Chapter 19, Problem 19.113QP

How many faradays are required for each of the following processes? How many coulombs arc required?

1. a Reduction of 5.0 mol Na⁺ to Na
2. b Reduction of 5.0 mol Cu²⁺ to Cu
3. c Oxidation of 6.0 g H₂O to O₂
4. d Oxidation of 6.0 g Cl⁻ to Cl₂

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 specie loss it’s of 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 19.113QP

Reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$ required $\text{5}\text{.0}\text{F}$ current

Reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$ required $\text{4}{\text{.82×10}}^{\text{5}}\text{C}$ charge

Explanation of Solution

Given:

Reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$

In above reduction 1 mole of electron required to reduce ${\text{Na}}^{\text{+}}\text{ to Na}$

The amount of faradays required is,

$\begin{array}{l}\text{=5}\text{.0}\text{mol}{\text{Na}}^{\text{+}}\text{×}\frac{\text{1}\text{mol}{\text{e}}^{\text{-}}}{\text{1}\text{mol}{\text{Na}}^{\text{+}}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}\text{5}\text{.0}\text{F}\end{array}$

The mole of electron required reducing ${\text{Na}}^{\text{+}}\text{ to Na}$ and given mole of species are plugged in above equation to give a amount of faradays required reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$.

Reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$ required $\text{5}\text{.0}\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=5}\text{.0 F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}\text{4}{\text{.82×10}}^{\text{5}}\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give a coulombs charge reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$.

Reduction of $\text{5}{\text{.0 mol Na}}^{\text{+}}\text{ to Na}$ required $\text{4}{\text{.82×10}}^{\text{5}}\text{C}$ charge

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 specie loss it’s of 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 19.113QP

Reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$ required $\text{5}\text{.0}\text{F}$ current

Reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$ required $\text{9}{\text{.62×10}}^{\text{5}}\text{C}$ charge

Explanation of Solution

Given:

Reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$

In above reduction 2 moles of electrons required to reduce ${\text{Cu}}^{\text{2+}}\text{ to Cu}$

The amount of faradays required is,

$\begin{array}{l}\text{=5}\text{.0}\text{mol}{\text{Cu}}^{\text{2+}}\text{×}\frac{\text{2}\text{mol}{\text{e}}^{\text{-}}}{\text{1}\text{mol}{\text{Cu}}^{\text{2+}}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}\text{10}\text{.0}\text{F}\end{array}$

The moles of electrons required to reduce ${\text{Cu}}^{\text{2+}}\text{ to Cu}$ and given mole of species are plugged in above equation to give an amount of faradays required reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$.

Reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$ required $\text{5}\text{.0}\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=10}\text{.0 F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}9.64{\text{×10}}^{\text{5}}\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give a coulombs charge reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$.

Reduction of $\text{5}{\text{.0 mol Cu}}^{\text{2+}}\text{ to Cu}$ required $\text{9}{\text{.62×10}}^{\text{5}}\text{C}$ charge

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 specie loss it’s of 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 19.113QP

Oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$ required $0.67\text{F}$ current

Oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$ required $6.4{\text{×10}}^4\text{C}$ charge

Explanation of Solution

Given:

Oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$

In above Oxidation 4 moles of electrons required to reduce ${\text{H}}_{\text{2}}{\text{O to O}}_{\text{2}}$

The amount of faradays required is,

The molar mass of water is $\text{18}\text{.10}\text{g}$

$\begin{array}{l}\text{=6}\text{.0 g}{\text{H}}_{\text{2}}\text{O×}\frac{\text{1 mol}{\text{H}}_{\text{2}}\text{O}}{18.10{\text{H}}_{\text{2}}\text{O}}\times \frac{4\text{mol}{\text{e}}^{\text{-}}}{2\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}0.67\text{F}\end{array}$

The molar mass, mole of electron required oxidizing $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$ and given mole of species are plugged in above equation to give an amount of faradays required oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$.

Oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$ required $0.67\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=0}\text{.67 F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}6.4{\text{×10}}^4\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give a coulombs charge Oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$.

Oxidation of $\text{6}{\text{.0 g H}}_{\text{2}}{\text{O to O}}_{\text{2}}$ required $6.4{\text{×10}}^4\text{C}$ charge

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 specie loss it’s of 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 19.113QP

Oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$ required $0.17\text{F}$ current

Oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$ required $1.6{\text{×10}}^4\text{C}$ charge

Explanation of Solution

Given:

Oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$

In above Oxidation 2 moles of electrons required to oxidizing ${\text{ Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$

The amount of faradays required is,

The molar mass of Chlorine is $\text{35}\text{.45}\text{g}$

$\begin{array}{l}\text{=6}\text{.0 g}{\text{Cl}}^{-}\text{×}\frac{\text{1 mol}{\text{Cl}}^{-}}{35.45g{\text{Cl}}^{-}}\times \frac{2\text{mol}{\text{e}}^{\text{-}}}{2\text{mol}{\text{Cl}}^{-}}\text{×}\frac{\text{1}\text{F}}{\text{1}\text{mol}{\text{e}}^{\text{-}}}\\ \text{=}0.17\text{F}\end{array}$

The molar mass, mole of electron required oxidizing $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$ and given mole of species are plugged in above equation to give an amount of faradays required oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$.

Oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$ required $0.17\text{F}$ current

The amount of coulombs required is,

$\begin{array}{l}\text{=}0.17\text{F F×}\frac{\text{96485}\text{C}}{\text{1 F}}\\ \text{=}1.6{\text{×10}}^4\text{C}\end{array}$

The calculated Faraday is plugged in above equation to give a coulombs charge oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$.

Oxidation of $\text{6}{\text{.0 g Cl}}^{\text{-}}\text{to}{\text{ Cl}}_{\text{2}}$ required $1.6{\text{×10}}^4\text{C}$ charge.