The E cell should be calculated before the titration is carried out. Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential. Nernst equation gives the relationship between standard reduction potential, E cell o , reduction potential, E cell and the activities of species present in the electrochemical cell at temperature, T as: E cell = E cell o - RT nF ln ( Q ) Where Q is reaction quotient, F is Faraday constant, R is universal gas constant and n is amount in terms of mol (electrons transferred) This equation is specified at room temperature, T = 298 .15 K as: E cell = E cell o - 0 .0591 n log ( Q )

Question

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Answer 1

Question

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

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A small artisanal cheesemaker is testing the acidity of their milk before it coagulates. During fermentation, bacteria produce lactic acid (K₁ = 1.4 x 104), a weak acid that helps to curdle the milk and develop flavor. The cheesemaker has measured that the developing mixture contains lactic acid at an initial concentration of 0.025 M. Your task is to calculate the pH of this mixture and determine whether it meets the required acidity for proper cheese development. To achieve the best flavor, texture and reduce/control microbial growth, the pH range needs to be between pH 4.6 and 5.0. Assumptions: Lactic acid is a monoprotic acid H H :0:0: H-C-C H :0: O-H Figure 1: Lewis Structure for Lactic Acid For simplicity, you can use the generic formula HA to represent the acid You can assume lactic acid dissociation is in water as milk is mostly water. Temperature is 25°C 1. Write the K, expression for the dissociation of lactic acid in the space provided. Do not forget to include state symbols.…

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Answer 2

Question

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

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Answer 3

Question

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

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Answer 4

Question

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

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Answer 5

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

Answer 6

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

Answer 7

Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated before the titration is carried out.

Concept Introduction:The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

Answer 8

a.

Expert Solution

Answer to Problem 132MP

The $Ecell=1.58 V$ before the titration is carried out.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

The concentration of $[Cu2+]$ is $1.0 M$ . The compartment containing vanadium (1.0 L solution) is titrated with $0.08 M H2EDTA2−$ , resulting in the following reaction:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The observed $Ecell$ at stoichiometric point (occurred at a volume of $H2EDTA2− = 500.0 mL$ ) for the process is $1.98 V$ . The solution was buffered at pH = 10.0.

The half-reactions at which galvanic cell is based on:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV2+(aq)+2e−→V(s) Eo = -1.20 V$

Since, the reduction potential value of copper is greater than that of vanadium so, the copper will undergo reduction and vanadium will undergo oxidation so, the half-reactions are written as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V$

Adding both the reactions to get the overall reactions as:

$Cu(aq)2++2e−→Cu(s) Eo = 0.34 VV(s) → V2+(aq)+2e− Eo = 1.20 V _ Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello= (0.34 + 1.2) V = 1.54 V$

So, the overall balanced reaction for the galvanic cell is:

$Cu2+(aq)+V(s) → Cu(s)+V2+(aq) Ecello = 1.54 V$

Now, according to Nernst equation at room temperature, $T = 298.15 K$ ,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]$ - (1)

Now, the concentration of $V2+$ is calculated as:

The initial moles of $V2+$ is calculated as:

$Initial moles of V2+ = moles of H2EDTA2− addedInitial moles of V2+= (Molarity)H2EDTA2−VH2EDTA2−Initial moles of V2+=(0.0800 M)(500 mL)(1 L1000 mL)Initial moles of V2+=0.0400 mol$

The concentration of $V2+$ in solution of $1.0 L$ is calculated using formula:

$Molarity of V2+=Moles of V2+Litres solution$

Substituting the values:

$Molarity of V2+=0.0400 mol1.00 LMolarity of V2+=0.0400 M$

Now, substituting the values in equation (1) as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.0400 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.398)Ecell=1.54 V+0.04Ecell=1.58 V$

Hence, the $Ecell=1.58 V$ before the titration is carried out.

Answer 13

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

Answer 14

b.

Interpretation Introduction

Interpretation: The value of equilibrium constant for the titration reaction needs to be calculated.

Concept Introduction: The relationship between reactants and products of a reaction in equilibrium with respect to some unit is said to be equilibrium expression. It is the expression that gives ratio between products and reactants. The expression is:

$K = concentration of productsconcentration of reactants$

Answer 15

b.

Expert Solution

Answer to Problem 132MP

$K = 1.6×108$ .

