Chapter 17, Problem 95E

What mass of each of the following substances can be produced in 1.0 h with a current of 15 A?

a. Co from aqueous Co²⁺

b. Hf from aqueous Hf⁴⁺

c. I₂ from aqueous KI

d. Cr from molten CrO₃

Interpretation Introduction

Interpretation: The mass of each of the following substances produced in one hour with a current of $\text{15 A}$ is to be calculated.

Concept introduction: The mass of molten salt and ionic solution is calculated by the process of electrolysis in which an electric current is passed through a molten salt or either an ionic solution. In the process of electrolysis the ions are forced to undergo redox reactions or either an oxidation or reduction reaction.

To determine: The mass of Co from aqueous ${\text{Co}}^{2-}$ in one hour with a current of $\text{15 A}$.

Answer to Problem 95E

1. a. The mass of Co from aqueous ${\text{Co}}^{2-}$ is $\underset{\_}{16.4g}$.

Explanation of Solution

Explanation

The mass of Co from aqueous ${\text{Co}}^{2-}$ is $\underset{\_}{16.4g}$.

Given

Current (C) is $\text{15 A}$.

The conversion of A to $\text{C/s}$ is done as,

$\text{1A}=\text{1 C/s}$

Therefore, the conversion of $\text{15A to C/s}$ is done as,

$1\text{5 A}=1\text{5 C/s}$

Time (t) is $\text{1}\text{.0 hr}$.

The conversion of hr to sec is done as,

$\begin{array}{c}\text{1hr}=60\min \\ =360\text{0 s}ec\end{array}$

Therefore, the conversion of $\text{1 hr to sec}$ is done as,

$\text{1 hr}=3\text{600 sec}$

The atomic weight of Cobalt is $58.9\text{3 g/mol}$.

The equivalent weight of Cobalt is calculated by using the formula,

$\text{Equivalent weight of Cobalt }=\frac{\text{Atomic weight of Cobalt}}{\text{Number ofvalence electrons}}$

Substitute the value of atomic weight of Cobalt and number of valence electrons in the above expression.

$\begin{array}{c}\text{Equivalent weight of Cobalt }=\frac{\text{Atomic weight of Cobalt}}{\text{Number ofvalence electrons}}\\ =\frac{58.\text{93 g/mol}}{2{\text{e}}^{-}}\\ =29.4{\text{65 g/mol e}}^{-}\end{array}$

Hence, the electro chemical equivalent (Z) of Cobalt is calculated by using formula,

$Z=\frac{\text{Equivalent weight}}{\text{F}}$

Where,

• F is the faradays constant $\left(96,48{\text{5 C/mol e}}^{-}\right)$.

Substitute the values of equivalent weight and F in the above expression,

$\begin{array}{c}Z=\frac{\text{Equivalent weight}}{\text{F}}\\ =\frac{29.46{\text{5 g/mol e}}^{-}}{96,48{\text{5 C/mol e}}^{-}}\\ =3.0\times {10}^{-4}\text{ g/C}\end{array}$

Now, the mass is calculated by using the formula,

$W=Z.C.t$

Where,

• Z is the electro chemical equivalent.
• C is the current.
• $t$ is the time.

Substitute the values of Z, C and t in the above expression,

$\begin{array}{c}W=Z.C.t\\ =3.0\times {10}^{\text{-4}}\text{ g/C}\times \text{15 C/s }\times \text{3600 s}\\ =\underset{\_}{16.4g}\end{array}$

Therefore, the mass of Co from aqueous ${\text{Co}}^{2-}$ is $\underset{\_}{16.4g}$.

Interpretation Introduction

Interpretation: The mass of each of the following substances produced in one hour with a current of $\text{15 A}$ is to be calculated.

Concept introduction: The mass of molten salt and ionic solution is calculated by the process of electrolysis in which an electric current is passed through a molten salt or either an ionic solution. In the process of electrolysis the ions are forced to undergo redox reactions or either an oxidation or reduction reaction.

