Wt871rwp Electrochemistry 440 Electrochemistry-presentation 2012-03-31

7/24/2019 Wt871rwp Electrochemistry 440 Electrochemistry-presentation 2012-03-31

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Electrochemistry


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Electrochemistry deals with relationships between

reactions and electricity

In electrochemical reactions, electrons are transferred

from one species to another. Provide insight into batteries, corrosion, electroplating,

spontaneity of reactions

Electrochemistry


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Electrochemical Reactions

In electrochemical reactions, electrons are transferred

between various reactant and product species in reactions.

As a result, oxidation state/number of one or more

substances/species change

Oxidation number is the formal charge on the atom when it is

connected to other atoms.

In order to keep track of what species loses electrons and

what gains them, we assign oxidation numbers/oxidation statesto individual atoms.


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Oxidation Numbers

Zn(s) + 2H+(aq) Zn2++ H2(g)

0 +10+2

Take a look at this reaction between Zn metal and acid withassigned oxidation numbers.How do we know what number goes with each atom?Where do these numbers came from?


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Rules for Assigning Oxidation Numbers

ElementsElements in their elemental form have anoxidation number of 0.

Compounds The sum of the oxidation numbers in aneutral compound is 0.

Monoatomic

ions

The oxidation number of a monatomicion is the same as its charge.

Polyatomic

ions

The sum of the oxidation numbers in apolyatomic ion is the charge on the ion.

How do we assign oxidation numbers ?


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Hydrogen-1 when bonded to a metal+1 when bonded to a nonmetal

Fluorine Fluorine always has an oxidationnumber of -1.

Other

halogens

Usually -1.May have positive oxidation numbers inoxyanions.


For example, Cl has an oxidation number of +5 in ClO3-.


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NonmetalsNonmetals tend to have negativeoxidation numbers although some are

positive in certain compounds or ions.

OxygenOxygen has a oxidation number of -2,except in the peroxide ion, when itsoxidation number is -1.



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1 What is the oxidation number of each oxygen

atom in the compound MnO2?

A -2

B -1

C 0

D +1

E +2


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2 What is the oxidation number of the manganese

atom in the compound MnO2?

A +3

B +2

C +1

D +4

E +7


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3 What is the oxidation number of oxygen atom in

MnO41-, the permanganate ion?

A -2

B -1

C 0

D +2

E +4


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4 What is the oxidation number of the manganese

atom in MnO41-, the permanganate ion?

A +1

B +2

C +5

D +4

E +7


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5 What is the oxidation number of sulfur in HSO41-,

the hydrogen sulfate ion?

A -2

B +1

C +2

D +4

E +6


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Oxidation-loss of electronsA species is oxidized when it loses electrons.Here, zinc loses two electrons to go from neutral Zn metal to

the Zn2+ion.Zn is also a reducing agent- provides electrons (reductant)Reducing agent loses electrons.

Zn(s) + 2H+(aq) Zn2++ H2(g)

0 +10+2

Oxidation and Reduction

LEO

The lion says

GER

OILRIG


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Reduction- gaining of electronsA species is reducedwhen it gains electrons.

Here, each of the H+gains an electron, and they combine toform H2.H is an oxidizing agent- accepts electrons (oxidant)

An oxidizing agent gains electrons.

Zn(s) + 2H+

(aq)

Zn2+

+ H2(g)

0+1

0+2


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What is reduced is the oxidizing agent.

H+oxidizes Zn by taking electrons from it.

What is oxidized is the reducing agent.

Zn reduces H+by giving it electrons.

Zn(s) + 2H+(aq) Zn2++ H2(g)

0 +10+2


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Zn(s) + 2H+(aq) Zn2++ H2(g)

0 +10+2

An electrochemical reaction in which oxidation and

reduction occurs is known as a REDOXreaction

Redox Reactions


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6 Which of the following is/are an oxidation-reduction (redox) reactions?

(a) K2CrO

4+ BaCl

2 KCl + BaCrO

4

(b) Pb2++ 2 Br1-PbBr2

(c) Cu + S CuS

A a only

B b only

C c only

D a and c

E b and c


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7 Which substance is oxidizedin the following

reaction? (First, assign oxidation numbers.)

Cu + S CuS

A Cu

B S

C Cu and S

D CuS

E This is not a redox reaction.


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8 Which substance is the reducing agent below?

Cu + S CuS

A Cu

B S

C Cu and S

D CuS



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9 Which substance is oxidizedin the following

reaction? (First, assign oxidation numbers.)

Ca + Fe3+ Ca2+ + Fe

A Ca

B Fe3+

C Ca2+

D Fe



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10 Which substance is the oxidizing agentbelow?

Ca + Fe3+ Ca2+ + Fe

A Ca

B Fe3+

C Ca2+

D Fe



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11 Which substance is reducedin the following reaction?(First, assign oxidation numbers.)

