Chapter 11, Problem 132MP

a.

Interpretation Introduction

Interpretation: The ${\text{E}}_{\text{cell}}$ 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, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

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

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 132MP

The ${\text{E}}_{\text{cell}}=1.58\text{ V}$ before the titration is carried out.

Explanation of Solution

Given:

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

  $\begin{array}{l}\text{Cu}(_{\text{aq)}}{\text{+2e}}^{-}\to {\text{Cu(s) E}}^o\text{ = 0}\text{.34 V}\\ {\text{V}}^{2+}({\text{aq)+2e}}^{-}\to {\text{V(s) E}}^o\text{ = -1}\text{.20 V}\end{array}$

The concentration of $\left[\text{Cu}\right]$ is $1.0\text{ M}$ . The compartment containing vanadium (1.0 L solution) is titrated with $\text{0}{\text{.08 M H}}_{\text{2}}{\text{EDTA}}^{2-}$ , resulting in the following reaction:

  ${\text{H}}_2{\text{EDTA}}^{2-}({\text{aq)+V}}^{2+}\text{(aq)}\rightleftharpoons {\text{VEDTA}}^{2-}({\text{aq)+2H}}^{+}(\text{aq)}$

The observed ${\text{E}}_{\text{cell}}$ at stoichiometric point (occurred at a volume of ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}\text{ = 500}\text{.0 mL}$ ) for the process is $\text{1}\text{.98 V}$ . The solution was buffered at pH = 10.0.

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

  $\begin{array}{l}\text{Cu}(_{\text{aq)}}{\text{+2e}}^{-}\to {\text{Cu(s) E}}^o\text{ = 0}\text{.34 V}\\ {\text{V}}^{2+}({\text{aq)+2e}}^{-}\to {\text{V(s) E}}^o\text{ = -1}\text{.20 V}\end{array}$

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:

  $\begin{array}{l}\text{Cu}(_{\text{aq)}}{\text{+2e}}^{-}\to {\text{Cu(s) E}}^o\text{ = 0}\text{.34 V}\\ \text{V(s) }\to {\text{ V}}^{2+}({\text{aq)+2e}}^{-}{\text{ E}}^o\text{ = 1}\text{.20 V}\end{array}$

Adding both the reactions to get the overall reactions as:

  $\begin{array}{l}\text{Cu}(_{\text{aq)}}{\text{+2e}}^{-}\to {\text{Cu(s) E}}^o\text{ = 0}\text{.34 V}\\ \underset{\_}{\text{V(s) }\to {\text{ V}}^{2+}({\text{aq)+2e}}^{-}{\text{ E}}^o\text{ = 1}\text{.20 V }}\\ {\text{Cu}}^{2+}(\text{aq)+V(s) }\to {\text{ Cu(s)+V}}^{2+}({\text{aq) E}}_{cell}^o\text{= }\left(\text{0}\text{.34 + 1}\text{.2}\right)\text{ V = 1}\text{.54 V}\end{array}$

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

  ${\text{Cu}}^{2+}(\text{aq)+V(s) }\to {\text{ Cu(s)+V}}^{2+}({\text{aq) E}}_{cell}^o\text{ = 1}\text{.54 V}$

Now, according to Nernst equation at room temperature, $\text{T = 298}\text{.15 K}$ ,

  ${\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0591 V}}{\text{n}}\log \frac{\left[\text{V}\right]}{\left[{\text{Cu}}^{2+}\right]}$ - (1)

Now, the concentration of ${\text{V}}^{2+}$ is calculated as:

The initial moles of ${\text{V}}^{2+}$ is calculated as:

  $\begin{array}{l}{\text{Initial moles of V}}^{2+}{\text{ = moles of H}}_2{\text{EDTA}}^{2-}\text{ added}\\ {\text{Initial moles of V}}^{2+}{\text{= (Molarity)}}_{{\text{H}}_2{\text{EDTA}}^{2-}}{\text{V}}_{{\text{H}}_2{\text{EDTA}}^{2-}}\\ {\text{Initial moles of V}}^{2+}=\text{(0}\text{.0800 M)(500 mL)}\left(\frac{\text{1 L}}{\text{1000 mL}}\right)\\ {\text{Initial moles of V}}^{2+}=0.0400\text{ mol}\end{array}$

