Chapter 20, Problem 1DE

(a)

Interpretation Introduction

To determine: Sketch of a labeled voltaic cell and the concentration of the solution.

Answer to Problem 1DE

Solution:

The labeled sketch of the voltaic cell is,

The concentration of the copper and the manganese ion is $1\text{M}$ and the mass of manganese and copper present in the $50\text{mL}$ of their ionic solution is $\text{2}\text{.747}\text{g}$ and $\text{3}\text{.177}\text{g}$ , respectively.

Explanation of Solution

Given

The electrical potential output of the cell is $1.50\text{V}$ .

The external device draws the current of $0.50\text{A}$ in $2\text{h}=2\times 3600\text{s}$ .

The electric potential of the cell is calculated as,

$\text{Electric}\text{potential}\text{of}\text{the}\text{cell}=\text{Oxidation}\text{potential}+\text{Reduction}\text{potential}$

According to the above equation, the difference in the oxidation and the reduction potential should be $1.50\text{V}$ .

The standard oxidation potential of manganese is $+1.18\text{V}$ and the standard reduction potential of copper is $+0.337\text{V}$ . The sum of these two values of standard oxidation and standard reduction potential gives,

$+1.18\text{V}++0.337\text{V}=1.51\text{V}$

Therefore, the pair of manganese and copper is used to construct a voltaic cell with manganese being anode at which oxidation takes place and copper being cathode at which reduction takes place.

The standard reduction potential is the potential of the electrode dipped in the aqueous solution of its ion of concentration $1\text{M}$ .

Therefore, the concentration of the copper and the manganese ion is $1\text{M}$ .

We are given the beaker of the capacity of $100\text{mL}$ .

The volume of the solution of manganese and copper ion is assumed to be $50\text{mL}=\text{0}\text{.05}\text{L}$ .

The labeled sketch of the voltaic cell is,

Figure. 1

The atomic mass of manganese and copper is $54.938\text{g}/\text{mol}$ and $63.546\text{g}/\text{mol}$ , respectively.

The mass of the manganese in the $50\text{mL}$ solution is calculated by the formula,

$\text{Mass}=\text{Molar}\text{concentration}\times \text{Molar}\text{mass}\times \text{Volume}\text{of}\text{solution}\text{in}\text{liter}$

Substitute the value of molar concentration, molar mass and the volume of the solution of manganese in the above equation.

$\begin{array}{c}\text{Mass}=1\text{M}\times 54.938\text{g}/\text{mol}\times \text{0}\text{.05}\text{L}\\ \text{=2}\text{.747}\text{g}\end{array}$

The mass of the copper in the $50\text{mL}$ solution is calculated by the formula,

$\text{Mass}=\text{Molar}\text{concentration}\times \text{Molar}\text{mass}\times \text{Volume}\text{of}\text{solution}\text{in}\text{liter}$

Substitute the value of molar concentration, molar mass and the volume of the solution of copper in the above equation.

$\begin{array}{c}\text{Mass}=1\text{M}\times 63.546\text{g}/\text{mol}\times \text{0}\text{.05}\text{L}\\ \text{=3}\text{.177}\text{g}\end{array}$

Therefore, the mass of manganese and copper present in the $50\text{mL}$ of their ionic solution is $\text{2}\text{.747}\text{g}$ and $\text{3}\text{.177}\text{g}$ , respectively.

Conclusion

The concentration of the copper and the manganese ion is $1\text{M}$ . The mass of manganese is $\text{2}\text{.747}\text{g}$ and copper is $\text{3}\text{.177}\text{g}$ present in the $50\text{mL}$ of their ionic solution.

(b)

Interpretation Introduction

To determine: The concentration of manganese and copper ion in the solution after $2$ hour.

Answer to Problem 1DE

Solution: The concentration of manganese and copper ion in the solution after $2$ hour is $1.37\text{M}$ and $0.63\text{M}$ , respectively.

Explanation of Solution

Given

The electrical potential output of the cell is $1.50\text{V}$ .

The external device draws the current of $0.50\text{A}$ in $2\text{h}=2\times 3600\text{s}$ .

The charge transport is calculated by the formula,

$\text{Charge}\text{transfer}=\text{Current}\times \text{time}$

Substitute the value of current and time in the above formula,

$\begin{array}{c}\text{Charge}\text{transfer}=0.50\text{A}\times 2\times 3600\text{s}\\ =3600\text{C}\end{array}$

Charge on each electron is $1.602\times {10}^{-19}\text{C}$ and the value of Avogadro’s number is $6.023\times {10}^{23}$ .

The moles of electron present in $3600\text{C}$ is calculated as,

$\text{Number}\text{of}\text{electrons}=\frac{\text{Charge}\text{transfer}}{\text{Charge}\text{on}\text{electron}\times \text{Avogadro's}\text{number}}$

Substitute the value of charge transfer, charge on each electron and Avogadro’s number in the above formula.

$\begin{array}{c}\text{Number}\text{of}\text{electrons}=\frac{3600\text{C}}{1.602\times {10}^{-19}\text{C}\times 6.023\times {10}^{23}}\\ =0.037\end{array}$

Two electron transfer takes place by oxidation and reduction of manganese and copper, respectively.

