Solution Manual for Quantitative Chemical Analysis
Solution Manual for Quantitative Chemical Analysis
9th Edition
ISBN: 9781464175633
Author: Daniel Harris
Publisher: Palgrave Macmillan Higher Ed
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Chapter 14, Problem 14.38P

(a)

Interpretation Introduction

Interpretation:

The half-reaction and Nernst equation for each half-cell has to be written.

Concept introduction:

Nernst equation can be used to determine the cell potential at any instant, the difference from the standard state.

Ecell = Ecello - (RTnF)lnQ

Where:

Ecell = cell potential under nonstandard conditions

Ecello = cell potential under standard conditions

R = gas constant, which is 8.31 (volt-coulomb)/(mol-K)

T = temperature (kelvin), which is generally 298°K (77°F/25°C) 

n = number of moles of electrons exchanged in the electrochemical reaction

F = Faraday's constant, 96500 coulombs/mol

Q = reaction quotient, which is the equilibrium expression with initial concentrations rather than equilibrium concentrations

(a)

Expert Solution
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Explanation of Solution

The half-reaction and Nernst equation for each half-cell can be calculated using below formula

E(right)=Eo(right) -RT2FlnAF-2(right)PO21/2(right)

E(left)=Eo(left) -RT2FlnAF-2(left)PO21/2(left)

Both half left and right are added

MgF2(s)+Al2O3(s)+12O2(g)+2e-MgAl2O4(s)+2F-MgO(s)+2F-MgF2(s)+12O2(g)+2e-_netreaction:Al2O3(s)+MgO(s)MgAl2O4(s)

Nernst equation for net reaction

E(right)-E(left)=Eo(right)-Eo(left) -RT2FlnAF-2(right)PO21/2(right)+RT2FlnAF-2(left)PO21/2(left)

F- have both side same activities of O2 .  The ln terms cancel, leaving

E(cell)=Eo(right)-Eo(left)=Eo(cell)

(b)

Interpretation Introduction

Interpretation:

The value of ΔGo has to be calculated.

Concept introduction:

The value of ΔGo can be calculated as,

Eo=-ΔGonF

Where

Eo = The standard reduction potential

n = number of electrons

The equilibrium constant for the reaction can be calculated as,

Finding K from Eo

K=10nEo0.05916at25oC

Where K is equilibrium constant

(b)

Expert Solution
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Answer to Problem 14.38P

The value of ΔGo is -29.51kJ/mol

Explanation of Solution

ΔGo=-nFEo=-(2)(9.6485×104C/mol)(0.1529J/C)=-29.51kJ/mol

The fact that a volt is equivalent to one joule/coulomb

(c)

Interpretation Introduction

Interpretation:

The electron flow direction has to be found.

Concept introduction:

The value of ΔGo can be calculated as,

Eo=-ΔGonF

Where

Eo = The standard reduction potential

n = number of electrons

ΔGo=ΔHo-TΔSowhere,TistemperatureΔGoisstandfreeenergychangeΔHoisstandenthalpychange

(c)

Expert Solution
Check Mark

Explanation of Solution

ΔGo=ΔHo-TΔSo-nFEo=ΔHo-TΔSo=-nF(0.1223V+3.06×10-5T)=-nF(0.1223V)-nF(3.06×10-5T)=-nF(0.1223VΔHo)-T{nF(3.06×10-5V/KΔSo)}

ΔHo=-nF(0.1223V)=-(2)(9.6485×104C/mol)(0.1223J/C)=-23.60kJ/mol

ΔSo=-nF(3.06×10-5V/K)=-(2)(9.6485×104C/mol)(3.06×10-5V/K)=5.90C.V/(K.mol)=5.90J/(K.mol)

We made use of the conversion coulomb volt =joule

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Chapter 14 Solutions

Solution Manual for Quantitative Chemical Analysis

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