Chapter 18, Problem 159MP

A galvanic cell is based on the following half-reactions:

In this cell, the copper compartment contains a copper electrode and [Cu²⁺] = 1.00 M, and the vanadium compartment contains a vanadium electrode and V²⁺ at an unknown concentration. The compartment containing the vanadium (1.00 L of solution) was titrated with 0.0800 M H₂EDTA²⁻, resulting in the reaction

${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}\left(aq\right)+{\text{V}}^{2+}\left(aq\right)\rightleftharpoons {\text{VEDTA}}^{2-}\left(aq\right)+{\text{2H}}^{+}\left(aq\right)K=?$

The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred at a volume of 500.0 mL H₂EDTA²⁻ solution added. At the stoichiometric point,  was observed to be 1 .98 V. The solution was buffered at a pH of 10.00.

a. Calculate  before the titration was carried out.

b. Calculate the value of the equilibrium constant, K, for the titration reaction.

c. Calculate  at the halfway point in the titration.

Interpretation Introduction

Interpretation: The ${\text{E}}_{\text{cell}}$  before the titration was carried out is to be calculated. The value of the equilibrium constant (K).for the titration reaction and the value of ${\text{E}}_{\text{cell}}$ at the halfway point in the titration are to be calculated.

Concept introduction: Electrochemical cell is a combination of two half cells in which the two electrodes are joined by a wire and a salt bridge. Electrons flow from anode where oxidation occurs to cathode where reduction takes place. Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. Equilibrium constant is defined as the ratio of the concentration of products and the concentration of the reactants.

To determine: The ${\text{E}}_{\text{cell}}$ before the titration was carried out.

Answer to Problem 159MP

The cell potential before titration is $\text{-1}\text{.5}\text{V}$.

Explanation of Solution

Explanation

The concentration of $\left[{\text{Cu}}^{\text{2+}}\right]$ is $1.0\text{0 M}$.

The volume of vanadium solution is $1.0\text{0 L}$.

The concentration of ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}$ is $0.080\text{0 M}$.

The volume of ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}$ solution added is $500.0\text{ mL}$.

The $E_{\text{cell}}$ is $1.9\text{8 V}$.

The value of $\text{pH}$ is $10.00$.

The titration reaction of $1.0\text{0 L}$ vanadium with $0.080\text{0 M}$ ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}$ is shown as follows,

${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}\left(aq\right)+{\text{V}}^{\text{2+}}\left(aq\right)\underset{}{\rightleftharpoons }{\text{VEDTA}}^{2-}\left(aq\right)+{\text{2H}}^{\text{+}}\left(aq\right)$

When $500.0\text{ mL}$ ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}$ is added, the initial moles of ${\text{V}}^{\text{2+}}$ present in the solution is same as that of added moles of ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}$. So, the initial moles of ${\text{V}}^{\text{2+}}$ are determined as follows,

$\begin{array}{c}{\text{Initial moles of V}}^{2+}={\text{moles of H}}_{\text{2}}{\text{EDTA}}^{2-}\text{ added}\\ ={\left(\text{Molarity}\right)}_{{\text{H}}_{\text{2}}{\text{EDTA}}^{2-}}\times {\text{V}}_{{\text{H}}_{\text{2}}{\text{EDTA}}^{2-}}\end{array}$

Substitute the values of molarity and volume of ${\text{H}}_{\text{2}}{\text{EDTA}}^{2-}$ in the above expression to calculate the moles of ${\text{V}}^{\text{2+}}$.

$\begin{array}{c}{\text{Initial moles of V}}^{2+}={\text{moles of H}}_{\text{2}}{\text{EDTA}}^{2-}\text{ added}\\ ={\left(\text{Molarity}\right)}_{{\text{H}}_{\text{2}}{\text{EDTA}}^{2-}}\times {\text{V}}_{{\text{H}}_{\text{2}}{\text{EDTA}}^{2-}}\\ =\left(0.080\text{0 M}\right)\left(50\text{0 mL}\right)\left(\frac{\text{1 L}}{10\text{00 mL}}\right)\\ =0.040\text{0 mol}\end{array}$

Thus, the initial moles of ${\text{V}}^{\text{2+}}$ are $0.040\text{0 mol}$.