3 K + Al(NO3)3 Al + 3 KNO3

A K

B Al

C N

D O



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12 Which substance is the reducing agent?

3 K + Al(NO3)3 Al + 3 KNO3

A K

B Al(NO3)3

C KNO3

D This is not a redox reaction.


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H2S (g) + Cl2(g) --> 2HCl (g) + S (s)

a) Assign oxidation numbers to each element above.

b) Which element is oxidized?

c) Which element is reduced?

d) Name the reducing agent.

e) Name the oxidizing agent.

Redox Practice 1


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SnCl2(aq) + 2HgCl2(aq) --> SnCl4(aq) + Hg2Cl2(s)

a) Assign oxidation numbers to each element above.

b) Which element is oxidized?

c) Which element is reduced?

d) Name the reducing agent.

e) Name the oxidizing agent.

Redox Practice 2


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13 Which element is oxidized in thereaction below?

Fe2++ H++ Cr2O

72- Fe3++ Cr3++ H

2O


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14 H2S (g) + Cl2(g) --> 2HCl (g) + S (s)

Which is oxidized?

Which is reduced?


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15 SnCl2(aq) + 2HgCl2 (aq) --> SnCl4(aq) + Hg2Cl2(s)

Which is oxidized?

Which is reduced?


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Redox reactions in aqueous solutions

A large number of redox reactions occur in aqueous

solutions.

Unlike acid base nutralization and precipitation

reactions,most of the reaction proceed slowly.

Each redox reaction is the sum of two half reactions:

Consider the reaction of iodide ions and hydrogen

peroxide.

2I- (aq) + H2O2(aq) + 2e- I2 + 2OH- (aq) + 2e-


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2I- (aq) + H2O2(aq) + 2e- I2 + 2OH- (aq) + 2e-

1. Oxidation half reaction

2. Reduction half reaction.

2I-(aq) I2 + 2e- oxidation

H2O2(aq) + 2e- 2OH-(aq) reduction

Add the two half reactions to get the overall reaction.

2I- (aq) + H2O2(aq) + 2e- I2 + 2OH- (aq) + 2e-

How do we balance a redox reaction?

Redox reactions in aqueous solutions

This reaction involves two parts as represented below.


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Balancing Redox reactions

Half-reaction method (oxidation # method)

Assign oxidation numbers to determine what is

oxidized and what is reduced.

Identify the oxidation and reduction process.

Write down the individual oxidation and reduction

equations.

Balance these half reactions

Combine them to attain the balanced equation forthe overall reaction.

This method can be used in general to balance any

redox reaction unless any specific condition such as

acidic or basic is mentioned


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Let us consider the simple replacement reaction of Mgwith AgCl

0 +1-1-1 +20

Mg + AgCl Ag + MgCl2

Oxidation: Mg --> Mg2+ + 2 e- --------(1)

Reduction: Ag++ 1 e---> Ag ---------(2)

Since all the atoms are balanced, we need to balance

only electrons. Multiply equation (2) x 2

Oxidation: Mg --> Mg2+ + 2 e- --------(1)

Reduction: 2Ag++ 2 e- --> 2Ag ---------(3)


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Oxidation: Mg --> MgCl2 + 2e-

Reduction: 2AgCl + 2e---> 2Ag

Adding the half-reactions (1) and (3) yields the following:

Overall: Mg+ 2AgCl + 2e---> MgCl2+ 2e- + 2Ag

Overall: Mg+ 2AgCl + 2e---> MgCl2 + 2e- + 2Ag

and we cancel out electrons from both sides:

Net equation: Mg + 2AgCl --> MgCl2+ 2Ag


Since the original equation is given with chlorine you would keep ithere in the final balanced equation too.


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Redox reactions -balancing

Fe3O4+C --> Fe + CO

Fe3O4 + 4C --> 3Fe + 4CO

Practice


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Practice:SnO2+ C --> Sn + CO

Redox reactions -balancing

SnO2+ 2C --> Sn + 2CO


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This diagram shows the steps involved inbalancing half-reactions.

Write down the individual half reaction.

First balance atoms other than H and O. Balance oxygen atoms by adding H2O. Balance hydrogen atoms by adding H+. Balance charge by adding electrons.

Multiply the half-reactions by integers sothat the electrons gained and lost are thesame.

The Half-Reaction Method

Other atoms

O

H

e-

In acidic medium:


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Add the half-reactions, subtracting things

that appear on both sides.

Make sure the equation is balanced

according to mass.

Make sure the equation is balanced

according to charge.

In acidic medium: Continued


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Consider the reaction between MnO4and C2O42:

MnO4(aq) + C2O42 (aq) Mn2+(aq) + CO2(aq)

In acidic medium:

First, we assign oxidation numbers.

