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.CE

(a)

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons 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 ).

(a)

Expert Solution
Check Mark

Answer to Problem 14.CE

The cell voltage  of the battery is 1.65 V .  Electrons flow from left hand electrode to right hand electrode that is iron to platinum.

Explanation of Solution

To determine: The cell voltage and direction flow of electrons.

Right hand cell:Br(l)+2e-2Br-E+0=1.078 V-Left hand cell:Fe2++2e-Fe(s)E-0=- 0.44 VNet reaction:Br2(l) + Fe(s) 2Br- + Fe2+

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

E+=(1.078-0.059162log(0.050)2)-(-0.44-0.059162log10.010)=1.155 - (- 0.50) = 1.65 V

The cell potential is positive one.  Thus, electrons flow from left hand electrode to right hand electrode that is iron to platinum.

(b)

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons 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 ).

(b)

Expert Solution
Check Mark

Answer to Problem 14.CE

The cell voltage of the battery is - 0.77 V .  Electrons flow from right hand electrode to left hand electrode that is iron to copper.

Explanation of Solution

To determine: The cell voltage and direction flow of electrons.

Right hand cell:Fe2++2e-2Fe(s)E+0=-0.44 V-Left hand cell:Cu2++2e-Cu(s)E-0=0.339 VNet reaction:Fe2+ + Cu(s) Fe(s) + Cu2+

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

E+=(-0.44-0.059162log10.050)-(0.339-0.059162log10.020)=- 0.48 - (0.289) = - 0.77 V

The cell potential is negative one.  Thus, electrons flow from right hand electrode to left hand electrode that is iron to copper.

(c)

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons 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

Answer to Problem 14.CE

The cell voltage  of the battery is - 0.77 V .  Electrons flow from right hand electrode to left hand electrode that is iron to copper.

Explanation of Solution

To determine: The cell voltage and direction flow of electrons.

Right hand cell:Cl2(g)+2e-2Cl -(s)E+0=1.360V-Left hand cell:Hg2Cl2+2e-2Hg(s)+2Cl-E-0=0.268VNet reaction:Cl2(g) + 2Hg(l) Hg2Cl2(s)

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

E+=(1.360-0.059162log(0.040)20.050)-(0.268-0.059162log(0.060)2)= 1.434 - (0.340) = 1.094V

The cell potential is positive one.  Thus, electrons flow from left hand electrode to right hand electrode that is mercury to platinum.

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Chapter 14 Solutions

Quantitative Chemical Analysis

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