Electrochemistry Slide 2 / 144 Electrochemistry Electrochemistry - - PowerPoint PPT Presentation
Slide 1 / 144 Electrochemistry Slide 2 / 144 Electrochemistry Electrochemistry deals with relationships between reactions and electricity In electrochemical reactions, electrons are transferred from one species to another. Provide
Slide 1 / 144
Slide 2 / 144
· 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
Slide 3 / 144
· 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 states to individual atoms.
Slide 4 / 144
Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)
+1 +2
Take a look at this reaction between Zn metal and acid with assigned oxidation numbers. How do we know what number goes with each atom? Where do these numbers came from?
Slide 5 / 144
Elements
Elements in their elemental form have an
Compounds The sum of the oxidation numbers in a
neutral compound is 0.
Monoatomic ions
The oxidation number of a monatomic ion is the same as its charge.
Polyatomic ions
The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.
Slide 6 / 144
Hydrogen -1 when bonded to a metal
+1 when bonded to a nonmetal
Fluorine
Fluorine always has an oxidation number of -1.
Other halogens
Usually -1. May have positive oxidation numbers in
For example, Cl has an oxidation number of +5 in ClO3
Slide 7 / 144
Nonmetals
Nonmetals tend to have negative
positive in certain compounds or ions.
Oxygen has a oxidation number of -2, except in the peroxide ion, when its
Slide 8 / 144
1 What is the oxidation number of each oxygen atom in the compound MnO2 ?
A
B
C D
+1
E
+2
Slide 9 / 144
2 What is the oxidation number of the manganese atom in the compound MnO2 ?
A
+3
B
+2
C
+1
D
+4
E
+7
Slide 10 / 144
3 What is the oxidation number of oxygen atom in MnO4
1-, the permanganate ion?
A
B
C D +2 E +4
Slide 11 / 144
4 What is the oxidation number of the manganese atom in MnO4
1-, the permanganate ion?
A +1
B
+2
C
+5
D
+4
E
+7
Slide 12 / 144
5 What is the oxidation number of sulfur in HSO4
1-,
the hydrogen sulfate ion?
A
B
+1
C
+2
D
+4
E
+6
Slide 13 / 144
Oxidation-loss of electrons A 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)
+1 +2
Oxidation and Reduction
LEO
The lion says
GER
OIL RIG
Slide 14 / 144
Reduction- gaining of electrons A species is reduced when it gains electrons. Here, each of the H+ gains an electron, and they combine to form H2. H is an oxidizing agent- accepts electrons (oxidant) An oxidizing agent gains electrons.
Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)
+1 +2
Slide 15 / 144
· 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)
+1 +2
Slide 16 / 144
Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)
+1 +2
An electrochemical reaction in which oxidation and reduction occurs is known as a REDOX reaction
Slide 17 / 144
6 Which of the following is/are an oxidation- reduction (redox) reactions? (a) K2CrO4 + BaCl2 ➝ KCl + BaCrO4 (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
Slide 18 / 144
7 Which substance is oxidized in 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.
Slide 19 / 144
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.
Slide 20 / 144
9 Which substance is oxidized in 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.
Slide 21 / 144
10 Which substance is the oxidizing agent below? Ca + Fe3+ ➝ Ca2+ + Fe
A
Ca
B
Fe3+
C
Ca2+
D
Fe
E
This is not a redox reaction.
Slide 22 / 144
11 Which substance is reduced in 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.
Slide 23 / 144
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.
Slide 24 / 144
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.
Slide 25 / 144
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.
Slide 26 / 144
13 Which element is oxidized in the reaction below?
Fe2+ + H+ + Cr2O7
2- ➝ Fe3+ + Cr3+ + H2O
Slide 27 / 144
14
H2S (g) + Cl2 (g) --> 2HCl (g) + S (s)
Which is oxidized? Which is reduced?
Slide 28 / 144
15 SnCl2(aq) + 2HgCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)
Which is oxidized? Which is reduced?