We only assign oxidation numbers to elements whose

oxidation numbers CHANGES. Here, oxygen's oxidation number remains constant at -2.

MnO4(aq) + C2O42 (aq) Mn2+(aq) + CO2(aq)

+7 +3 +2 +4


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Which substance gets reduced? Which substance gets oxidized?

Which substance is the reducing agent?

Which substance is the oxidizing agent?

MnO4

(aq) + C2O

42 (aq) Mn2+(aq) + CO

2(aq)

+7 +3 +2 +4

In acidic medium:


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Since the manganese goes from +7 to +2, it is reduced.

The MnO4-ion is the oxidizing agent.

Since the carbon goes from +3 to +4, it is oxidized.

The C2O42-ion is the reducing agent.

MnO4(aq) + C

2O

42 (aq) Mn2+(aq) + CO

2(aq)

+7 +3 +2 +4

In acidic medium:


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Oxidation Half-Reaction

C2O

42CO

2

To balance the carbon, we add a coefficient of 2:

C2O

422 CO

2

In acidic medium:


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Oxidation Half-Reaction

C2O422 CO2

The oxygen is now balanced as well. To balance

the charge, we must add 2 electrons to the rightside.

C2O422 CO2+ 2 e

In acidic medium:


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Reduction Half-Reaction

MnO4Mn2+

The manganese is balanced; to balance the

oxygen, we must add 4 waters to the right side.

MnO4Mn2++ 4 H2O

In acidic medium:


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MnO4Mn2++ 4 H

2O

To balance the hydrogen, we add 8 H+to the left side.

8 H++ MnO4Mn2++ 4 H

2O

In acidic medium:


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8 H++ MnO4Mn2++ 4 H

2O

To balance the charge, we add 5 eto the left side.

5 e+ 8 H++ MnO4

Mn2++ 4 H2O

In acidic medium:


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Combining the Half-Reactions

Now we evaluate the two half-reactions together:

C2O422 CO2+ 2 e

5 e

+ 8 H+

+ MnO4

Mn2+

+ 4 H2O

To attain the same number of electrons on eachside, we will multiply the first reaction by 5 and thesecond by 2.

In acidic medium:


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5 C2O4210 CO2+ 10 e

10 e+ 16 H++ 2 MnO42 Mn2++ 8 H2O

When we add these together, we get:

10 e+ 16 H++ 2 MnO4+ 5 C2O42 -->2 Mn2++ 8 H2O + 10 CO2+10 e

In acidic medium:


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10 e+ 16 H++ 2 MnO4+ 5 C2O422 Mn2++ 8 H2O + 10 CO2+10 e

The only thing that appears on both sides are theelectrons. Subtracting them, we are left with:

16 H++ 2 MnO4+ 5 C2O422 Mn2++ 8 H2O + 10 CO2

In acidic medium:


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a) Write the oxidation half reaction.b) Write the reduction half reaction.c) Write the balanced net reaction.d) Identify the oxidizing agent.e) Identify the reducing agent.

Cd(s) + NiO2(s) --> Cd(OH)2(s) + Ni(OH)2(s )0 +4 -2 +2 -2 +1 +2 -2 +1

Practice 1

The Half-Reaction MethodIn acidic medium:


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Practice 2 Cu + NO3-

--> NO2 + Cu

2+

In acidic medium:


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Practice 3

Cr2O72-+ Fe2++ H+ --> Cr3+ + Fe 3+ + H2O


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Practice 4 :MnO4-

+ Br

-

--> Mn

2+

+ Br2 in acidic solution


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Practice 5 : Cr2O72-+ C2H4O --> C2H4O2 + Cr3+in acidIcsolution

Cr2O72-

+ 8H+ + 3C2H4O --> 3C2H4O2 + 2Cr3+

+ 4H2O


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Redox reaction in basic medium

Some redox reactions requires basic medium to

occur.In this case the following steps need to beperformed to balance the reaction.

1- Assign the oxidation numbers2- Balance the "other atoms" involved3- Separate the half reactions

4- Add water molecules to balance oxygen atom whatever sidedeficient in O atoms5- Add water molecules equal in number to the deficiency of Hatoms.6- Add same number of OH- to the other side.7- Balance the charge by adding electrons on the appropriateside

8- Balance the electrons lost /gained by multiplying the reactionsby integers9- Add the two reactions removing any duplication if any ofcommon species on either side.

Can also be performed without splitting the two equations.