The concentration of ${\text{V}}^{2+}$ in solution of $\text{1}\text{.0 L}$ is calculated using formula:

  ${\text{Molarity of V}}^{2+}=\frac{{\text{Moles of V}}^{2+}}{\text{Litres solution}}$

Substituting the values:

  $\begin{array}{l}{\text{Molarity of V}}^{2+}=\frac{0.0400\text{ mol}}{\text{1}\text{.00 L}}\\ {\text{Molarity of V}}^{2+}=0.0400\text{ M}\end{array}$

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

  $\begin{array}{l}{\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0591 V}}{\text{n}}\log \frac{\left[\text{V}\right]}{\left[{\text{Cu}}^{2+}\right]}\\ {\text{E}}_{\text{cell}}=\text{1}\text{.54 V-}\frac{0.0591\text{ V}}{\text{2}}\text{log}\frac{\left[\text{0}\text{.0400 M}\right]}{\left[\text{1}\text{.00 M}\right]}\\ {\text{E}}_{\text{cell}}=\text{1}\text{.54 V-}\left(0.02955\right)\left(\text{-1}\text{.398}\right)\\ {\text{E}}_{\text{cell}}=\text{1}\text{.54 V+}0.04\\ {\text{E}}_{\text{cell}}=1.58\text{ V}\end{array}$

Hence, the ${\text{E}}_{\text{cell}}=1.58\text{ V}$ 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:

  $\text{K = }\frac{\text{concentration of products}}{\text{concentration of reactants}}$

Answer to Problem 132MP

  $\text{K = 1}\text{.6}\times {\text{10}}^8$ .

Explanation of Solution

The given titration reaction is:

  ${\text{H}}_2{\text{EDTA}}^{2-}({\text{aq)+V}}^{2+}\text{(aq)}\rightleftharpoons {\text{VEDTA}}^{2-}({\text{aq)+2H}}^{+}(\text{aq)}$

The expression for equilibrium constant is:

  $\text{K=}\frac{\left[{\text{VEDTA}}^{2\text{-}}\right]{\left[{\text{H}}^{+}\right]}^2}{\left[{\text{H}}_2{\text{EDTA}}^{2-}\right]\left[{\text{V}}^{2+}\right]}$ - (2) The concentration of the species is:

  $\left[{\text{V}}^{2+}\right]$ is calculated using Nernst equation as the ${\text{E}}_{cell}$ value at stoichiometric point is $1.98\text{ V}$ . So,

  $\begin{array}{l}{\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0591 V}}{\text{n}}\log \frac{\left[\text{V}\right]}{\left[{\text{Cu}}^{2+}\right]}\\ 1.98\text{ V}=\text{1}\text{.54 V-}\frac{0.0591\text{ V}}{\text{2}}\text{log}\frac{\left[{\text{V}}^{2+}\right]}{\left[\text{1}\text{.00 M}\right]}\\ 0.44\text{ = -0}\text{.02955log}\frac{\left[{\text{V}}^{2+}\right]}{\left[\text{1}\text{.00 M}\right]}\\ \text{-log}\frac{\left[{\text{V}}^{2+}\right]}{\left[\text{1}\text{.00 M}\right]}\text{ = 14}\text{.89}\\ \left[{\text{V}}^{2+}\right]=(1.00{\text{ M)10}}^{-14.89}\\ \left[{\text{V}}^{2+}\right]\text{= 1}\text{.3}\times {\text{10}}^{-15}\text{ M}\end{array}$

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 $\left[{\text{V}}^{\text{2+}}\right]\text{= }\left[{\text{H}}_{\text{2}}{\text{EDTA}}^{\text{2-}}\right]\text{ = 1}{\text{.3×10}}^{\text{-15}}\text{ M}$ .

