Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 8, Problem 8.19E
Interpretation Introduction

(a)

Interpretation:

The values of E° and ΔG° for the given reaction are to be calculated.

Concept introduction:

Standard Gibbs free energy of a reaction is used to check whether the reaction is spontaneous or not. If the value of ΔG° is positive, then the reaction is non spontaneous. If the value of ΔG° is negative, then the reaction is spontaneous.

Expert Solution
Check Mark

Answer to Problem 8.19E

The value of E° and ΔG° for the given reaction is 0.0005V and 96.485J respectively.

Explanation of Solution

The given reaction is represented as,

I2+II3

From Table 8.2, the reduction half reaction of I3 and the standard reduction potential of I3 is represented as,

I3+2e3I      E°=0.5360V

The above equation is reversed and the value of E° is multiplied by 1 to form an oxidation half reaction. The oxidation half reaction is represented as,

3II3+2e      E°=0.5360V …(1)

The number of moles of electrons transferred in the above reaction is 2mol.

From Table 8.2, the reduction half reaction of I2 and the standard reduction potential of I2 is represented as,

I2+2e2I      E°=0.5355V …(2)

The number of moles of electrons transferred in the above reaction is 2mol.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

ΔG°=nFE° …(3)

Where,

ΔG° represents the standard Gibbs free energy of the reaction.

n represents the number of moles.

F represents the Faraday’s constant with value 96,485 C/mol.

E° represents the standard electrical potential.

Substitute the values of the standard oxidation potential of I, F and n in the equation (3).

ΔG°=(2 mol)(96,485 C/mol)(0.5360V)(1J/C1 V)=103431.92J

The value ΔG° for the reaction (1) is 103431.92J.

Substitute the values of the standard reduction potential of I2, F and number of moles of electrons transfer in the equation (3).

ΔG°=(2 mol)(96,485 C/mol)(0.5355V)(1J/C1 V)=103335.435J

The value ΔG° for the reaction (2) is 103.335kJ.

The value of ΔG° of overall reaction is calculated by adding the ΔG° of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

ΔG°=ΔG°1+ΔG°2

Where,

ΔG°1 represents the standard Gibbs free energy of the reaction (1).

ΔG°2 represents the standard Gibbs free energy of the reaction (2).

Substitute the value of ΔG°1 and ΔG°2 in the above equation.

ΔG°=103431.92J+(103335.435J)=96.485J

Therefore, the value ΔG° for the given reaction is 96.485J.

The number of electrons transferred in the overall reaction is 2 mol.

Rearrange the equation (3) for the value of E°.

E°=ΔG°nF

Substitute the values of ΔG°, F and number of moles of electrons transfer in the equation (3).

E°=(96.485J)(2 mol)(96,485 C/mol)(1J/C1 V)=0.0005V

The value of E° for the given reaction is 0.0005V.

Conclusion

The value of E° and ΔG° for the given reaction is 0.0005V and 96.485J respectively.

Interpretation Introduction

(b)

Interpretation:

The values of E° and ΔG° for the given reaction are to be calculated.

Concept introduction:

Standard Gibbs free energy of a reaction is used to check whether the reaction is spontaneous or not. If the value of ΔG° is positive, then the reaction is non spontaneous. If the value of ΔG° is negative, then the reaction is spontaneous.

Expert Solution
Check Mark

Answer to Problem 8.19E

The value of E° and ΔG° for the given reaction is 0.912V and 176.1kJ respectively.

Explanation of Solution

The given reaction is represented as,

Cr2++2eCr

From Table 8.2, the reduction half reaction of Cr3+ into Cr2+ and the standard reduction potential of Cr3+ into Cr2+ is represented as,

Cr3++eCr2+      E°=0.407V

The above equation is reversed and the value of E° is multiplied by 1 to form an oxidation half reaction. The oxidation half reaction is represented as,

Cr2+Cr3++e      E°=0.407V …(4)

The number of moles of electrons transferred in the above reaction is 1mol.

From Table 8.2, the reduction half reaction of Cr3+ into Cr  and the standard reduction potential of Cr3+ into Cr  is represented as,

Cr3++3eCr     E°=0.744V …(5)

The number of moles of electrons transferred in the above reaction is 3mol.

The relation between standard Gibbs free energy and standard electrical potential is represented as,

ΔG°=nFE° …(3)

Where,

ΔG° represents the standard Gibbs free energy of the reaction.

n represents the number of moles.

F represents the Faraday’s constant with value 96,485 C/mol.

E° represents the standard electrical potential.

Substitute the values of the standard oxidation potential of Cr2+, F and number of moles of electrons transfer in the equation (3).

ΔG°=(1 mol)(96,485 C/mol)(0.407V)(1J/C1 V)=(39269.395J)(1 kJ1000 J)=39.269kJ

The value ΔG° for the reaction (4) is 39.269kJ.

Substitute the values of the standard reduction potential of Cr3+, F and number of moles of electrons transfer in the equation (3).

ΔG°=(3 mol)(96,485 C/mol)(0.744V)(1J/C1 V)=(215354.53J)(1 kJ1000 J)=215.355kJ

The value ΔG° for the reaction (5) is 215.355kJ.

The value of ΔG° of overall reaction is calculated by adding the ΔG° of two half reactions. The standard Gibbs free energy of the given reaction is calculated as,

ΔG°=ΔG°1+ΔG°2

Where,

ΔG°1 represents the standard Gibbs free energy of the reaction (4).

ΔG°1 represents the standard Gibbs free energy of the reaction (5).

Substitute the value of ΔrxnG°1 and ΔrxnG°1 in the above equation.

ΔG°=39.269kJ+215.355kJ=176.086kJ176.1kJ

Therefore, the value ΔG° for the given reaction is 176.1kJ.

The number of electrons transferred in the overall reaction is 2 mol.

Rearrange the equation (3) for the value of E°.

E°=ΔG°nF

Substitute the values of ΔG°, F and number of moles of electrons transferred in the equation (3).

E°=(176.1kJ)(1000 J1 kJ)(2 mol)(96,485 C/mol)(1J/C1 V)=0.912V

The value of E° for the given reaction is 0.912V.

Conclusion

The value of E° and ΔG° for the given reaction is 0.912V and 176.1kJ respectively.

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Chapter 8 Solutions

Physical Chemistry

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