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Zn + NO3---> Zn2++ NH4

+in basic medium:

Balancing in Basic Solution

Oxidation half reaction: Zn ----> Zn2++ 2e-

Reduction half reaction: NO3- ---> NH4

+

NO3----> NH4+ + 3H2O

10H2O + NO3----> NH4

++ 3H2O

10H2O + NO3----> NH4

++ 3H2O + 10OH-

8e- 10H2O + NO3-

---> NH4+

+ 3H2O + 10OH-

4Zn ----> 4 Zn2+ + 8e-

4Zn + 1NO3-+ 7H2O--> 4Zn

2++ 1NH4+ + 10 OH-


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Zn + NO3---> Zn2++ NH4

+in basic medium:

1. Assigh oxidation #s:Zn + NO3

---> Zn

2++ NH4

+

0 +5 2- +2 -3 +1

2. Balance the change in Oxidation # change on either side.

4Zn + 1NO3

---> 4Zn

2++ 1NH4

+

Increases by 2

decreases by 8

increases by 8

decreases by 8

**Can also be performed without splitting the two equations.


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4Zn + 1NO3---> 4Zn

2++ 1NH4

+

3. Balance O atoms by adding H2O molecules to the side deficient in O atoms.3 O atoms on the LHS so add 3 water on the RHS

4Zn + 1NO3---> 4Zn

2++ 1NH4

+ + 3H2O

4. The H atoms are then balanced by adding H2O to the side lacks H.10 H on the RHS, so add 10 water on the LHS.

4Zn + 1NO3-+ 10H2O--> 4Zn

2++ 1NH4

+ + 3H2O

5. Add 10 OH- on the other side of the reaction to balance the extra H and O.

4Zn + 1NO3-+ 10H2O--> 4Zn

2++ 1NH4+ + 3H2O + 10 OH

-

6. If this produces water on both sides, you might have to subtract water fromeachside. 4Zn + 1NO3

-+ 7H2O--> 4Zn

2++ 1NH4

+ + 10 OH

-


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Fe(OH)2 + H2O2--> Fe(OH)3+ H2O in basic solution

+2 +3-1 -2

increase by 1, 1 e-given

decrease by 1 for each O atom, total 2 e- taken

2Fe(OH)2 + H2O2--> 2Fe(OH)3+ H2O in basic solution

2Fe(OH)2 + 2H2O--> 2Fe(OH)3

Balance O atoms by adding 2 H2Oto LHS

2Fe(OH)2 + 2H2O--> 2Fe(OH)3+ 2H2O

Balance H atoms by adding2 H2O to RHS

Add 2 OH-on the LHS

2Fe(OH)2 + 2H2O +2OH---> 2Fe(OH)3+ 2H2O


Practice: 1 Fe(OH)2 + H2O2--> Fe(OH)3+ H2O in basic solution

Oxidation:


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H2O2--> H2O

Add 1 H2Oon RHS

H2O2--> H2O + H2O

Add 2H2Oon LHS to balance H atoms

2H2O + H2O2--> H2O + H2O

Add 2 OH-to RHS

2H2O + H2O2--> H2O + H2O +2OH-

A

Add the two equations: 2Fe(OH)2 + H2O2--> 2Fe(OH)3

Practice 2: Fe(OH)2 + H2O2--> Fe(OH)3+ H2O in basic solution

Reduction


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Practice 3 : Bi(OH)3+ SnO2

--> Bi + SnO3

2Bi(OH)3+ 3SnO2--> 2Bi + 3SnO3 + 3H2O


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Practice 4: Cr(OH)4-1 + H2O2 --> (CrO4) 2- + H2O

2Cr(OH)-1

+ 2OH-+ 3H2O2 --> 2CrO4

2- +8H2O

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Voltaic Cells

The energy released in a spontaneous reaction canbe used to perform electrical work.

Such a set up through which we can transferelectrons is called a voltaic cellor galvanic cellorelectrochemical cell

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Voltaic Cells

In spontaneous oxidation-reduction (redox)reactions, electrons are transferred and energy isreleased.

In the above stup, lectron transfer takes place insidethe beaker

Zn + Cu2+Zn2++ Cu

Zn metal strip

placed in CuSO4

Zn metalCu

2+

2e-Cu atom

Zn2+

Cu2+

Zn metal

Note that the blue color fades as more Cu is reduced to metallic copper

single replacement

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Voltaic Cells

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/

ZnCutrans fer.html

This shows what is occurring on an atomic level at the anode

and the cathode.

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Voltaic Cells

Here the Cu and Zn strips are in two different beakers

Zn/ZnNO3

Zn Zn2++ 2e-

OXD- Half reaction

Cu/CuNO3

Cu2++ 2e- Cu

RED- Half reaction

We can use the energy to do work if we make the electronsflow through an external device.

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Voltaic Cells

Here the Cu and Zn strips are in two different beakers

The salt bridge allows the migration of the ions tokeep electrical neutrality Electrons are generated at the anode and flowsthrough the external line to the cathode.

