Chapter 19, Problem 19.114QP

(a)

Interpretation Introduction

Interpretation:

The amount of Faraday currents and coulombs charges are required to reduction of given mole of species should be calculated.

Concept introduction:

Faraday:

Faraday is a unit for electric charge, which is equal to the Faraday’s constant and it is given by $\text{(F)}$ in physics and chemistry.

Faraday’s constant:

One mole (Avogadro's number) of electrons containing electrical charge is given by the ratio of number of electrons per mole with number of electrons per coulomb and it is,

$\begin{array}{l}\text{=}\frac{\text{6}{\text{.23×10}}^{\text{23}}}{\text{6}{\text{.24×10}}^{\text{18}}}\\ \\ \text{=}\text{96485}\frac{\text{Coulombs}}{\text{mole}}\end{array}$

In the redox reaction, a species gains electron from its reaction mixture to get reduced while during the reaction is known as reduction.

In the redox reaction, a species lose its electron to get oxidized while during the reaction is known as reduction.

The charge of the species where decreases, while reduction occurs.

In the electro reduction process, the coulombs is given by the amount of charge required to reduction of given mole of species.

In the electro oxidation process, the coulombs is given by the amount of charge required to oxidation of given mole of species.

Answer to Problem 19.114QP

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$ required $2.\text{0}\text{F}$ current

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$ required $\text{1}{\text{.9×10}}^{\text{5}}\text{C}$ charge

Explanation of Solution

Given:

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$

In above reduction 1 mole of electron is required to reduce ${\text{Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$

The amount of faradays required is,

$\begin{array}{l}\text{=2}\text{.0}\text{mol}{\text{Fe}}^{\text{3+}}\text{×}\frac{\text{1}\text{mol}{\text{e}}^{\text{-}}}{\text{1}\text{mol}{\text{Fe}}^{\text{3+}}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}\text{2}\text{.0}\text{F}\end{array}$

The mole of electron required reducing $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$ and given mole of species are plugged in above equation to give a amount of faradays required reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$.

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$ required $2.\text{0}\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=2}\text{.0 F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}\text{1}{\text{.9×10}}^{\text{5}}\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give an amount of coulombs charge reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$.

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}{\text{to Fe}}^{\text{2+}}$ required $\text{1}{\text{.9×10}}^{\text{5}}\text{C}$ charge

(b)

Interpretation Introduction

Interpretation:

The amount of Faraday currents and coulombs charges are required to reduction of given mole of species should be calculated.

Concept introduction:

Faraday:

Faraday is a unit for electric charge, which is equal to the Faraday’s constant and it is given by $\text{(F)}$ in physics and chemistry.

Faraday’s constant:

One mole (Avogadro's number) of electrons containing electrical charge is given by the ratio of number of electrons per mole with number of electrons per coulomb and it is,

$\begin{array}{l}\text{=}\frac{\text{6}{\text{.23×10}}^{\text{23}}}{\text{6}{\text{.24×10}}^{\text{18}}}\\ \\ \text{=}\text{96485}\frac{\text{Coulombs}}{\text{mole}}\end{array}$

In the redox reaction, a species gains electron from its reaction mixture to get reduced while during the reaction is known as reduction.

In the redox reaction, a species lose its electron to get oxidized while during the reaction is known as reduction.

The charge of the species where decreases, while reduction occurs.

In the electro reduction process, the coulombs is given by the amount of charge required to reduction of given mole of species.

In the electro oxidation process, the coulombs is given by the amount of charge required to oxidation of given mole of species.

Answer to Problem 19.114QP

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$ required $6.\text{0}\text{F}$ current

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$ required $5.8{\text{×10}}^{\text{5}}\text{C}$ charge

Explanation of Solution

Given:

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$

In above reduction 3 mole of electron is required to reduce ${\text{Fe}}^{\text{3+}}\text{to Fe}$

The amount of faradays required is,

$\begin{array}{l}\text{=2}\text{.0}\text{mol}{\text{Fe}}^{\text{3+}}\text{×}\frac{3\text{mol}{\text{e}}^{\text{-}}}{\text{1}\text{mol}{\text{Fe}}^{\text{3+}}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}6.\text{0}\text{F}\end{array}$

The mole of electron required reducing $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$ and given mole of species are plugged in above equation to give a amount of faradays required reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$.

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$ required $6.\text{0}\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=6}\text{.0 F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}5.8{\text{×10}}^{\text{5}}\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give a coulombs charge reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$.

Reduction of $\text{2}{\text{.0 mol Fe}}^{\text{3+}}\text{to Fe}$ required $5.8{\text{×10}}^{\text{5}}\text{C}$ charge

(c)

Interpretation Introduction

Interpretation:

The amount of Faraday currents and coulombs charges are required to reduction of given mole of species should be calculated.