The concentration of ${\text{V}}^{\text{2+}}$ in $1.0\text{0 L}$ solution is calculated by using the expression,

${\text{Molarity of V}}^{\text{2+}}=\frac{{\text{Moles of V}}^{\text{2+}}}{\text{Liters Solution}}$

Substitute the value of moles of ${\text{V}}^{\text{2+}}$ and volume of solution in the above expression to calculate the molarity (concentration) of ${\text{V}}^{\text{2+}}$.

$\begin{array}{c}{\text{Molarity of V}}^{\text{2+}}=\frac{{\text{Moles of V}}^{\text{2+}}}{\text{Liters Solution}}\\ =\frac{0.040\text{0 mol}}{1.0\text{0 L}}\\ =0.040\text{0 M}\end{array}$

Hence, the concentration of ${\text{V}}^{\text{2+}}$ is $0.04\text{00 M}$.

The half cell reaction for galvanic cell containing $1.0\text{0 M}$ ${\text{Cu}}^{\text{2+}}\left(aq\right){\text{ and V}}^{\text{2+}}\left(aq\right)$ is written as follows,

$\text{Cu}\left(s\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(aq\right)+2{\text{e}}^{-}{E}_{\text{anode}}^{\text{ο}}=0.3\text{4 V}$ (1)

${\text{V}}^{\text{2+}}\left(aq\right)+2{\text{e}}^{-}\underset{}{\to }\text{V}\left(s\right)E_{\text{cathode}}^{\text{ο}}=-1.2\text{0 V}$ (2)

The overall reaction is,

$\text{Cu}\left(s\right)+{\text{V}}^{\text{2+}}\left(aq\right)\stackrel{}{\to }{\text{Cu}}^{\text{2+}}\left(aq\right)+\text{V}\left(s\right)$ (3)

Where,

• $E_{\text{anode}}^{\text{ο}}$ is the standard reduction potential of anode.
• $E_{\text{cathode}}^{\text{ο}}$ is the standard reduction potential of cathode.

Cell potential of the reaction is the difference in voltage between the cathode and anode. The cell potential is calculated by using the expression,

$E^{\text{ο}}=E^{\text{ο}}\left(\text{cathode}\right)-E^{\text{ο}}\left(\text{anode}\right)$

Substitute the value of standard reduction potential of anode and cathode in the above expression to calculate the value of cell potential of the reaction.

$\begin{array}{c}{E}^{\text{ο}}=E^{\text{ο}}\left(\text{cathode}\right)-E^{\text{ο}}\left(\text{anode}\right)\\ =-1.20\text{ V}-\left(\text{0}\text{.34 V}\right)\\ =-1.5\text{4 V}\end{array}$

Hence, the standard cell potential of reaction is $-1.5\text{4 V}$.

The number of electrons involved in the reaction is two. The cell potential is calculated by using the Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}$

Where,

• $n$ is the number of electrons in the cell reaction.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.

Substitute the values of standard cell potential, $n$ and molar concentrations of vanadium and copper ions in the above expression to calculate the cell potential.

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ =-1.5\text{4 V}-\frac{0.0591}{2}\log \frac{\left[0.040\text{0 M}\right]}{\left[1.0\text{0 M}\right]}\\ =-\underset{\_}{1.5V}\end{array}$

Hence, the cell potential before titration is $-\underset{\_}{1.5V}$.

Interpretation Introduction

Interpretation: The $E_{\text{cell}}$ before the titration was carried out is to be calculated. The value of the equilibrium constant $\left(K\right)$ for the titration reaction and the value of $E_{\text{cell}}$ at the halfway point in the titration are to be calculated.