To determine: The mass of $\text{Hf}$ from aqueous ${\text{Hf}}^{4+}$ in one hour with a current of $\text{15 A}$.

Answer to Problem 95E

1. b. The mass of $\text{Hf}$ from aqueous ${\text{Hf}}^{4+}$ is $\underset{\_}{25.0g}$.

Explanation of Solution

Explanation

The mass of $\text{Hf}$ from aqueous ${\text{Hf}}^{4+}$ is $\underset{\_}{25.0g}$.

The atomic weight of $\text{Hf}$ is $\text{178}\text{.49 g/mol}$.

The equivalent weight of $\text{Hf}$ is calculated by using the formula,

$\text{Equivalent weight of Hf }=\frac{\text{Atomic weight of Hf}}{\text{Number ofvalence electrons}}$

Substitute the value of atomic weight of $\text{Hf}$ and number of valence electrons in the above expression.

$\begin{array}{c}\text{Equivalent weight of Hf }=\frac{\text{Atomic weight of Hf}}{\text{Number ofvalence electrons}}\\ =\frac{\text{178}\text{.49 g/mol}}{{\text{4e}}^{-}}\\ =44.62{\text{ g/mol e}}^{-}\end{array}$

Hence, the electro chemical equivalent (Z) of $\text{Hf}$ is calculated by using formula,

$Z=\frac{\text{Equivalent weight}}{\text{F}}$

Where,

• F is the faradays constant $\left(96,48{\text{5 C/mol e}}^{-}\right)$.

Substitute the values of equivalent weight and F in the above expression,

$\begin{array}{c}Z=\frac{\text{Equivalent weight}}{\text{F}}\\ =\frac{\text{44}{\text{.62g/mol e}}^{-}}{96,48{\text{5 C/mol e}}^{-}}\\ =4.6\times {10}^{-4}\text{ g/C}\end{array}$

Now, the mass is calculated by using the formula,

$W=Z.C.t$

Where,

• Z is the electro chemical equivalent.
• C is the current.
• $t$ is the time.

Substitute the values of Z, C and t in the above expression,

$\begin{array}{c}W=Z.C.t\\ =4.6\times {10}^{\text{-4}}\text{ g/C}\times \text{15 C/s }\times \text{3600 s}\\ =\underset{\_}{24.97g}\\ \approx \underset{\_}{25.0g}\end{array}$

The mass of $\text{Hf}$ from aqueous ${\text{Hf}}^{4+}$ is $\underset{\_}{25.0g}$.

Interpretation Introduction

Interpretation: The mass of each of the following substances produced in one hour with a current of $\text{15 A}$ is to be calculated.

Concept introduction: The mass of molten salt and ionic solution is calculated by the process of electrolysis in which an electric current is passed through a molten salt or either an ionic solution. In the process of electrolysis the ions are forced to undergo redox reactions or either an oxidation or reduction reaction.

To determine: The mass of ${\text{I}}_2$ from aqueous $\text{KI}$ in one hour with a current of $\text{15 A}$.

Answer to Problem 95E

1. c. The mass of ${\text{I}}_2$ from aqueous $\text{KI}$ is $\underset{\_}{71.0g}$.

Explanation of Solution

Explanation

The mass of ${\text{I}}_2$ from aqueous $\text{KI}$ is $\underset{\_}{71.0g}$.

The atomic weight of Iodine is $\text{126}\text{.90 g/mol}$

The equivalent weight of Iodine is calculated by using the formula,

$\text{Equivalent weight of Iodine }=\frac{\text{Atomic weight of Iodine}}{\text{Number ofvalence electrons}}$

Substitute the value of atomic weight of Iodine and number of valence electrons in the above expression.

$\begin{array}{c}\text{Equivalent weight of Iodine }=\frac{\text{Atomic weight of Iodine}}{\text{Number ofvalence electrons}}\\ =\frac{\text{126}\text{.90 g/mol}}{{\text{1e}}^{-}}\\ =126.90{\text{ g/mol e}}^{-}\end{array}$

Hence, the electro chemical equivalent (Z) of Iodine is calculated by using formula,

$Z=\frac{\text{Equivalent weight}}{\text{F}}$

Where,

• F is the faradays constant $\left(96,48{\text{5 C/mol e}}^{-}\right)$.