Concept introduction:

Faraday:

Faraday is a unit for electric charge, which is equal to the Faraday’s constant and it is given by $\text{(F)}$ in physics and chemistry.

Faraday’s constant:

One mole (Avogadro's number) of electrons containing electrical charge is given by the ratio of number of electrons per mole with number of electrons per coulomb and it is,

$\begin{array}{l}\text{=}\frac{\text{6}{\text{.23×10}}^{\text{23}}}{\text{6}{\text{.24×10}}^{\text{18}}}\\ \\ \text{=}\text{96485}\frac{\text{Coulombs}}{\text{mole}}\end{array}$

In the redox reaction, a species gains electron from its reaction mixture to get reduced while during the reaction is known as reduction.

In the redox reaction, a species lose its electron to get oxidized while during the reaction is known as reduction.

The charge of the species where decreases, while reduction occurs.

In the electro reduction process, the coulombs is given by the amount of charge required to reduction of given mole of species.

In the electro oxidation process, the coulombs is given by the amount of charge required to oxidation of given mole of species.

Answer to Problem 19.114QP

Oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$ required $0.042\text{F}$ current

Oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$ required $4.1{\text{×10}}^3\text{C}$ charge

Explanation of Solution

Given:

Oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$

In above Oxidation 2 mole of electron is required to reduce ${\text{Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$

The amount of faradays required is,

The molar mass of tin is $\text{118}\text{.7 g }$

$\begin{array}{l}\text{=2}\text{.5}\text{.0 g}{\text{Sn}}^{\text{2+}}\text{×}\frac{\text{1 mol}{\text{Sn}}^{\text{2+}}}{\text{118}{\text{.7 g Sn}}^{\text{2+}}}\times \frac{2\text{mol}{\text{e}}^{\text{-}}}{1\text{mol}{\text{Sn}}^{\text{2+}}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}0.042\text{F}\end{array}$

The molar mass, mole of electron required to oxidize $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$ and given mole of species are plugged in above equation to give an amount of faradays required oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$.

Oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$ required $0.042\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=}0.042\text{F F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}4.1{\text{×10}}^3\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give an amount of coulombs charge oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$.

Oxidation of $\text{2}{\text{.5 g Sn}}^{\text{2+}}\text{to }{\text{Sn}}^{\text{4+}}$ requires $4.1{\text{×10}}^3\text{C}$ charge

(d)

Interpretation Introduction

Interpretation:

The amount of Faraday currents and coulombs charges are required to reduction of given mole of species should be calculated.

Concept introduction:

Faraday:

Faraday is a unit for electric charge, which is equal to the Faraday’s constant and it is given by $\text{(F)}$ in physics and chemistry.

Faraday’s constant:

One mole (Avogadro's number) of electrons containing electrical charge is given by the ratio of number of electrons per mole with number of electrons per coulomb and it is,

$\begin{array}{l}\text{=}\frac{\text{6}{\text{.23×10}}^{\text{23}}}{\text{6}{\text{.24×10}}^{\text{18}}}\\ \\ \text{=}\text{96485}\frac{\text{Coulombs}}{\text{mole}}\end{array}$

In the redox reaction, a species gains electron from its reaction mixture to get reduced while during the reaction is known as reduction.

In the redox reaction, a species lose its electron to get oxidized while during the reaction is known as reduction.

The charge of the species where decreases, while reduction occurs.

In the electro reduction process, the coulombs is given by the amount of charge required to reduction of given mole of species.

In the electro oxidation process, the coulombs is given by the amount of charge required to oxidation of given mole of species.

Answer to Problem 19.114QP

Reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$ required $0.17\text{F}$ current

Reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$ required $3.7{\text{×10}}^3\text{C}$ charge

Explanation of Solution

Given:

Reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$

In above reduction 3 moles of electrons required to reduce ${\text{ Au}}^{\text{3+}}\text{to Au}$

The amount of faradays required is,

The molar mass of Gold is $\text{197}\text{.0}\text{g}$

$\begin{array}{l}\text{=6}\text{.0 g}{\text{Au}}^{\text{3+}}\text{×}\frac{\text{1 mol}{\text{Au}}^{\text{3+}}}{\text{197}\text{.0}\text{g}{\text{Au}}^{\text{3+}}}\text{×}\frac{\text{3}\text{mol}{\text{e}}^{\text{-}}}{\text{1}\text{mol}{\text{Au}}^{\text{3+}}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}\text{0}\text{.038}\text{F}\end{array}$

The molar mass, mole of electron required reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$ and given mole of species are plugged in above equation to give an amount of faradays required reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$.

Reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$ required $0.17\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=}0.038\text{F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}3.7{\text{×10}}^3\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give a coulombs charge reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$.

Reduction of $\text{2}{\text{.5 g Au}}^{\text{3+}}\text{to Au}$ required $3.7{\text{×10}}^3\text{C}$ charge.