Slide 29 / 144
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-
Slide 30 / 144
2I- (aq) + H2O2(aq) + 2e- ⇒ I2 + 2OH- (aq) + 2e-
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?
This reaction involves two parts as represented below.
Slide 31 / 144
Balancing Redox reactions
Half-reaction method (oxidation # method) · Assign oxidation numbers to determine what is
· 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 for the 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
Slide 32 / 144
Half-reaction method (oxidation # method) Let us consider the simple replacement reaction of Mg with AgCl +1
+2
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
Oxidation: Mg --> Mg2+ + 2 e- --------(1) Reduction: 2Ag+ + 2 e- --> 2Ag ---------(3)
Slide 33 / 144
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 it here in the final balanced equation too.
Slide 34 / 144
Redox reactions -balancing Fe3O4 +C --> Fe + CO
Fe3O4 + 4C --> 3Fe + 4CO
Practice
Slide 35 / 144
Practice: SnO2 + C --> Sn + CO
Redox reactions -balancing
SnO2 + 2C --> Sn + 2CO
Slide 36 / 144
This diagram shows the steps involved in balancing 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 so that the electrons gained and lost are the same.
Other atoms O
H e-
In acidic medium:
Slide 37 / 144
· 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
Slide 38 / 144
Consider the reaction between MnO4
− and C2O4
2− :
MnO4
− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)
In acidic medium:
· First, we assign oxidation numbers. · We only assign oxidation numbers to elements whose
· Here, oxygen's oxidation number remains constant at -2.
MnO4
− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)
+7 +3 +2 +4
Slide 39 / 144
· Which substance gets reduced? · Which substance gets oxidized? · Which substance is the reducing agent? · Which substance is the oxidizing agent?
MnO4
− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)
+7 +3 +2 +4
In acidic medium:
Slide 40 / 144
Since the manganese goes from +7 to +2, it is reduced. The MnO4
Since the carbon goes from +3 to +4, it is oxidized. The C2O4 2- ion is the reducing agent.
MnO4
− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)
+7 +3 +2 +4
In acidic medium:
Slide 41 / 144
C2O4
2− ➝ CO2
To balance the carbon, we add a coefficient of 2: C2O4
2− ➝ 2 CO2
In acidic medium:
Slide 42 / 144
C2O4
2− ➝ 2 CO2
The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.
C2O4
2− ➝ 2 CO2 + 2 e−
In acidic medium:
Slide 43 / 144
MnO4
− ➝ Mn2+
The manganese is balanced; to balance the
MnO4
− ➝ Mn2+ + 4 H2O
In acidic medium:
Slide 44 / 144
MnO4
− ➝ Mn2+ + 4 H2O
To balance the hydrogen, we add 8 H+ to the left side.
8 H+ + MnO4
− ➝ Mn2+ + 4 H2O
In acidic medium:
Slide 45 / 144
8 H+ + MnO4
− ➝ Mn2+ + 4 H2O
To balance the charge, we add 5 e− to the left side.
5 e− + 8 H+ + MnO4
− ➝ Mn2+ + 4 H2O
In acidic medium:
Slide 46 / 144
Now we evaluate the two half-reactions together:
C2O4
2− ➝ 2 CO2 + 2 e−
5 e− + 8 H+ + MnO4
− ➝ Mn2+ + 4 H2O
To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.
In acidic medium:
Slide 47 / 144
5 C2O4
2− ➝ 10 CO2 + 10 e−
10 e− + 16 H+ + 2 MnO4
− ➝ 2 Mn2+ + 8 H2O
When we add these together, we get: 10 e− + 16 H+ + 2 MnO4
− + 5 C2O4 2− -->
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
In acidic medium:
Slide 48 / 144
10 e− + 16 H+ + 2 MnO4
− + 5 C2O4 2− ➝
2 Mn2+ + 8 H2O + 10 CO2 +10 e−
The only thing that appears on both sides are the
16 H+ + 2 MnO4
− + 5 C2O4 2− ➝
2 Mn2+ + 8 H2O + 10 CO2
In acidic medium:
Slide 49 / 144
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
In acidic medium:
Slide 50 / 144
Practice 2
Cu + NO3
In acidic medium:
Slide 51 / 144
Practice 3
Cr2O7
2- + Fe2+ + H+ --> Cr3+ + Fe 3+ + H2O
Slide 52 / 144
Practice 4 :MnO4
Slide 53 / 144
Practice 5 : Cr2O7
2- + C2H4O --> C2H4O2 + Cr3+ in acidIc
solution
Cr2O7
2- + 8H+ + 3C2H4O --> 3C2H4O2 + 2Cr3+ + 4H2O
Slide 54 / 144
Some redox reactions requires basic medium to
performed to balance the reaction.
