wt871rwp electrochemistry 440 electrochemistry-presentation 2012-03-31
TRANSCRIPT
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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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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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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)
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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.
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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 =
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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]
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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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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
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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 ___________________.
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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
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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
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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)
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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
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Electrochemistry Applied
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Electrochemistry -Applied
Batteries
Hydrogen fuel cells
Corrosion
Corrosion prevention
Biology
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Batteries
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Batteries
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Alkaline Batteries
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Alkaline Batteries
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Hydrogen Fuel Cells
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Corrosion and
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Corrosion Prevention
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In Biology
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electron transport in Mitochondria ......
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