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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

A lithium-ion camera battery is rated at 7500 mAh. That is, it can deliver 7500 milliamps (mA) or 7.5 amps of steady current for an hour.

  1. (a) How many moles of electrons can the battery deliver in one hour?
  2. (b) What mass of lithium is oxidized under these conditions in 1.0 hour?

(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 (m) deposited at any electrode is directly proportional to the charge (Q) passed. The mathematical form of the Fraday’s first law is written as’

m=(QF)(MZ)

Here,

The symbol F is the Faraday’s constant.

The symbol M is the molar mass of the substance in grams per mol.

The symbol Z is the valency number of ions of the substance (electrons transferred per ion).

In the simple case of constant current electrolysis, Q=I×t leading to

m=(I×tF)(MZ) (1)

The above formula is written in terms of the number of moles (n),

n=(I×tF)(1Z) (2)

Here, t is the total time the constant current (I) is applied.

Explanation

Given:

The constant current I=7.5A.

The time t is 3600s.

The faraday’s constant F is equal to 96500Cmol1.

The reaction of a lithium-ion battery is written as,

    Li(s)LiCoO2(s)+e

Here, Z=1

(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 (m) deposited at any electrode is directly proportional to the charge (Q) passed. The mathematical form of the Fraday’s first law is written as’

m=(QF)(MZ)

Here,

The symbol F is the Faraday’s constant.

The symbol M is the molar mass of the substance in grams per mol.

The symbol Z is the valence number of ions of the substance (electrons transferred per ion).

In the simple case of constant current electrolysis, Q=I×t leading to

m=(I×tF)(MZ) (1)

The above formula is written in terms of the number of moles (n),

n=(I×tF)(1Z) (2)

Here, t is the total time the constant current (I) is applied.

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Chapter 19 Solutions

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Sect-19.8 P-19.11CYUSect-19.9 P-1.1ACPSect-19.9 P-1.2ACPSect-19.9 P-1.3ACPSect-19.9 P-2.1ACPSect-19.9 P-2.2ACPSect-19.9 P-2.3ACPSect-19.9 P-2.4ACPSect-19.9 P-2.5ACPCh-19 P-1PSCh-19 P-2PSCh-19 P-3PSCh-19 P-4PSCh-19 P-5PSCh-19 P-6PSCh-19 P-7PSCh-19 P-8PSCh-19 P-9PSCh-19 P-10PSCh-19 P-11PSCh-19 P-12PSCh-19 P-13PSCh-19 P-14PSCh-19 P-15PSCh-19 P-16PSCh-19 P-17PSCh-19 P-18PSCh-19 P-19PSCh-19 P-20PSCh-19 P-21PSCh-19 P-22PSCh-19 P-23PSCh-19 P-24PSCh-19 P-25PSCh-19 P-26PSCh-19 P-27PSCh-19 P-28PSCh-19 P-29PSCh-19 P-30PSCh-19 P-31PSCh-19 P-32PSCh-19 P-33PSCh-19 P-34PSCh-19 P-35PSCh-19 P-36PSCh-19 P-37PSCh-19 P-38PSCh-19 P-39PSCh-19 P-40PSCh-19 P-41PSCh-19 P-42PSCh-19 P-43PSCh-19 P-44PSCh-19 P-45PSCh-19 P-46PSCh-19 P-47PSCh-19 P-48PSCh-19 P-49PSCh-19 P-50PSCh-19 P-51PSCh-19 P-52PSCh-19 P-53PSCh-19 P-54PSCh-19 P-55PSCh-19 P-56PSCh-19 P-57GQCh-19 P-58GQCh-19 P-59GQCh-19 P-60GQCh-19 P-61GQCh-19 P-62GQCh-19 P-63GQCh-19 P-64GQCh-19 P-65GQCh-19 P-66GQCh-19 P-67GQCh-19 P-68GQCh-19 P-69GQCh-19 P-70GQCh-19 P-71GQCh-19 P-72GQCh-19 P-73GQCh-19 P-74GQCh-19 P-75GQCh-19 P-76GQCh-19 P-77GQCh-19 P-78GQCh-19 P-79GQCh-19 P-80GQCh-19 P-81GQCh-19 P-82GQCh-19 P-83GQCh-19 P-84GQCh-19 P-85GQCh-19 P-86GQCh-19 P-87GQCh-19 P-88GQCh-19 P-89GQCh-19 P-90GQCh-19 P-91GQCh-19 P-92GQCh-19 P-93GQCh-19 P-94GQCh-19 P-95GQCh-19 P-96GQCh-19 P-97GQCh-19 P-98GQCh-19 P-99GQCh-19 P-100GQCh-19 P-101GQCh-19 P-102GQCh-19 P-103GQCh-19 P-104GQCh-19 P-105ILCh-19 P-106ILCh-19 P-107ILCh-19 P-108ILCh-19 P-109ILCh-19 P-110ILCh-19 P-111SCQCh-19 P-112SCQCh-19 P-113SCQCh-19 P-114SCQCh-19 P-115SCQ

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