Chapter 19, Problem 98GQ

(a)

Interpretation Introduction

Interpretation:

The number of moles of electrons a battery can deliver in one hour has to be determined.

Concept introduction:

The Faraday’s first law of electrolysis state that the mass of the substance $\left(\text{m}\right)$ deposited at any electrode is directly proportional to the charge $\left(\text{Q}\right)$ passed. The mathematical form of the Fraday’s first law is written as’

$\text{m}=\left(\frac{\text{Q}}{\text{F}}\right)\left(\frac{\text{M}}{\text{Z}}\right)$

Here,

The symbol $\text{F}$ is the Faraday’s constant.

The symbol $\text{M}$ is the molar mass of the substance in grams per mol.

The symbol $\text{Z}$ is the valency number of ions of the substance (electrons transferred per ion).

In the simple case of constant current electrolysis, $\text{Q}=\text{I}\times \text{t}$ leading to

$\text{m}=\left(\frac{\text{I}\times \text{t}}{\text{F}}\right)\left(\frac{\text{M}}{\text{Z}}\right)$ (1)

The above formula is written in terms of the number of moles $\left(\text{n}\right)$,

$\text{n}=\left(\frac{\text{I}\times \text{t}}{\text{F}}\right)\left(\frac{1}{\text{Z}}\right)$ (2)

Here, $\text{t}$ is the total time the constant current $\left(\text{I}\right)$ is applied.

(b)

Interpretation Introduction

Interpretation:

The mass of lithium oxidized under given conditions in one hour has to be calculated.

Concept introduction:

The Faraday’s first law of electrolysis state that the mass of the substance $\left(\text{m}\right)$ deposited at any electrode is directly proportional to the charge $\left(\text{Q}\right)$ passed. The mathematical form of the Fraday’s first law is written as’

$\text{m}=\left(\frac{\text{Q}}{\text{F}}\right)\left(\frac{\text{M}}{\text{Z}}\right)$

Here,

The symbol $\text{F}$ is the Faraday’s constant.

The symbol $\text{M}$ is the molar mass of the substance in grams per mol.

The symbol $\text{Z}$ is the valence number of ions of the substance (electrons transferred per ion).

In the simple case of constant current electrolysis, $\text{Q}=\text{I}\times \text{t}$ leading to

$\text{m}=\left(\frac{\text{I}\times \text{t}}{\text{F}}\right)\left(\frac{\text{M}}{\text{Z}}\right)$ (1)

The above formula is written in terms of the number of moles $\left(\text{n}\right)$,

$\text{n}=\left(\frac{\text{I}\times \text{t}}{\text{F}}\right)\left(\frac{1}{\text{Z}}\right)$ (2)

Here, $\text{t}$ is the total time the constant current $\left(\text{I}\right)$ is applied.