Zn/ZnNO3

Zn Zn2++ 2e-OXD- Half reaction

Cu/CuNO3Cu2++ 2e- Cu

RED- Half reaction

salt bridge

e-

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http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/

animations/CuZncell.html

Voltaic Cells

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Voltaic Cells

A typical cell looks like this. The oxidation occurs at the anode. The reduction occurs at the cathode.

Zn2+

NO3-

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Voltaic Cells

Once even one electron flows from the anode to thecathode, the charges in each beaker would not bebalanced and the flow of electrons would stop.

more Zn2+

areproduced

more NO3-are

created in solution

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Voltaic Cells

Therefore, we use a salt bridge, usually a U-

shaped tube that contains a gel of a salt solution,to keep the charges balanced. Cations move toward the cathode.

Anions move toward the anode.

more Zn2+

areproduced more NO3

-are in

solution

The increase in Zn2+and NO3-ions in the two compartment create

electrical imbalance.The salt bridge ions will neutralize these ions and create neutrality.

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Voltaic Cells

In the cell, then,electrons leave theanode and flowthrough the wire to the

cathode. As the electrons

leave the anode, thecations formed

dissolve into the

solution in the anode

Zn metalCu

2+

2e-Cu atom

Zn2+

Cu2+

Zn metal

e-

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Voltaic Cells

As the electrons reachthe cathode, cations inthe cathode are

attracted to the nownegative cathode.

Zn metalCu

2+

2e-Cu atom

Zn2+

Cu2+

Zn metal

e-

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Voltaic Cells

The electrons are taken

by the cation, and theneutral metal atomsare deposited onto thecathode.

Zn metalCu

2+

2e-Cu atom

Zn2+

Cu2+

Zn metal

e-

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Voltaic Cells

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/

flashfiles/electroChem/volticCell.html

This shows how a typical voltaic cell works

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16 The electrode at whichoxidation occurs is called

the _______________.

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17 In a voltaic cell, electronsflow from the ______ to the

________.

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18 Which element is oxidized in thereaction below?

Fe2++ H++ Cr2O72- Fe3++ Cr3++ H2O

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19 Fe2++ H++ Cr2O72- Fe

3++ Cr3++ H2O

If a voltaic cell is made with Fe and Cr electrode in

contact with their own solution, the electrons willflow from ------ to --------- electrode.

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A) maintain electrical neutrality in the half-cells viamigration of ions.B) provide a source of ions to react at the anode and

cathode.C) provide oxygen to facilitate oxidation at the anode.D) provide a means for electrons to travel from the anodeto the cathode.E) provide a means for electrons to travel from thecathode to the anode.

20 The purpose of the salt bridge inan electrochemical cell is to

________________.

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21 Acell was made with Mg and Cu as two electrodes. The

electrons will flow from ------- to --------- electrode.

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22 The electrode where reduction is taking place is the ----------

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23 The cation concentration increases in the solution where

oxidation occurs.

Yes

No

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24 the cations move towards the anode and anions move

towards the cathode in a voltaic cell.

True

False

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25 The salt bridge ions may react with the Ions in the cell

compartments to form a precipitate.

Yes

No

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26 Which of the following substanceswould NOT provide a suitable salt

bridge?

A KNO3

B Na2SO4

C LiC2H3O2

D PbCl2

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27 Which of the following substanceswould provide a suitable salt

bridge?

A AgBr

B KCl

C BaF2

D CuS

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28 In a Cu-Zn voltaic cell, which of the following is

true?

A Both strips of metal will increase in mass.

B Both strips of metal will decrease in mass.

C Cu will increase in mass; Zn will decrease.

D Cu will decrease in mass; Zn will increase.

ENeither metal will change its mass, since

electrons have negligible mass.

Zn/ZnNO3

Zn Zn2++ 2e-OXD- Half reaction

Cu/CuNO3

Cu2++ 2e- Cu

RED- Half reaction

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29 In any voltaic cell, which of the following is true?

A The cathode will always increase in mass.

B The anode strip will always decrease in mass.

C The anode strip will always increase in mass.

D Both A and B

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Water only spontaneously flows one way in awaterfall.

Likewise, electrons only spontaneously flow one

way in a redox reactionfrom higher to lower

potential energy.

The accumulation of large number of electrons atthe anode create higher potential at the anode.

Natural flow will occur to cathode where there is

less potential

Higher - to - lower

Electro motive force

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The potential difference between the anode and

cathode in a cell is called the electromotive force(emf).

It is also called the cell potential and is designated

Ecell.


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The difference in potential energy /electon charge ismeasured in volts.

1 volt is the potential required to impart 1joule energy to acharge of 1coulomb

1v = 1J / 1C

The potential difference between the electrodes is thedriving force that pushes the electrons - so called EMF

In a voltaic cell, EMF = Ecell


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Electromotive Force (emf)

In a spontaneous reaction, Ecellis positive

EMF depends on the cell reaction involved

Standard condition: 1M, 1atm and 25C

Ecell = standard cell potential

Cell potential is measured in volts (V).

