General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 19, Problem 19.24QP

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Eocell is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: ΔG0=-nFEcell0

Relation between Ecell0 and equilibrium constant (K) of a redox reaction: Ecell0=RTnFlnK

Relation between ΔG0 and K:ΔG0=-RTlnK

(a)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is,

Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)

Lead ion is reduced and Magnesium is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Mg (+2) as follows,

Mg2+(aq)+2eMg(s)E0=2.37V

The standard reduction potential of Pb (+2) as follows,

Pb2+(aq)+2ePb(s)E0=0.13V

Calculated standard emf for galvanic cell as follows,

oxidation:Mg(s)Mg2+(aq)+2eEanode0=+2.37V(half-reactionchangethesignbut not magnitude)Reduction:Pb2+(aq)+2ePb(s)Ecathode0=0.13V_overall:Pb2+(aq)+Mg(s)Pb(s)+Mg2+(aq)_

Ecello=Ecathodeo+Eanodeo=(-0.13)V+(+2.37)V=+2.24V

Therefore, Standard emf of a galvanic cell is +2.24V

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

ΔG0=-nFEcell0=(2e)(96,500J/V.mol)(+2.24V)=432kJ/mol

Using equation as follows,

ΔG0=-RTlnK432kJ/mol=(8.314J/K.mol)(298K)lnKlnK=432×103J/mol(8.314J/K.mol)(298K)=174.4K=e174.4=5×1075

Therefore, the equilibrium constant obtained is 5×1075

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Eocell is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: ΔG0=-nFEcell0

Relation between Ecell0 and equilibrium constant (K) of a redox reaction: Ecell0=RTnFlnK

Relation between ΔG0 and K:ΔG0=-RTlnK

(b)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is,

Br2(l)+2I(aq)2Br(aq)+I2(s)

Bromium is reduced and Iodide ion is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Br (-1) as follows,

Br2(l) + 2e2Br(aq)E0=+1.07V

The standard reduction potential of Iodide ion as follows,

I2(s) + 2e2I(aq)E0=+0.53V

Calculated standard emf for galvanic cell as follows,

oxidation:2I(aq)I2(s) + 2eEanode0=0.53V(half-reactionchangethesignbut not magnitude)Reduction:Br2(l) + 2e2Br(aq)Ecathode0=+1.07V_overall:Br2(l)+2I(aq)2Br(aq)+I2(s)_

Ecello=Ecathodeo+Eanodeo=(+1.07)V+(-0.53)V=+0.54V

Therefore, Standard emf of a galvanic cell is +0.54V

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

ΔG0=-nFEcell0=(2e)(96,500J/V.mol)(+0.54V)=104kJ/mol

Using equation as follows,

ΔG0=-RTlnK104kJ/mol=(8.314J/K.mol)(298K)lnKlnK=104×103J/mol(8.314J/K.mol)(298K)=41.98K=e41.98=1.8×1018

Therefore, the equilibrium constant obtained is 2×1018

c)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Eocell is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: ΔG0=-nFEcell0

Relation between Ecell0 and equilibrium constant (K) of a redox reaction: Ecell0=RTnFlnK

Relation between ΔG0 and K:ΔG0=-RTlnK

c)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is,

O2(g)+4H+(aq)+4Fe2+(aq)2H2O(l)+4Fe3+(aq)

Oxygen and proton are reduced and Iron ion is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Oxygen as follows,

O2(g)+4H+(aq)+4e2H2O(l)E0=+1.23V

The standard reduction potential of Iron ion as follows,

Fe3+(aq)+eFe2+(aq)E0=+0.77V

Calculated standard emf for galvanic cell as follows,

Reversing half-reaction changes the sign of E0 but not the magnitude.

oxidation:4Fe2+(aq)4Fe3+(aq)+4eEanode0=0.77VReduction:O2(g)+4H+(aq)+4e2H2O(l)Ecathode0=+1.23V_overall:O2(g)+4H+(aq)+4Fe2+(aq)2H2O(l)+4Fe3+(aq)_

Ecello=Ecathodeo+Eanodeo=(+1.23)V+(-0.77)V=+0.46V

Therefore, Standard emf of a galvanic cell is +0.46V

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

ΔG0=-nFEcell0=(4e)(96,500J/V.mol)(+0.46V)=177kJ/mol

Using equation as follows,

ΔG0=-RTlnK177kJ/mol=(8.314J/K.mol)(298K)lnKlnK=177×103J/mol(8.314J/K.mol)(298K)=71.4K=e71.4=1×1031

Therefore, the equilibrium constant obtained is 1×1031

d)

Interpretation Introduction

Interpretation:

The equilibrium constant and Gibb’s free energy for the following cell reaction has to be calculated.

Concept introduction:

Standard reduction potential: The voltage associated with a reduction reaction at an electrode when all solutes are 1M and all gases are at 1 atm. The hydrogen electrode is called the standard hydrogen electrode (SHE).

Standard emf: Eocell is composed of a contribution from the anode and a contribution from the cathode is given by,

Ecello=EcathodeoEanodeo

Where both Ecathodeo and Eanodeo are the standard reduction potentials of the electrodes.

Thermodynamics of redox reactions:

The change in free-energy represents the maximum amount of useful work that can be obtained in a reaction: ΔG0=-nFEcell0

Relation between Ecell0 and equilibrium constant (K) of a redox reaction: Ecell0=RTnFlnK

Relation between ΔG0 and K:ΔG0=-RTlnK

d)

Expert Solution
Check Mark

Explanation of Solution

The given reaction is,

2Al(s)+3I2(s)2Al3+(aq)+6I(aq)

Iodine is reduced and Aluminum is oxidized; hence oxidation occurs at anode electrode and reduction occurs at cathode electrode.

Calculation of Standard emf:

The standard reduction potential of Al (+3) as follows,

Al3+(aq)+3eAl(s)E0=1.66V

The standard reduction potential of Iodide ion as follows,

I2(s) + 2e2I(aq)E0=+0.53V

Calculated standard emf for galvanic cell as follows,

Reversing oxidation half-reaction changes the sign E0 but not the magnitude.

oxidation:2Al(s)2Al3+(aq)+6eEanode0=+1.66VReduction:3I2(s) + 6e6I(aq)Ecathode0=+0.53V_overall:2Al(s)+3I2(s)2Al3+(aq)+6I(aq)_

Ecello=Ecathodeo+Eanodeo=(+0.53)V+(+1.66)V=+2.19V

Therefore, Standard emf of a galvanic cell is +2.19V

Equilibrium constant calculation:

Using Gibb’s free energy equation as follows,

ΔG0=-nFEcell0=(6e)(96,500J/V.mol)(+2.19V)=1268kJ/mol

Using equation as follows,

ΔG0=-RTlnK1268kJ/mol=(8.314J/K.mol)(298K)lnKlnK=1268×103J/mol(8.314J/K.mol)(298K)=511.8K=e511.8=1.8×10222

Therefore, the equilibrium constant obtained is 1.8×10222

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Chapter 19 Solutions

General Chemistry

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