Therefore, the change in moles of manganese and copper is:

$\frac{0.037}{2}=0.0185\text{mol}$ .

The initial concentration of the ions is $1\text{M}$ .

Thus, one mole of ion is present in one liter of the solution and $\text{0}\text{.05}$ mole ion is present in $\text{0}\text{.05}$ liter of the solution.

The number of moles of manganese ion increase due to oxidation of metal. Therefore, the molar concentration of manganese ion in $50\text{mL}=\text{0}\text{.05}\text{L}$ of the solution is:

$\frac{0.05+0.0185}{0.05}=1.37\text{M}$ .

The number of moles of copper ion increase due to oxidation of metal. Therefore, the molar concentration of copper ion in $50\text{mL}=\text{0}\text{.05}\text{L}$ of the solution is:

$\frac{0.05-0.0185}{0.05}=0.63\text{M}$ .

Thus, the concentration of manganese and copper ion after two hours is $1.37\text{M}$ and $0.63\text{M}$ , respectively.

Conclusion

The concentration of manganese and copper ion after two hours is $1.37\text{M}$ and $0.63\text{M}$ , respectively.

(c)

Interpretation Introduction

To determine: The voltage that a cell register at the end of the discharge.

Answer to Problem 1DE

Solution: The voltage that a cell register at the end of the discharge is $1.405\text{V}$ .

Explanation of Solution

Given

The electrical potential output of the cell is $1.50\text{V}$ .

The external device draws the current of $0.50\text{A}$ in $2\text{h}=2\times 3600\text{s}$ .

The concentration of manganese and copper ion after two hours is $1.37\text{M}$ and $0.63\text{M}$ , respectively at discharge.

The standard oxidation potential of manganese is $+1.18\text{V}$ and the standard reduction potential of copper is $+0.337\text{V}$ .

The standard reduction potential is the potential of the electrode dipped in the aqueous solution of its ion of concentration $1\text{M}$ .

Thus, one mole of manganese ion gives the oxidation potential of $+1.18\text{V}$ .

Therefore, $1.37\text{M}$ of manganese ion gives the oxidation potential of $+1.18\text{V}\times \text{1}\text{.37}=+\text{1}\text{.617}\text{V}$ .

One mole of copper ion gives the reduction potential of $+0.337\text{V}$ .

Therefore, $0.63\text{M}$ of manganese ion gives the oxidation potential of $+0.337\text{V}\times 0.63=+\text{0}\text{.212}\text{V}$ .

The electric potential of the cell at discharge is calculated as,

$\text{Electric}\text{potential}\text{of}\text{the}\text{cell}=\text{Oxidation}\text{potential}-\text{Reduction}\text{potential}$

Substitute the value of oxidation and reduction potential in the above equation.

$\begin{array}{c}\text{Electric}\text{potential}\text{of}\text{the}\text{cell}=+\text{1}\text{.617}\text{V}-\text{0}\text{.212}\text{V}\\ =1.405\text{V}\end{array}$

Thus, the end cell potential is $1.405\text{V}$ .

Conclusion

The end cell potential is $1.405\text{V}$ .

(d)

Interpretation Introduction

To determine: The time taken for the reactant of one cell to get completely consumed.

Answer to Problem 1DE

Solution: The time taken for the reactant of one cell to get completely consumed is $108.11$ .

Explanation of Solution

Given

The electrical potential output of the cell is $1.50\text{V}$ .

The external device draws the current of $0.50\text{A}$ in $2\text{h}=2\times 3600\text{s}$ .

The charge transport is calculated by the formula,

$\text{Charge}\text{transfer}=\text{Current}\times \text{time}$

Substitute the value of current and time in the above formula,

$\begin{array}{c}\text{Charge}\text{transfer}=0.50\text{A}\times 2\times 3600\text{s}\\ =3600\text{C}\end{array}$

Charge on each electron is $1.602\times {10}^{-19}\text{C}$ and the value of Avogadro’s number is $6.023\times {10}^{23}$ .

The moles of electron present in $3600\text{C}$ is calculated as,

$\text{Number}\text{of}\text{electrons}=\frac{\text{Charge}\text{transfer}}{\text{Charge}\text{on}\text{electron}\times \text{Avogadro's}\text{number}}$

Substitute the value of charge transfer, charge on each electron and Avogadro’s number in the above formula.

$\begin{array}{c}\text{Number}\text{of}\text{electrons}=\frac{3600\text{C}}{1.602\times {10}^{-19}\text{C}\times 6.023\times {10}^{23}}\\ =0.037\end{array}$

Two electron transfer takes place by oxidation and reduction of manganese and copper, respectively.

Therefore, the change in moles of manganese and copper is $\frac{0.037}{2}=0.0185\text{mol}$ .

Charge is directly proportional to time at constant current flow.

The charge transferred is $0.0185\text{M}$ in two hours.

The total initial concentration of the reactant of one half cell is one molar.

Therefore, the charge transfer of one molar takes place in $\frac{2}{0.0185}=108.11$ hours.

Conclusion

The charge transfer of one molar takes place in $108.11$ hours.