1V = 1J/C

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Standard Reduction Potentials

Reduction potentials for many electrodes have been

measured and tabulated.

By convention, the process is viewed as a reduction and the

values are reported as reduction potential

Li+

(aq) + e-

Li(s) -3.05Na+(aq) + e- Na(s) -2.71Al3+(aq) + 3e- Al(s) -1.662H+(aq) + 2e- H2(g 0Cu2+(aq) + 2e- Cu(s) + 0.34F2(g) + 2e- 2F-(aq) + 2.87

The more negative value indicate that, reduction is unlikely atthat electrodeThemore positivethe value is, reduction is highly likelyat thatelectrode.This parallels their activity in single replacement reaction.

Electrode potential: The tendency of an electrode to lose or gain

electrons is called electrode potential ( oxidation or reduction potential)

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Standard Hydrogen Electrode ( SHE)

By definition, the reduction potential for hydrogen is 0 V:2 H+(aq, 1M) + 2 eH2(g, 1 atm)

Pt

H2, 1 atm

HCl, 1M


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How did we measure the reduction potential of all lements?

Pt

H2, 1 atm

HCl, 1M

Zn

Zn(NO3)2

Their values are referenced to a Standard HydrogenElectrode (SHE).The metal electrode will be connected to the SHE

By definition, the reduction potential for hydrogen is 0 V:

The reduction potential measured will be that of the

metal


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30 If a volatic cell is made with iron andzinc, which metal will be reduced?

Use the reduction potential table and compare the values.The more positive the value is, that is where reduction takesplace, is the cathode.Oxidation - at Anode (vowels)Reduction - at Cathode (consonants)

FeZn

0.1M Zn(NO3)2 0.1M Fe(NO3)2

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31 If a volatic cell is made with Cu and Na,which metal will be the cathode?

A Cu

B Na

C Cu and Na cannot make a voltaic cell.

F2(g) + 2e- 2F-(aq) + 2.87Cu2+(aq) + 2e- Cu(s) + 0.342H+(aq) + 2e- H2(g 0Al3+(aq) + 3e- Al(s) -1.66

Na+(aq) + e- Na(s) -2.71Li+(aq) + e- Li(s) -3.05

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32 If a volatic cell is made with Li and Al,which metal will be the anode?

A Li

B Al

C Li and Al cannot make a voltaic cell.

F2(g) + 2e- 2F-(aq) + 2.87Cu2+(aq) + 2e- Cu(s) + 0.342H+(aq) + 2e- H2(g 0Al3+(aq) + 3e- Al(s) -1.66

Na+(aq) + e- Na(s) -2.71Li+(aq) + e- Li(s) -3.05

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The cell potential at standard conditions can be foundthrough this equation:

Ecell = Eo

red.pot(cathode) Eo

red.pot(anode)

Because cell potential is based on the potential

energy per unit of charge, it is an intensive property.

This means that it does not depend on the amount of

substance (e.g. mass or moles).

Cell Potentials

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Cell Potentials

For the oxidation in this cell,

Ered= - 0.76v

For the reduction,

Ered= + 0.34v

A cell with Cu and Zn electrodes

1M Zn(NO3)2 1M Cu(NO3)2

CuZn

Zn(s) Zn2+

+ 2e- Cu2+

(aq)+ 2e-Cu(s)

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Cell Potentials

Ecell

= Ered(cathode)

- Ered(anode)

= +0.34V - (-0.76V)

Ered =+1.10V

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The greater the difference between the two

electrode potential, the greater the voltage ofthe cell.

Cu2+

+ 2e- --> Cu

Zn --> Zn2+

+ 2e-

More positive

-0.76

+ 0.34

E0cell = 0.34 - (-0.76)

= + 1.10vE

cell

= +0.34V - (-0.76V)

=+1.10V

Cell Potentials

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33 Which of the following volatic cells wouldyield the greatest voltage (Eocell)?

A Cu-Al

B Cu-Na

C Al-Li

F2(g) + 2e- 2F-(aq) + 2.87Cu2+(aq) + 2e- Cu(s) + 0.342H+(aq) + 2e- H2(g 0

Al3+

(aq) + 3e- Al(s) -1.66Na+(aq) + e- Na(s) -2.71Li+(aq) + e- Li(s) -3.05

D F2 - Cu

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34 Which of the following volatic cells wouldyield the lowest voltage (Eocell)?

