1 Electrochemical Electrochemical Cell Voltaic or Galvanic Cells D8 c34

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ElectrochemicElectrochemicalal Cell

ElectrochemicElectrochemicalal Cell

Voltaic orGalvanic Cells

D8 c34

21.3 Electrolytic Cells >21.3 Electrolytic Cells >

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Electrolytic vs. Voltaic CellsElectrolytic vs. Voltaic Cells

Electrolytic vs. Voltaic Cells

How do voltaic and electrolytic cells differ?




The process in which electrical energy is used to bring about a chemical change is called electrolysis.

• You are already familiar with some results of electrolysis, such as gold-plated jewelry, chrome-plated automobile parts, and silver-plated dishes.



The apparatus in which electrolysis is carried out is an electrolytic cell.

• An electrolytic cell is an electrochemical cell used to cause

a chemical change through the application of electrical energy.

• An electrolytic cell uses electrical energy (direct current) to make a non-spontaneous

redox reaction proceed to completion.

55

ElectrolysisElectrolysisUsing electrical energy to produce chemical change.

Sn2+(aq) + 2 Cl-(aq) ---> Sn(s) + Cl2(g)

SnSnClCl22

SnClSnCl22(aq)(aq)




In both voltaic and electrolytic cells, electrons flow from the anode to the

cathode in the external circuit.

Anode (oxidation)

Anode (oxidation)

Cathode (reduction)

Cathode (reduction)

Energy

Energy

BatteryVoltaic Cell Electrolytic Celle–

e– e–

e–




The key difference between voltaic and electrolytic cells is that in a voltaic cell, the flow of electrons is the result of a spontaneous redox reaction, whereas in an electrolytic cell, electrons are caused to flow by an outside power source, such as a battery.

Anode (oxidation)

Anode (oxidation)

Cathode (reduction)

Cathode (reduction)

Energy

Energy


e– e–

e–




• In a voltaic cell, the anode is the negative electrode and the cathode is the positive electrode.

• In an electrolytic cell, the cathode is considered the negative electrode.

Anode (oxidation)

Anode (oxidation)

Cathode (reduction)

Cathode (reduction)

Energy

Energy


e– e–

e–



In electrolysis, an electric current is used to do which of the following?

A. Cause a chemical change

B. Produce a battery

C. Generate heat

D. Run a motor



In electrolysis, an electric current is used to do which of the following?

A. Cause a chemical change

B. Produce a battery

C. Generate heat

D. Run a motor



Driving Nonspontaneous Driving Nonspontaneous ProcessesProcesses

Driving Nonspontaneous Processes

What are some applications that use electrolytic cells?




Electrolysis of a solution or of a melted, or molten, ionic compound can result in the separation of elements from compounds.

Electrolytic cells are also commonly used in the plating, purifying, and refining of metals.

1313Electrolysis Electrolysis

of of waterwater

Cathode (-) Cathode (-)

4 H2O + 4e- --->

2H2 + 4 OH-

Anode (+) Anode (+)

2 H2O --->

O2(g) + 4 H+ +

4e-



The overall cell reaction is obtained by adding the half-reactions (after doubling the reduction half-reaction equation to balance electrons).

Electrolysis of Water

Oxidation: 2H2O(l) → O2(g) + 4H+(aq) + 4e–

Reduction: 2 [2H2O(l) + 2e– → H2(g) + 2OH–(aq)]

6H2O(l) → 2H2(g) + O2(g) + 4H+(aq) + 4OH–(aq)




Electrolysis of Molten Sodium Chloride

The electrolytic cell in which this commercial process is carried out is called the Downs cell.• The cell operates at

a temperature of 801°C so that the sodium chloride is maintained in the molten state.




Sodium and chlorine are produced through the electrolysis of pure molten sodium chloride, rather than an aqueous solution of NaCl.

• Chlorine gas is produced at the anode.

• Molten sodium collects at the cathode.



• The overall equation is the sum of the two half-reactions:


Reduction: 2Na+(l) + 2e– → 2Na(l)

2NaCl(l) → 2Na(l) + Cl2(g)

Oxidation: 2Cl–(l) → Cl2(g) + 2e–



Electroplating

Electroplating is the deposition of a thin layer of a metal on an object in an electrolytic cell.

• An object that is to be silver-plated is made the cathode in an electrolytic cell.

• The anode is the metallic silver that is to be deposited.

• The electrolyte is a solution of a silver salt, such as silver cyanide.



Electroplating

When direct current is applied, silver ions move from the anode to the object to be

plated.

Reduction: Ag+(aq) + e– → Ag(s) (at cathode)

This figure shows statuettes that were electroplated with copper, nickel, and 24-carat gold.