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The given titration reaction is:

$H2EDTA2−(aq)+V2+(aq)⇌VEDTA2−(aq)+2H+(aq)$

The expression for equilibrium constant is:

$K=[VEDTA2-][H+]2[H2EDTA2−][V2+]$ - (2) The concentration of the species is:

$[V2+]$ is calculated using Nernst equation as the $Ecell$ value at stoichiometric point is $1.98 V$ . So,

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]1.98 V=1.54 V-0.0591 V2log[V2+][1.00 M]0.44 = -0.02955log[V2+][1.00 M]-log[V2+][1.00 M] = 14.89[V2+]=(1.00 M)10−14.89[V2+]= 1.3×10−15 M$

From the balanced reaction, the number of moles of reactants are same that is 1 mole for each so, the equilibrium concentration of both the reactants is same that is $[V2+]= [H2EDTA2-] = 1.3×10-15 M$ .

At equilibrium, the number of moles of $VEDTA2−$ is same as that of $V2+$ that is $0.04 mol$ so, the concentration of $VEDTA2−$ is:

$Molarity of VEDTA2−=Moles of VEDTA2−Litres solution$

Substituting the values:

$Molarity of VEDTA2−=0.0400 mol1.00 L+0.500 LMolarity of VEDTA2−=0.0267 M$

Now, the concentration of $H+$ is calculated using relation:

$pH = -log[H+]$

Substituting the values:

$10=−log[H+][H+]=1.0×10−10 M$

The values of concentrations are substitute in equation (2) as:

$K = [VEDTA2-][H+]2[H2EDTA2−][V2+]K = (0.0267)(1.0×10−10)2(1.3×10−15)(1.3×10−15)K = 1.6×108$

Answer 20

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

Answer 21

c.

Interpretation Introduction

Interpretation: The $Ecell$ should be calculated at the halfway point the titration.

Concept Introduction: The measure of energy per unit charge which is available from the redox reactions to carry out the reaction is said to be cell potential.

Nernst equation gives the relationship between standard reduction potential, $Ecello$ , reduction potential, $Ecell$ and the activities of species present in the electrochemical cell at temperature, $T$ as:

$Ecell = Ecello- RTnFln(Q)$

Where $Q$ is reaction quotient, $F$ is Faraday constant, $R$ is universal gas constant and $n$ is amount in terms of mol (electrons transferred)

This equation is specified at room temperature, $T = 298.15 K$ as:

$Ecell = Ecello- 0.0591nlog(Q)$

Answer 22

c.

Expert Solution

Answer to Problem 132MP

The $Ecell= 1.59 V$ at the halfway point the titration.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The volume added to $1.0 L$ solution becomes half at the hallway equivalence point. So, the total volume of the solution is calculated as:

$VTotal = Vvandium+VaddedVTotal = 1.00 L+0.500 L2VTotal =1.25 L$

At hallway equivalence point, the initial moles of $V2+$ becomes half that is:

$Moles of V2+ = 0.042Moles of V2+ = 0.02$

The concentration of $V2+$ is calculated as:

$Molarity of V2+=Moles of V2+ Litres solution Molarity of V2+=0.0200 mol1.25 LMolarity of V2+=0.016 M$

Substituting this value of concentration is Nernst equation as:

$Ecell=Ecello−0.0591 Vnlog[V2+][Cu2+]Ecell=1.54 V-0.0591 V2log[0.016 M][1.00 M]Ecell=1.54 V-(0.02955)(-1.80)Ecell=1.54 V + 0.05Ecell= 1.59 V$

Hence, the $Ecell= 1.59 V$ at the halfway point the titration.

Answer 27

Ch. 11 - Prob. 1DQ Ch. 11 - Prob. 2DQ Ch. 11 - You want to “plate out” nickel metal from a nickel...Ch. 11 - A copper penny can be dissolved in nitric acid but...Ch. 11 - Sketch a cell that forms iron metal from iron(II)...Ch. 11 - Which of the following is the best reducing agent:...Ch. 11 - You are told that metal A is a better reducing...Ch. 11 - Explain the following relationships: G and w, cell...Ch. 11 - Explain why cell potentials are not multiplied by...Ch. 11 - What is the difference between andWhen is equal to...

Answer to Problem 132MP

Explanation of Solution

Answer to Problem 132MP

Explanation of Solution

Answer to Problem 132MP

Explanation of Solution

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