Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
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Chapter 14, Problem 14.30P

(a)

Interpretation Introduction

Interpretation:

The net reaction for given solutions has to be written.

Concept introduction:

Identifying relevant half-reaction:

Reduction reaction of each half cell have to be identified.  In order to identify it, element with two oxidation number should be taken.   From the elements reduction reaction for both right and left hand cell is given.

(a)

Expert Solution
Check Mark

Explanation of Solution

To write: The net reaction for given solutions.

Elements with two oxidation states are Ce3+ , Ce4+ and MnO4- , Mn2+ .  The reduction reaction for both half-cell is given as

Right hand cell:5Ce4++5e-5Ce3+E+0=1.92V-Left hand cell:MnO4-+8H++5e-Mn2++4H2OE-0=1.229VNetreaction5Ce4++Mn2++4H2O5Ce3++MnO4-+8H+E0=0.193V

(b)

Interpretation Introduction

Interpretation:

The standard free energy and equilibrium constant of given cell has to be determined.

Concept introduction:

Relation between E0andK is given as

E0=0.05916nlogKK=10nE0/0.05916ΔG0=-nFE0E0isstandardreductionpotentialKisequilibriumconstant

(b)

Expert Solution
Check Mark

Explanation of Solution

To determine: The standard free energy and equilibrium constant of given cell.

When the potentiometer is replaced with a wire in a cell there would be more current flows and concentration varies until the cell attains equilibrium.  At this situation there is nothing to drive reaction.  The cell voltage becomes zero.

E0=0.193VΔG0=-nFE0=-1(96485C/mol)(0.193V)=-93.1KJK=10nE0/0.05916=101(0.193V)/0.05916=2×1016

(c)

Interpretation Introduction

Interpretation:

The cell voltage of given cell has to be determined.

Concept introduction:

Nernst Equation:

For Half-reaction,

aA+ne-bB

The Nernst equation results in the half-cell potential E as,

E=E°-RTnFlnABbAAa

Here,

= standard reduction potential ( AA=AB=1 )

R = gas constant (8.314J/(K.mol))=8.314((V.C)/(K.mol))

T =Temperature (in K)

N = number of electrons in half-reaction

F = Faraday constant ( 9.649×104C/mol )

A = Activity of species, i

The voltage of a battery is calculated as

Cell voltage = potential of right hand electrode ( E+ ) – potential of left hand electrode( E ).

(c)

Expert Solution
Check Mark

Explanation of Solution

To determine: The cell voltage of given cell.

Right hand cell:5Ce4++5e-5Ce3+E+0=1.92V-Left hand cell:MnO4-+8H++5e-Mn2++4H2OE-0=1.229VNetreaction5Ce4++Mn2++4H2O5Ce3++MnO4-+8H+

Applying the electrode potential in Nernst equation, the cell voltage is calculated as

E+=(1.70-0.059162log[0.100]5[1×10-4]5)-(-0.44-0.059162log[1×10-4][0.1][1.0]8)=1.5223 - 1.5425 = -0.020 V

(d)

Interpretation Introduction

Interpretation:

The change in free energy of given cell has to be determined.

Concept introduction:

Relation between change in free energy and change in electric potential:

ΔG=-n.N.F.E 

Here,

E is volts

n = unit charge per molecule

N = number of electrons in half-reaction

F = Faraday constant ( 9.649×104C/mol )

(d)

Expert Solution
Check Mark

Explanation of Solution

To determine: The change in free energy of given cell.

The free energy difference is calculated as

ΔG=-n.N.F.E =-5mol×96485C/mol×-0.020V=+10KJ

(e)

Interpretation Introduction

Interpretation:

The pH at which given solution attains equilibrium has to be determined.

Concept introduction:

When the potentiometer is replaced with a wire in a cell there would be more current flows and concentration varies until the cell attains equilibrium.  At this situation there is nothing to drive reaction.  The cell voltage becomes zero.

Relation between E0andK is given as

E0=0.05916nlogKK=10nE0/0.05916ΔG0=-nFE0E0isstandardreductionpotentialKisequilibriumconstant

(e)

Expert Solution
Check Mark

Explanation of Solution

To determine: The pH at which given solution attains equilibrium.

When the solution is at equilibrium the cell voltage E becomes zero.

E0=0.05916nlogKE0=0.059165log[Ce3+]5[MnO4-][H+]8[Ce4+][Mn2+][H+]=0.62pH=-log[H+]=-log(0.62)=0.21

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Chapter 14 Solutions

Quantitative Chemical Analysis

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