1- Assign the oxidation numbers 2- Balance the "other atoms" involved 3- Separate the half reactions 4- Add water molecules to balance oxygen atom whatever side deficient in O atoms 5- Add water molecules equal in number to the deficiency of H atoms. 6- Add same number of OH- to the other side. 7- Balance the charge by adding electrons on the appropriate side 8- Balance the electrons lost /gained by multiplying the reactions by integers 9- Add the two reactions removing any duplication if any of common species on either side. Can also be performed without splitting the two equations.
Slide 55 / 144
Zn + NO3
+ in basic medium:
Balancing in Basic Solution
Oxidation half reaction: Zn ----> Zn2+ + 2e- Reduction half reaction: NO3
+
NO3
10H2O + NO3
+ + 3H2O
10H2O + NO3
+ + 3H2O + 10OH-
8e- 10H2O + NO3
+ + 3H2O + 10OH-
4Zn ----> 4 Zn2+ + 8e-
4Zn + 1NO3
Slide 56 / 144
Balancing in Basic Solution
Zn + NO3
+ in basic medium:
Zn + NO3
+
0 +5 2- +2 -3 +1
4Zn + 1NO3
+ Increases by 2 decreases by 8 increases by 8 decreases by 8
Can also be performed without splitting the two equations.
Slide 57 / 144
Balancing in Basic Solution
4Zn + 1NO3
+
3 O atoms on the LHS so add 3 water on the RHS
4Zn + 1NO3
+ + 3H2O
10 H on the RHS, so add 10 water on the LHS.
4Zn + 1NO3
+ + 3H2O
4Zn + 1NO3
+ + 3H2O + 10 OH-
each side.
4Zn + 1NO3
Slide 58 / 144
Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O in basic solution
+2
+3
increase by 1, 1 e- given decrease by 1 for each O atom, total 2 e- taken
2 Fe(OH)2 + H2O2 --> 2Fe(OH)3 + H2O in basic solution 2 Fe(OH)2 + 2H2O --> 2Fe(OH)3
Balance O atoms by adding 2 H2O to LHS
2 Fe(OH)2 + 2H2O --> 2Fe(OH)3 + 2H2O
Balance H atoms by adding 2 H2O to RHS Add 2 OH- on the LHS
2 Fe(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:
Slide 59 / 144
Balancing in Basic Solution H2O2 --> H2O
Add 1 H2O on RHS
H2O2 --> H2O + H2O
Add 2H2O on 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
Slide 60 / 144
Practice 3 : Bi(OH)3 + SnO2
2Bi(OH)3 + 3SnO2
Balancing in Basic Solution
Slide 61 / 144
Balancing in Basic Solution Practice 4: Cr(OH)4
2Cr(OH)
2- + 8H2O
Slide 62 / 144
Voltaic Cells
The energy released in a spontaneous reaction can be used to perform electrical work. Such a set up through which we can transfer electrons is called a voltaic cell or galvanic cell or electrochemical cell
Slide 63 / 144
Voltaic Cells
In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. In the above stup, lectron transfer takes place inside the beaker Zn + Cu2+ ➝ Zn2+ + Cu Zn metal strip placed in CuSO4
Zn metal Cu
2+2e- Cu atom Zn2+
Cu2+ Zn metal
Note that the blue color fades as more Cu is reduced to metallic copper
single replacement
Slide 64 / 144
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/ ZnCutransfer.html
This shows what is occurring on an atomic level at the anode and the cathode.