Concept introduction: Electrochemical cell is a combination of two half cells in which the two electrodes are joined by a wire and a salt bridge. Electrons flow from anode where oxidation occurs to cathode where reduction takes place. Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. Equilibrium constant is defined as the ratio of the concentration of products and the concentration of the reactants.

To determine: The value of the equilibrium constant $\left(K\right)$ for the titration reaction.

Answer to Problem 159MP

The equilibrium constant is $\underset{\_}{1.6\times 10^8}$.

Explanation of Solution

Explanation

Given

Given

$E_{\text{cell}}=1.9\text{8 V}$

The concentration of ${\text{V}}^{\text{2+}}\left(aq\right)$ is calculated by using Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}$

Substitute the values of $E_{\text{cell}}$, $E_{\text{cell}}^{\text{ο}}$  and $n$ in the above expression to calculate the concentration of ${\text{V}}^{\text{2+}}\left(aq\right)$.

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ 1.9\text{8 V}=-1.\text{5 V}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[1.0\text{0 M}\right]}\\ \left[{\text{V}}^{\text{2+}}\right]=\left(1.0\text{0 M}\right){10}^{-14.89}\\ =1.3\times {10}^{-15}\text{ M}\end{array}$

Thus, the concentration of ${\text{V}}^{\text{2+}}\left(aq\right)$ is $1.3\times {10}^{-15}\text{ M}$.

The titration reaction shows that equal moles of reactant are reacted. So, the equilibrium concentration of reactants ${\text{VEDTA}}^{2-}$ and ${\text{V}}^{\text{2+}}$ is same. Hence, the concentration of ${\text{H}}_2{\text{EDTA}}^{2-}$ at equilibrium is $1.3\times {10}^{-15}\text{ M}$.

${\text{V}}^{\text{2+}}$ concentration is very less in comparison to its initial concentration. Also, the moles of ${\text{VEDTA}}^{2-}$ produced is same in number as that of initial moles of ${\text{V}}^{\text{2+}}$($0.04\text{00 M}$). Thus the concentration of ${\text{VEDTA}}^{2-}$ after addition of $500.0\text{ mL}$$\left(\text{or 0}\text{.500 L}\right)$ ${\text{H}}_2{\text{EDTA}}^{2-}$ is calculated as,

${\text{Molarity of VEDTA}}^{2-}=\frac{{\text{Moles of VEDTA}}^{2-}}{\text{Liters Solution}}$

Substitute the value of moles of ${\text{VEDTA}}^{2-}$ and volume of solution in the above expression to calculate the molarity (concentration) of ${\text{V}}^{\text{2+}}$.

$\begin{array}{c}{\text{Molarity of VEDTA}}^{2-}=\frac{{\text{Moles of VEDTA}}^{2-}}{\text{Liters Solution}}\\ =\frac{0.040\text{0 mol}}{1.0\text{0 L}+0.50\text{0 L}}\\ =0.02\text{67 M}\end{array}$

Hence, the concentration of ${\text{VEDTA}}^{2-}$  is $0.0267\text{ M}$.

The buffered solution $\text{pH}=10.00$. So, $\text{pH}$ does not change. The concentration of ${\text{H}}^{\text{+}}$ is calculated by using the $\text{pH}$ formula,

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{\text{+}}\right]\\ 10=-\log \left[{\text{H}}^{\text{+}}\right]\\ \left[{\text{H}}^{\text{+}}\right]=1.0\times {10}^{-10}\text{ M}\end{array}$

Thus, the concentration of ${\text{H}}^{\text{+}}$ is $1.0\times {10}^{-10}\text{ M}$.

The equilibrium constant is calculated by using the given expression,

$K=\frac{\left[{\text{VEDTA}}^{2-}\right]{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{H}}_{\text{2}}{\text{EDTA}}^{2-}\right]\left[{\text{V}}^{\text{2+}}\right]}$

Substitute the values of given concentrations in the above expression to calculate the equilibrium constant.