Substitute the values of equivalent weight and F in the above expression,

$\begin{array}{c}Z=\frac{\text{Equivalent weight}}{\text{F}}\\ =\frac{\text{126}{\text{.90 g/mol e}}^{-}}{96,48{\text{5 C/mol e}}^{-}}\\ =13.1\times {10}^{-4}\text{ g/C}\end{array}$

Now, the mass is calculated by using the formula,

$W=Z.C.t$

Where,

• Z is the electro chemical equivalent.
• C is the current.
• $t$ is the time.

Substitute the values of Z, C and t in the above expression,

$\begin{array}{c}W=Z.C.t\\ =13.1\times {10}^{\text{-4}}\text{ g/C}\times \text{15 C/s }\times \text{3600 s}\\ =\underset{\_}{71.0g}\end{array}$

The mass of ${\text{I}}_2$ from aqueous $\text{KI}$ is $\underset{\_}{71.0g}$.

Interpretation Introduction

Interpretation: The mass of each of the following substances produced in one hour with a current of $\text{15 A}$ is to be calculated.

Concept introduction: The mass of molten salt and ionic solution is calculated by the process of electrolysis in which an electric current is passed through a molten salt or either an ionic solution. In the process of electrolysis the ions are forced to undergo redox reactions or either an oxidation or reduction reaction.

To determine: The mass of $\text{Cr}$ from molten ${\text{CrO}}_{\text{3}}$ in one hour with a current of $\text{15 A}$.

Answer to Problem 95E

1. d. The mass of $\text{Cr}$ from molten ${\text{CrO}}_{\text{3}}$ is $\underset{\_}{4.84g}$.

Explanation of Solution

Explanation

The mass of $\text{Cr}$ from molten ${\text{CrO}}_{\text{3}}$ is $\underset{\_}{4.84g}$.

The atomic weight of $\text{Cr}$ is $\text{51}\text{.996 g/mol}$

The equivalent weight of $\text{Cr}$ is calculated by using the formula,

$\text{Equivalent weight of Cr }=\frac{\text{Atomic weight of Cr}}{\text{Number ofvalence electrons}}$

Substitute the value of atomic weight of $\text{Cr}$ and number of valence electrons in the above expression.

$\begin{array}{c}\text{Equivalent weight of Cr }=\frac{\text{Atomic weight of Cr}}{\text{Number ofvalence electrons}}\\ =\frac{\text{51}\text{.996 g/mol}}{{\text{6e}}^{-}}\\ =8.666{\text{ g/mol e}}^{-}\end{array}$

Hence, the electro chemical equivalent (Z) of Iodine is calculated by using formula,

$Z=\frac{\text{Equivalent weight}}{\text{F}}$

Where,

• F is the faradays constant $\left(96,48{\text{5 C/mol e}}^{-}\right)$.

Substitute the values of equivalent weight and F in the above expression,

$\begin{array}{c}Z=\frac{\text{Equivalent weight}}{\text{F}}\\ =\frac{\text{8}{\text{.666 g/mol e}}^{-}}{96,48{\text{5 C/mol e}}^{-}}\\ =8.9817\times {10}^{-5}\text{ g/C}\end{array}$

Now, the mass is calculated by using the formula,

$W=Z.C.t$

Where,

• Z is the electro chemical equivalent.
• C is the current.
• $t$ is the time.

Substitute the values of Z, C and t in the above expression,

$\begin{array}{c}W=Z.C.t\\ =8.9817\times {10}^{\text{-5}}\text{ g/C}\times \text{15 C/s }\times \text{3600 s}\\ =\underset{\_}{4.85g}\end{array}$

The mass of $\text{Cr}$ from molten ${\text{CrO}}_{\text{3}}$ is $\underset{\_}{4.84g}$.