Slide 65 / 144
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 electrons flow through an external device.
Slide 66 / 144
Here the Cu and Zn strips are in two different beakers
· The salt bridge allows the migration of the ions to keep electrical neutrality · Electrons are generated at the anode and flows through the external line to the cathode. Zn/ZnNO3 Zn ➝ Zn2+ + 2e- OXD- Half reaction Cu/CuNO3 Cu2+ + 2e- ➝ Cu RED- Half reaction
salt bridge
e-
Slide 67 / 144
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ animations/CuZncell.html
Slide 68 / 144
Voltaic Cells
· A typical cell looks like this. · The oxidation occurs at the anode. · The reduction occurs at the cathode.
Zn
2+
NO3
Slide 69 / 144
Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.
more Zn
2+ are
produced more NO3
created in solution
Slide 70 / 144
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+ are produced more NO3
solution
The increase in Zn2+ and NO3
electrical imbalance. The salt bridge ions will neutralize these ions and create neutrality.
Slide 71 / 144
Voltaic Cells
· In the cell, then, electrons leave the anode and flow through the wire to the cathode. · As the electrons leave the anode, the cations formed dissolve into the solution in the anode
Zn metal Cu2+ 2e- Cu atom Zn
2+Cu
2+Zn metal
e-
Slide 72 / 144
Voltaic Cells
· As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.
Zn metal Cu2+ 2e- Cu atom Zn2+ Cu2+ Zn metal
e-
Slide 73 / 144
Voltaic Cells
· The electrons are taken by the cation, and the neutral metal atoms are deposited onto the cathode.
Zn metal Cu2+ 2e- Cu atom Zn
2+Cu
2+Zn metal
e-
Slide 74 / 144
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ flashfiles/electroChem/volticCell.html
This shows how a typical voltaic cell works
Slide 75 / 144
16 The electrode at which
the _______________.
Slide 76 / 144
17 In a voltaic cell, electrons flow from the ______ to the ________.
Slide 77 / 144
18 Which element is oxidized in the reaction below?
Fe2+ + H+ + Cr2O7
2- ➝ Fe3+ + Cr3+ + H2O
Slide 78 / 144
19 Fe2+ + H+ + Cr2O7
2- ➝ Fe3+ + Cr3+ + H2O
If a voltaic cell is made with Fe and Cr electrode in contact with their own solution, the electrons will flow from ------ to --------- electrode.
Slide 79 / 144
A) maintain electrical neutrality in the half-cells via migration 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 anode to the cathode. E) provide a means for electrons to travel from the cathode to the anode.
20 The purpose of the salt bridge in an electrochemical cell is to ________________.
Slide 80 / 144
21 A cell was made with Mg and Cu as two electrodes. The
electrons will flow from ------- to --------- electrode.
Slide 81 / 144
22 The electrode where reduction is taking place is the ----------
Slide 82 / 144
23 The cation concentration increases in the solution where
Yes No
Slide 83 / 144
24 the cations move towards the anode and anions move towards the cathode in a voltaic cell. True False
Slide 84 / 144
25 The salt bridge ions may react with the Ions in the cell compartments to form a precipitate. Yes No
Slide 85 / 144
26 Which of the following substances would NOT provide a suitable salt bridge?
A KNO3
B
Na2SO4
C
LiC2H3O2
D
PbCl2
Slide 86 / 144
27 Which of the following substances would provide a suitable salt bridge?
A AgBr
B
KCl
C
BaF2
D
CuS
Slide 87 / 144
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.
E
Neither 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
Slide 88 / 144
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
Slide 89 / 144
· Water only spontaneously flows one way in a waterfall. · Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy. · The accumulation of large number of electrons at the anode create higher potential at the anode. · Natural flow will occur to cathode where there is less potential · Higher - to - lower
Slide 90 / 144
· 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.