Electrorefining

In the process of electrorefining, a piece of impure metal is made into the anode of the cell.

• It is oxidized to the cation and then reduced to the pure metal at the cathode.

• This technique is used to obtain ultrapure silver, lead, and copper.



Electrowinning

In a process called electrowinning, impure metals can be purified in electrolytic cells.

• The cations of molten salts or aqueous solutions are reduced at the cathode to give very pure metals.

2222

Producing AluminumProducing Aluminum2 Al2 Al22OO33 + 3 C ---> 4 Al + 3 CO + 3 C ---> 4 Al + 3 CO22

Charles Hall (1863-1914) developed Charles Hall (1863-1914) developed electrolysis process. Founded Alcoa.electrolysis process. Founded Alcoa.



Key Concepts Key Concepts

The key difference between voltaic and electrolytic cells is that in a voltaic cell, the flow of electrons is the result of a spontaneous redox reaction, whereas in an electrolytic cell, electrons are caused to flow by an outside power source, such as a battery.

Electrolysis of a solution or of a melted, or molten, ionic compound can result in the separation of elements from compounds. Electrolytic cells are also commonly used in the plating, purifying, and refining of metals.



Glossary TermsGlossary Terms

• electrolysis: a process in which electrical energy is used to bring about a chemical change; the electrolysis of water produces hydrogen and oxygen

• electrolytic cell: an electrochemical cell used to cause a chemical change through the application of electrical energy



• The two types of electrochemical cells are voltaic cells and electrolytic cells.

• In an electrolytic cell, a nonspontaneous redox reaction is driven by the application of electrical energy.

• Electrolytic cells are used to produce commercially important chemicals and to plate, purify, and refine metals.

BIG IDEABIG IDEA

Matter and Energy

2626

QuanQuantitative Aspects of titative Aspects of ElectrochemistryElectrochemistry

electrolysis of aqueous silver ion.

Ag+ (aq) + e- ---> Ag(s)

1 mol e- ---> 1 mol Ag

If measure the moles of e-, can know the quantity of Ag formed.

But how to measure moles of e-?

2727

QuanQuantitative Aspects of titative Aspects of ElectrochemistryElectrochemistry

• Measure the electrical current

Current = charge passing

timeCurrent =

charge passingtime

I (amps) = coulombsseconds


Units

2828

But how is charge related to moles of But how is charge related to moles of electrons?electrons?

Current = charge passing

timeCurrent =

charge passingtime



Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

= = 96,500 C/mol e- 96,500 C/mol e- = = 1 Faraday1 Faraday

Charge on 1 mol e -

= 1.60 x 10-19 Ce -

6.02 x 1023

e -mol

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Michael FaradayMichael Faraday1791-18671791-1867

Originated the terms anode, cathode, anion, cation, electrode.

Discoverer of • electrolysis• magnetic props. of matter• electromagnetic induction• benzene and other organic

chemicals

3232Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

1.50 A flow thru a Ag+(aq) solution for 15.0 min. What mass of Ag metal is deposited?

Solution: Solution: Calc. charge

Charge (C) = current (A) x time (t)

= (1.5 A)(15.0 min)(60 s/min)

= 1350 C



3333Quantitative Aspects of Quantitative Aspects of ElectrochemistryElectrochemistry

Charge = 1350 C

NEXT: Calculate moles of e- used

1350 C • 1 mol e -96, 500 C

0.0140 mol e -1350 C • 1 mol e -96, 500 C

0.0140 mol e -

0.0140 mol e - • 1 mol Ag1 mol e -

0.0140 mol Ag or 1.51 g Ag0.0140 mol e - • 1 mol Ag1 mol e -

0.0140 mol Ag or 1.51 g Ag

(c)(c) Calc. quantity of Ag

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If a battery delivers 1.50 A, and you have 454 g of Pb, how long will the battery last?

SolutionSolution

a) 454 g Pb = 2.19 mol Pb

b) Calculate moles of e-

2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -2.19 mol Pb • 2 mol e -1 mol Pb

= 4.38 mol e -

c)c) Calculate chargeCalculate charge 4.38 mol e- • 96,500 C/mol e- = 423,000 C4.38 mol e- • 96,500 C/mol e- = 423,000 C

Pb(s) + HSO4-(aq) ---> PbSO4(s) + H+(aq) + 2e-

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NEXT: NEXT:

Use Charge = 423,000 C

Time (s) = Charge (C)

I (amps)Time (s) =

Charge (C)I (amps)

Time (s) = 423,000 C1.50 amp

= 282,000 sTime (s) = 423,000 C1.50 amp

= 282,000 s

About 78 hoursAbout 78 hours

d)d) Calculate timeCalculate time

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1 Electrochemical Electrochemical Cell Voltaic or Galvanic Cells D8 c34