A Cu-Al

B Al-Na

C Na-Li

F2(g) + 2e- 2F-(aq) + 2.87Cu2+(aq) + 2e- Cu(s) + 0.342H+(aq) + 2e- H2(g 0Al3+(aq) + 3e- Al(s) -1.66Na+(aq) + e- Na(s) -2.71Li+(aq) + e- Li(s) -3.05

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Oxidizing and Reducing Agents

The strongest oxidizers have the most positive

reduction potentials.

The strongest reducers have the most negativereduction potentials.

F is a strong oxidizing agent than Cl

F2(g) + 2e- --> 2F- 2.87v

Cl2(g) + 2e- --> 2Cl- 1.36v

I2(s) + 2e- --> 2I- 0.53v

Rb++ e- --> Rb(s) -2.92v

Most positive values

Most negative valuesIncreasings

trength

ofoxidizing

age

nt

Increasing

strength

ofreducing

agen

t

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35 The more _______ the value of

Ered

, the greater the driving

force for reduction.

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Class Practice:


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Class Practice:

Ni Sn

1M Ni (NO3)2 1M Sn(NO3)2

Identify:

CathodeAnodeOxidation half reactionReduction half reactionCombined cell reaction (balanced)

Ecell =

Slide 108 / 144

Cl P ti 2


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Class Practice 2

Fe Sn

1M Fe(NO3)3 1M Sn(NO3)2

Identify:

CathodeAnodeOxidation half reactionReduction half reactionCombined cell reaction (balanced)

Ecell

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36 Calculate E for the following reaction:Sn4+(aq) + 2K(s) --> Sn2+(aq) + 2K+(aq) A) +6.00 V

B) -3.08 VC) +3.08 V

D) +2.78 V E) -2.78 V

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Free Energy

Gfor a redox reaction can be found by using the

equation

G= nFE

where nis the number of moles of electrons

transferred, and Fis a constant, the Faraday.

1 F= 96,500 C/mol = 96,485 J/V-mol

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Free Energy

Under standard conditions,

G= -nFE

Standard condition: 250C, 1 atm and 1M

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Nernst Equation

Remember that

G= G+ RTln Q

This means

nFE= nFE+ RTln Q

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Nernst Equation

Dividing both sides by nF, we get the Nernstequation:

or, using base-10 logarithms,

E = E- [8.31 x 298] x 2.303 log Q

[n x 96500 ]

E = E- lnQRT

nF

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Nernst Equation


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Nernst Equation

At room temperature (298 K),

Thus, the equation becomes

E = E- logQ0.0592

n

= 0.0592 V2.303 RT

F

E = E- [8.31 x 298] x 2.303log Q

[n x 96500]

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Equilibrium Constant

When E=0, -nFE = 0

0= E logK (0.0592/n)

E = logK (0.0592/n)

logK = nE/0.0592

Equilibrium constant for a redox reaction can becalculated

using the above .

G= G+ RTln Q

E = E- logQ0.0592n

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37 The relationship between the


116/144

A) G = -nF/E

B) G = -E/nF

C) G = -nFE

D) G = -nRTF

E) G = -nF/ERT

37 The relationship between thechange in Gibbs free energy and theemf of an electrochemical cell is

given by __________________.

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Concentration Cells


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Notice that the Nernst equation implies that a cell could becreated that has the same substance at both electrodes.

For such a cell, E would be 0, but Q would not.

Therefore, as long as the concentrations are different,E will not be 0.

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/

voltaicCe llEMF.html

Ni Ni

1M [Ni2+

]0.001M [Ni2+

]

Ni Ni

0.5M [Ni2+

]0.5M [Ni2+

]

E = E- logQ0.0592n

**

[dilute] log Q = log ------------- [concent]

Slide 118 / 144

38 A cadmium rod is placed in a 0.010M solution of
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/voltaicCellEMF.htmlhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/voltaicCellEMF.htmlhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/voltaicCellEMF.html


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A cadmium rod is placed in a 0.010M solution of

cadmium sulfate at 298K. Calculate the potential of the

electrode at the is temperature.

E = E- logQ0.0592

n

= -0.0529/2 log0.01= - 0.0591

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/


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Elcectrolytic cell/ Electrolysis

Voltaic cells work as a result of a spontaneous

reaction

We can use electricity from outside source to make a

nonspontaneous reaction to become spontaneous.

A chemical reaction by using outside electricity is

known as electrolysis. Such a cell is known as anelectrolytic cell

Slide 120 / 144

Elcectrochemical/voltaic cell


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Electrolytic Cell Voltaic (Electrochemical) Cell

Energy is absorbed to drive

a nonspontaneous redox

reaction

Energy is released from a

spontaneous redox reaction

Surroundings (battery orpower supply) do work on

the system (cell)

System (cell) does work on thesurroundings (e.g. light bulb)

Elcectrochemical/voltaic cell

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O f th diff b t


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A) an electric current is produced by a chemical reactionB) electrons flow toward the anodeC) a nonspontaneous reaction is forced to occurD) gas is produced at the cathodeE) oxidation occurs at the cathode

39 One of the differences between avoltaic cell and an electrolytic cell

is that in an electrolytic cell,_____________________.