At equilibrium, the number of moles of ${\text{VEDTA}}^{2-}$ is same as that of ${\text{V}}^{\text{2+}}$ that is $0.04\text{ mol}$ so, the concentration of ${\text{VEDTA}}^{2-}$ is:

  ${\text{Molarity of VEDTA}}^{2-}=\frac{{\text{Moles of VEDTA}}^{2-}}{\text{Litres solution}}$

Substituting the values:

  $\begin{array}{l}{\text{Molarity of VEDTA}}^{2-}=\frac{0.0400\text{ mol}}{\text{1}\text{.00 L+0}\text{.500 L}}\\ {\text{Molarity of VEDTA}}^{2-}=\text{0}\text{.0267 M}\end{array}$

Now, the concentration of ${\text{H}}^{+}$ is calculated using relation:

  $\text{pH = -log}\left[{\text{H}}^{+}\right]$

Substituting the values:

  $\begin{array}{l}10=-\log \left[{\text{H}}^{+}\right]\\ \left[{\text{H}}^{+}\right]\text{=1}\text{.0}\times {\text{10}}^{-10}\text{ M}\end{array}$

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

  $\begin{array}{l}\text{K = }\frac{\left[{\text{VEDTA}}^{2\text{-}}\right]{\left[{\text{H}}^{+}\right]}^2}{\left[{\text{H}}_2{\text{EDTA}}^{2-}\right]\left[{\text{V}}^{2+}\right]}\\ \text{K = }\frac{\text{(0}\text{.0267)(1}\text{.0}\times {\text{10}}^{-10}{)}^2}{\text{(1}\text{.3}\times {\text{10}}^{-15})(1.3\times {\text{10}}^{-15})}\\ \text{K = 1}\text{.6}\times {\text{10}}^8\end{array}$

c.

Interpretation Introduction

Interpretation: The ${\text{E}}_{\text{cell}}$ 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, ${\text{E}}_{\text{cell}}^{\text{o}}$ , reduction potential, ${\text{E}}_{\text{cell}}$ and the activities of species present in the electrochemical cell at temperature, $\text{T}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{RT}}{\text{nF}}\text{ln}\left(\text{Q}\right)$

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

This equation is specified at room temperature, $\text{T = 298}\text{.15 K}$ as:

  ${\text{E}}_{\text{cell}}{\text{ = E}}_{\text{cell}}^{\text{o}}\text{- }\frac{\text{0}\text{.0591}}{\text{n}}\text{log}\left(\text{Q}\right)$

Answer to Problem 132MP

The ${\text{E}}_{\text{cell}}\text{= 1}\text{.59 V}$ at the halfway point the titration.

Explanation of Solution

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

  $\begin{array}{l}{\text{V}}_{\text{Total}}{\text{ = V}}_{\text{vandium}}{\text{+V}}_{\text{added}}\\ {\text{V}}_{\text{Total}}\text{ = 1}\text{.00 L+}\frac{0.500\text{ L}}{2}\\ {\text{V}}_{\text{Total}}\text{ =1}\text{.25 L}\end{array}$

At hallway equivalence point, the initial moles of ${\text{V}}^{2+}$ becomes half that is:

  $\begin{array}{l}{\text{Moles of V}}^{2+}\text{ = }\frac{0.04}{2}\\ {\text{Moles of V}}^{2+}\text{ = }0.02\end{array}$

The concentration of ${\text{V}}^{2+}$ is calculated as:

  $\begin{array}{l}{\text{Molarity of V}}^{2+}=\frac{{\text{Moles of V}}^{2+}}{\text{Litres solution }}\\ {\text{Molarity of V}}^{2+}=\frac{\text{0}\text{.0200 mol}}{\text{1}\text{.25 L}}\\ {\text{Molarity of V}}^{2+}=\text{0}\text{.016 M}\end{array}$

Substituting this value of concentration is Nernst equation as:

  $\begin{array}{l}{\text{E}}_{\text{cell}}={\text{E}}_{\text{cell}}^{\text{o}}-\frac{\text{0}\text{.0591 V}}{\text{n}}\log \frac{\left[\text{V}\right]}{\left[{\text{Cu}}^{2+}\right]}\\ {\text{E}}_{\text{cell}}=\text{1}\text{.54 V-}\frac{0.0591\text{ V}}{\text{2}}\text{log}\frac{\left[\text{0}\text{.016 M}\right]}{\left[\text{1}\text{.00 M}\right]}\\ {\text{E}}_{\text{cell}}=1.5\text{4 V-}(0.02955)\text{(-1}\text{.80)}\\ {\text{E}}_{\text{cell}}=1.5\text{4 V + 0}\text{.05}\\ {\text{E}}_{\text{cell}}\text{= 1}\text{.59 V}\end{array}$

Hence, the ${\text{E}}_{\text{cell}}\text{= 1}\text{.59 V}$ at the halfway point the titration.