Slide 91 / 144
The difference in potential energy /electon charge is measured in volts. 1 volt is the potential required to impart 1joule energy to a charge of 1coulomb 1v = 1J / 1C The potential difference between the electrodes is the driving force that pushes the electrons - so called EMF In a voltaic cell, EMF = Ecell
Slide 92 / 144
In a spontaneous reaction, Ecell is positive EMF depends on the cell reaction involved Standard condition: 1M, 1atm and 25°C E°cell = standard cell potential Cell potential is measured in volts (V).
Slide 93 / 144
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) -3.05 Na+(aq) + e- ➝ Na(s) -2.71 Al3+(aq) + 3e- ➝ Al(s) -1.66 2H+(aq) + 2e- ➝ H2(g 0 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 F2(g) + 2e- ➝ 2F-(aq) + 2.87
The more negative value indicate that, reduction is unlikely at that electrode The more positive the value is, reduction is highly likely at that electrode. 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 Hydrogen Electrode ( SHE)
· By definition, the reduction potential for hydrogen is 0 V: 2 H+ (aq, 1M) + 2 e− ➝ H2 (g, 1 atm)
Pt H2, 1 atm HCl, 1M
Standard Reduction Potentials
Slide 95 / 144
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 Hydrogen Electrode (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
Slide 96 / 144
30 If a volatic cell is made with iron and zinc, which metal will be reduced?
Use the reduction potential table and compare the values. The more positive the value is, that is where reduction takes place, is the cathode. Oxidation - at Anode (vowels) Reduction - at Cathode (consonants)
Fe Zn 0.1M Zn(NO3)2 0.1M Fe(NO3)2
Slide 97 / 144
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) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05
Slide 98 / 144
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) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05
Slide 99 / 144
The cell potential at standard conditions can be found through this equation:
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).
Slide 100 / 144
Cell Potentials
· For the oxidation in this cell, · E°
red = - 0.76v
· For the reduction, · E°
red = + 0.34v
A cell with Cu and Zn electrodes
1M Zn(NO3)2 1M Cu(NO3)2 Cu
Zn Zn(s) ➝ Zn2+ + 2e-
Cu2+ (aq) + 2e-➝ Cu(s)
Slide 101 / 144
E°
cell = E° red(cathode) - E° red(anode)
= +0.34V - (-0.76V)
red = +1.10V
Slide 102 / 144
The greater the difference between the two electrode potential, the greater the voltage of the cell.
Cu2+ + 2e- --> Cu Zn --> Zn2+ + 2e- More positive
+ 0.34
E0
cell = 0.34 - (-0.76)
= + 1.10v
E°
cell
= +0.34V - (-0.76V)
= +1.10V
Slide 103 / 144
33 Which of the following volatic cells would yield the greatest voltage (Eo
cell)?
A Cu-Al
B
Cu-Na
C
Al-Li F2(g) + 2e- ➝ 2F-(aq) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05
D
F2 - Cu
Slide 104 / 144
34 Which of the following volatic cells would yield the lowest voltage (Eo
cell)?
A Cu-Al
B
Al-Na
C
Na-Li F2(g) + 2e- ➝ 2F-(aq) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05
Slide 105 / 144
Oxidizing and Reducing Agents
· The strongest oxidizers have the most positive reduction potentials. · The strongest reducers have the most negative reduction 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 values
