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Chemistry & Chemical Reactivity

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

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Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

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

Electrolysis of molten NaCl is done in cells operating at 7.0 V and 4.0 × 104 A. What mass of Na(s) and Cl2(g) can be produced in 1 day in such a cell? What is the energy consumption in kilowatt-hours? (1 kWh = 3.6 × 106 J and 1 J = 1 C · V)

Interpretation Introduction

Interpretation:

The mass of sodium , chlorine and energy consumption during electrolysis of  molten NaCl has to be determined.

Concept introduction:

Electrolysis:

It is a decomposition of ionic compounds by passing electricity through molten compounds or aqueous solutions of compounds.

Electricity used to produce chemical changes. The apparatus used for electrolysis is called an electrolytic cell.

Current:

Rate of the charge changing in time.

The formula is as follows.

Current (Amperes, A) = Electric charge (Coloumbs, C)time, t(seconds,s)

Simplify as follows.

Current = Electric charge time

To rearrange the above formula is as follows.

Charge = Current(A) × time(s)

Explanation

During the electrolysis process molten NaCl dissociates into Na+and Cl ions. Na+ ions are moves towards cathode and Cl- ions are moves towards anode.

NaCl Na++Cl

At cathode Na+ gain one electron and for sodium and Cl- ion lose one electron to form chlorine gas .since chlorine atom is unstable.

The reactions occur at anode and cathode is as follows.

At anode:Oxidation: 2Cl  Cl2(g)+2e-At cathode:Reduction: 2Na++2e- 2Na(l)

The half cell reaction of the copper is as follows.

Cu2+(aq) + 2e-  Cu(s)

Let’s calculate the charge passing through the cell.

Charge(C)= Current (A) × time (s)= 4.0×104 A × 24 h × (60 min h-1)(60 s min-1)= 3.456×109 C

Let’s calculate the moles electrons:

1mol e = (3.456 ×109 C)(1 mol e96500 C)= 3.58×104 mol e-

Now, calculate the individual ions mass in electrolysis of molten sodium chloride.

Mass of sodium = (3.58 × 104mole-)(2 mole Na2 mol e-)(23

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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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