$\begin{array}{c}K=\frac{\left[{\text{VEDTA}}^{2-}\right]{\left[{\text{H}}^{\text{+}}\right]}^2}{\left[{\text{H}}_{\text{2}}{\text{EDTA}}^{2-}\right]\left[{\text{V}}^{\text{2+}}\right]}\\ =\frac{\left(0.0267\right){\left(1.0\times {10}^{-10}\right)}^2}{\left(1.3\times {10}^{-15}\right)\left(1.3\times {10}^{-15}\right)}\\ =\underset{\_}{1.6\times 10^8}\end{array}$

Hence, the equilibrium constant is $\underset{\_}{1.6\times 10^8}$.

Interpretation Introduction

Interpretation: The $E_{\text{cell}}$ before the titration was carried out is to be calculated. The value of the equilibrium constant $\left(K\right)$ for the titration reaction and the value of $E_{\text{cell}}$ at the halfway point in the titration are to be calculated.

Concept introduction: Electrochemical cell is a combination of two half cells in which the two electrodes are joined by a wire and a salt bridge. Electrons flow from anode where oxidation occurs to cathode where reduction takes place. Cell potential is defined as the measure of energy per unit charge available from the redox reaction to carry out the reaction. Equilibrium constant is defined as the ratio of the concentration of products and the concentration of the reactants.

To determine: The value of $E_{\text{cell}}$ at the halfway point in the titration.

Answer to Problem 159MP

Solution

The cell potential at halfway titration is $-\underset{\_}{1.49V}$.

Explanation of Solution

Explanation

Given

At halfway equivalence point, the volume of solution added to $1.0\text{0 L}$ solution becomes half. The total volume of solution is calculated as,

$\begin{array}{c}\text{Total volume}={\text{V}}_{\text{vanadium}}+{\text{V}}_{\text{added}}\\ =1.0\text{0 L}+\frac{0.50\text{0 L}}{2}\\ =1.2\text{5 L}\end{array}$

Thus, the total volume of solution is $1.\text{25 L}$.

The initial $0.04\text{00 M}$ of ${\text{V}}^{\text{2+}}$ also become half at equivalence point that is $0.02\text{00 M}$.Hence, the concentration of ${\text{V}}^{\text{2+}}$ at this equivalence point is calculated as,

${\text{Molarity of V}}^{\text{2+}}=\frac{{\text{Moles of V}}^{\text{2+}}}{\text{Liters Solution}}$

Substitute the value of moles of ${\text{V}}^{\text{2+}}$ and volume of solution in the above expression to calculate the molarity (concentration) of ${\text{V}}^{\text{2+}}$.

$\begin{array}{c}{\text{Molarity of V}}^{\text{2+}}=\frac{{\text{Moles of V}}^{\text{2+}}}{\text{Liters Solution}}\\ =\frac{0.020\text{0 mol}}{1.25\text{ L}}\\ =0.016\text{ M}\end{array}$

Hence, the concentration of ${\text{V}}^{\text{2+}}$ is $0.016\text{ M}$.

The number of electrons involved in the reaction is two. The cell potential is calculated by using the Nernst equation.

$E_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}$

Where,

• $n$ is the number of electrons in the cell reaction.
• $E_{\text{cell}}^{\text{ο}}$ is the standard cell potential.

Substitute the values of standard cell potential, $n$ and molar concentrations of vanadium and copper ions in the above expression to calculate the cell potential.

$\begin{array}{c}{E}_{\text{cell}}=E_{\text{cell}}^{\text{ο}}-\frac{0.0591}{n}\log \frac{\left[{\text{V}}^{\text{2+}}\right]}{\left[{\text{Cu}}^{\text{2+}}\right]}\\ =-1.5\text{4 V}-\frac{0.0591}{2}\log \frac{\left[0.016\text{ M}\right]}{\left[1.0\text{0 M}\right]}\\ =-\underset{\_}{1.49V}\end{array}$

Hence, the cell potential at halfway titration is $-\underset{\_}{1.49V}$.