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.

    Rules for Assigning Oxidation Numbers

    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.

    Rules for Assigning Oxidation Numbers

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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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    Oxidation and Reduction

    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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    Oxidation and Reduction

    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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    Oxidation and Reduction

    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

    E This is not a redox reaction.

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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

    E This is not a redox reaction.

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    10 Which substance is the oxidizing agentbelow?

    Ca + Fe3+ Ca2+ + Fe

    A Ca

    B Fe3+

    C Ca2+

    D Fe

    E This is not a redox reaction.

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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

    E This is not a redox reaction.

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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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    Half-reaction method (oxidation # method)

    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

    Half-reaction method (oxidation # method)

    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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    The Half-Reaction Method

    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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    The Half-Reaction Method

    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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    The Half-Reaction Method

    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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    The Half-Reaction Method

    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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    Reduction Half-Reaction

    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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    Reduction Half-Reaction

    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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    Combining the Half-Reactions

    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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    Combining the Half-Reactions

    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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    Balancing in Basic Solution

    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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    Balancing in Basic Solution

    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

    Balancing in Basic Solution

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

    Oxidation:

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    Balancing in Basic Solution

    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

    Balancing in Basic Solution

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    Balancing in Basic Solution

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

    2Cr(OH)-1

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

    2- +8H2O

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    Slide 62 / 144

    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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    Slide 63 / 144

    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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    Slide 64 / 144

    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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    Slide 65 / 144

    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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    Slide 66 / 144

    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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    Slide 67 / 144

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

    animations/CuZncell.html

    Voltaic Cells

    Slide 68 / 144

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    Slide 68 / 144

    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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    Slide 69 / 144

    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.

    Electro motive force

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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

    Electro motive force

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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)

    Slide 94 / 144

    Standard Reduction Potentials

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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

    Standard Reduction Potentials

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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

    Standard Reduction Potentials

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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.

    Slide 107 / 144

    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

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    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

    Slide 119 / 144

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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

    Slide 123 / 144

    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 ___________________.

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    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=

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    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

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    44 How many moles of electrons would be released

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    44 How many moles of electrons would be released

    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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    45 How many moles of electrons would be released

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    45 How many moles of electrons would be released

    by a charge of 48,250 C?

    A 0.25

    B 0.5

    C 1

    D 48,250

    E 96,500

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    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

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    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

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    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)

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    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

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    Electrochemistry -Applied

    Batteries

    Hydrogen fuel cells

    Corrosion

    Corrosion prevention

    Biology

    Slide 138 / 144

    Batteries

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    Batteries

    Slide 139 / 144

    Alkaline Batteries

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    Alkaline Batteries

    Slide 140 / 144

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    Hydrogen Fuel Cells

    Slide 141 / 144

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    Corrosion and

    Slide 142 / 144

    Corrosion Prevention

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    Slide 143 / 144

    In Biology

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    electron transport in Mitochondria ......

    Slide 144 / 144

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