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Electroplating


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Uses an active electrode to deposit a thin layer of onemetal to another metal object

Item to be coated is cathode (metal ions get reducedat the (-) electrode)

e-

Ag+

Ag+

Ag

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Electrolysis


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This flowchart shows the steps relating the quantity of electrical

charge used in electrolysis to the amounts of substances

oxidized or reduced.

Current(A)and time

Quantity ofcharge(Coulombs)

Moles ofelectrons(Faradays)

Moles ofsubstanceoxidizedor reduced

Grams ofsubstance

A typical problem will give the current (amperes) that is

applied for a specific amount of time (seconds). Youwould be asked to solve for the mass of metalthat can

be produced through electroplating.

Alternatively, you might be asked for either the time or

amount of current that is needed to produce a specific

amount (given mass) of metal.

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El t l i


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The quantity of charge passing through is measured incoulombs

1 mole of electrons passage = 96500C = 1Faraday

1coulomb = 1 ampere passing in 1 second

Coulombs (C ) = ampere x seconds

Electrolysis

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40 Th tit f h i


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A) joule

B) coulomb

C) calorie

D) NewtonE) Mole

40 The quantity of charge passing apoint in a circuit in one second

when the current is one ampereis called a ___________________.

Slide 126 / 144

41 H l b lt f t


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41 How many coulombs result from a current

of 50 amps (A) applied for 20 seconds?

A 2.5

B 10

C 70

D 700

E 1000 ACs=

Slide 127 / 144

42 How many seconds must a current of 25 A be


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42 How many seconds must a current of 25 A be

applied in order to produce a charge of 100 C?

A 0.25

B 0.4

C 4

D 75

E 125A Cs=

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43 What amount of charge is required to release one


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43 What amount of charge is required to release one

mole of electrons?

A 1 atm

B 25oC

C 0.0821 L-atm/mol-K

D 96,500 C

E 760 mm Hg

Slide 129 / 144

44 How many moles of electrons would be released


129/144


by a charge of 158,000 C?

A 96,500 / 158,000

B 158,000 / 96,500

C 158,000*96,500

D 158,000 - 96,500

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130/144


by a charge of 48,250 C?

A 0.25

B 0.5

C 1

D 48,250

E 96,500

Slide 131 / 144

El t l i


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If 10.0 A passes through molten AlCl3for 60 minutes, howmuch of Al will be deposited?

total charge C = 10.0 A x 60min x 60sec. = 3.6 x104C

Remember!! 1mole e- = 96500CHow many moles of e

- are we talking about in here?????

Moles of e- = 3.6 x 104/96500 = 0.373 moles of e

-

Al3+ + 3e- Al

1 mol Al = 3 mols e-

Moles of Al = 0.373 x 1 mole Al/3 mole e- = 0.124 mol Al

How many grams of Al ? = 27g x 0.124 mol = 3.36g Al

Quantitative aspect

Electrolysis

Slide 132 / 144

Electrolysis


132/144

Calculate the number of grams of aluminumproduced in 30.0 minutes by electrolysis of at acurrent of 12.0 A.

practice:1

Electrolysis

Slide 133 / 144

Electrolysis


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How many minutes will it take to plate out 6.36 g of Cu metalfrom a solution of Cu2+using a current of 12 amps in anelectrolytic cell?

practice:2

Slide 134 / 144

46 Plating out 1 mol of chromium requires of


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46 Plating out 1 mol of chromium requires _______ of

electrons.

A 0.33 mol

B 0.5 mol

C 1.0 mol

D 3.0 mol

Cr3+(aq) + 3 e---> Cr(s)

Slide 135 / 144

47 One mole of electrons would allow electroplating


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47 One mole of electrons would allow electroplating

of __________ mol of zinc.

A Zn cannot be electroplated.

B 0.5 mol

C 1.0 mol

D 2.0 mol

Zn2+(aq) + 2 e---> Zn(s)

Slide 136 / 144

48 How many minutes will it take to plate out 16 22 g of Al


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48 How many minutes will it take to plate out 16.22 g of Almetal from a solution of Al3+using a current of

12.9 amps in an electrolytic cell?

A 60.1 min

B 74.9 min

C 173 minD 225 min

E 13,480 min

Slide 137 / 144

Electrochemistry Applied


137/144

Electrochemistry -Applied

Batteries

Hydrogen fuel cells

Corrosion

Corrosion prevention

Biology

Slide 138 / 144

Batteries

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Wt871rwp Electrochemistry 440 Electrochemistry-presentation 2012-03-31