated as 0.82V_ .

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

The reaction involves the transfer of 2 moles of electrons.

Substitute the values in above equation,

ΔG°=(2mole)(96,485Cmole-)(0.82JC)=158,235.4J=-158.23kJ_

The value of ΔG° is calculated as -158.23kJ_

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

ΔG°=RTlnK=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • R is the universal gas constant.
  • T is the temperature in Kelvin.
  • K is the equilibrium constant.

Rearranged equation at 25°C is obtained as,

logK=((n)(E°)0.0591)

Substitute the obtained values in above equation,

logK=((2)(0.82)0.0591)=27.75K=1027.75=5.62×1027_

The value of K is 5.62×1027_ .

(c)

Answer:

c. There occurs no reaction when a silver wire is placed in a solution of CuCl2 .

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the E° , ΔG° and K for the possible reaction are to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

To determine: The reaction, if any, when a silver wire is placed into a solution of CuCl2

Answer to Problem 85E

c. There occurs no reaction when a silver wire is placed in a solution of CuCl2 .

Explanation of Solution

The reduction potential value for Ag+ is,

Ag++eAgE°=0.80V

Write it in reverse order,

AgAg++eE°ox=0.80V

The reduction potential value for Cu2+ is,

Cu2++2eCuE°red=0.34V

The value of E° is calculated as,

E°=E°ox+E°red=0.80V+0.34V=-0.64V_

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

As the value of E°cell is negative, therefore the value of ΔG° is positive. It means it corresponds to a non-spontaneous reaction.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Various questions based on the displacement of one ion from its solution by other are to be answered and the E° , ΔG° and K for the possible reaction are to be calculated.

Concept introduction:

The relationship between reduction potential and standard reduction potential value and activities of species present in an electrochemical cell at a given temperature is given by the Nernst equation.

The value of Ecell is calculated using Nernst formula,

E=E°(RTnF)ln(Q)

At room temperature the above equation is specifies as,

E=E°(0.0591n)log(Q)

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

To determine: The reaction, if any, an acidic solution of FeSO4 is exposed to air.

Answer to Problem 85E

d. The value of E° is 0.46V_ , the value of ΔG° is -177.53kJ_ and the value of K is 1.34×1031_ when an acidic solution of FeSO4 is exposed to air.

Explanation of Solution

The reduction potential value for O2 is,

O2+4H++4e2H2OE°=1.23V

The reduction potential value for Fe3+ is,

Fe3++eFe2+E°=0.77V

The reaction taking place at cathode is,

O2+4H++4e2H2OE°red=1.23V

The reaction taking place at anode is,

Fe2+Fe3++eE°ox=0.77V

Multiply the oxidation half-reaction with a coefficient of 4 and then add both the oxidation and reduction half-reaction,

O2+4H++4e2H2O4Fe2+4Fe3++4e

The final equation is,

4Fe2++O2+4H+4Fe3++2H2O

The value of E° is calculated as,

E°=Eox+Ered=0.77V+1.23V=0.46V_

The value of E° is calculated as 0.46V_ .

The relationship between cell potential and Gibbs free energy change is given by the formula,

ΔG°=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • n is the number of electrons involved in the reaction.
  • F is the Faraday’s constant.
  • E°cell is the cell potential at the standard condition.

The reaction involves the transfer of 4 moles of electrons.

Substitute the values in above equation,

ΔG°=(4mole)(96,485Cmole-)(0.46JC)=17,7532.4J=-177.53kJ_

The value of ΔG° is calculated as -177.53kJ_

The relationship between Gibbs free energy change and equilibrium constant is given by the formula,

ΔG°=RTlnK=nFE°cell

Where,

  • ΔG° is the Gibbs free energy change at the standard conditions.
  • R is the universal gas constant.
  • T is the temperature in Kelvin.
  • K is the equilibrium constant.

Rearranged equation at 25°C is obtained as,

logK=((n)(E°)0.0591)

Substitute the obtained values in above equation,

logK=((4)(0.46)0.0591)=31.13K=1031.13=1.34×1031_

The value of K is calculated as 1.34×1031_ .

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Chapter 18 Solutions

Chemistry

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