Increasing strength of oxidizing agent Increasing strength of reducing agent
Slide 106 / 144
35 The more _______ the value of Ered, the greater the driving force for reduction.
Slide 107 / 144
Class Practice:
Ni Sn 1M Ni (NO3)2 1M Sn(NO3)2
Identify: Cathode Anode Oxidation half reaction Reduction half reaction Combined cell reaction (balanced)
cell =
Slide 108 / 144
Fe Sn 1M Fe(NO3)3 1M Sn(NO3)2
Identify: Cathode Anode Oxidation half reaction Reduction half reaction Combined cell reaction (balanced)
cell
Slide 109 / 144
36 Calculate E° for the following reaction:
Sn4+(aq) + 2K(s) --> Sn2+(aq) + 2K+(aq) A) +6.00 V B) -3.08 V C) +3.08 V D) +2.78 V E) -2.78 V
Slide 110 / 144
ΔG for a redox reaction can be found by using the equation ΔG = −nFE where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96,500 C/mol = 96,485 J/V-mol
Slide 111 / 144
Under standard conditions,
ΔG° = -nFE°
Standard condition: 250C, 1 atm and 1M
Slide 112 / 144
· Remember that ΔG = ΔG° + RT ln Q · This means −nFE = −nFE° + RT ln Q
Slide 113 / 144
Dividing both sides by −nF, we get the Nernst equation:
E = E° - [8.31 x 298] x 2.303 log Q [n x 96500 ]
Slide 114 / 144
At room temperature (298 K),
Thus, the equation becomes
E = E° - [8.31 x 298] x 2.303 log Q [n x 96500 ]
Slide 115 / 144
· 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 be calculated using the above . ΔG = ΔG° + RT ln Q
E = E° - logQ 0.0592 n
Slide 116 / 144
A) ΔG = -nF/E B) ΔG = -E/nF C) ΔG = -nFE D) ΔG = -nRTF E) ΔG = -nF/ERT
37 The relationship between the change in Gibbs free energy and the emf of an electrochemical cell is given by __________________.
Slide 117 / 144
Concentration Cells
· Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, E°c 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/ voltaicCellEMF.html
Ni Ni
1M [Ni
2+]0.001M [Ni2+] Ni Ni 0.5M [Ni
2+]0.5M [Ni
2+]E = E° - logQ 0.0592 n **
[dilute] log Q = log ------------- [concent]
Slide 118 / 144
38
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° - logQ 0.0592 n = -0.0529/2 log0.01 = - 0.0591
Slide 119 / 144
· 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 an electrolytic cell
Slide 120 / 144
Electrolytic Cell Voltaic (Electrochemical) Cell
Energy is absorbed to drive a nonspontaneous redox reaction Energy is released from a spontaneous redox reaction Surroundings (battery or power supply) do work on the system (cell) System (cell) does work on the surroundings (e.g. light bulb)
Slide 121 / 144
A) an electric current is produced by a chemical reaction B) electrons flow toward the anode C) a nonspontaneous reaction is forced to occur D) gas is produced at the cathode E) oxidation occurs at the cathode
39 One of the differences between a voltaic cell and an electrolytic cell is that in an electrolytic cell, _____________________.
Slide 122 / 144
Electroplating
· Uses an active electrode to deposit a thin layer of one metal to another metal object · Item to be coated is cathode (metal ions get reduced at the (-) electrode)
e-
Ag+ Ag+
Ag
Slide 123 / 144
Electrolysis
This flowchart shows the steps relating the quantity of electrical charge used in electrolysis to the amounts of substances
Current (A) and time Quantity of charge (Coulombs) Moles of electrons (Faradays) Moles of substance
Grams of substance
A typical problem will give the current (amperes) that is applied for a specific amount of time (seconds). You would be asked to solve for the mass of metal that 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.
Slide 124 / 144
The quantity of charge passing through is measured in coulombs 1 mole of electrons passage = 96500C = 1Faraday 1coulomb = 1 ampere passing in 1 second Coulombs (C ) = ampere x seconds
Slide 125 / 144
A) joule B) coulomb C) calorie D) Newton E) Mole
40 The quantity of charge passing a point in a circuit in one second when the current is one ampere is called a ___________________.
Slide 126 / 144
41 How many coulombs result from a current
A 2.5
B
10
C
70
D
700
E
1000
Slide 127 / 144
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
125
Slide 128 / 144
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 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
Slide 130 / 144
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
Slide 131 / 144
If 10.0 A passes through molten AlCl3 for 60 minutes, how much of Al will be deposited? total charge C = 10.0 A x 60min x 60sec. = 3.6 x104 C Remember!! 1mole e- = 96500C How 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
Slide 132 / 144
Calculate the number of grams of aluminum produced in 30.0 minutes by electrolysis of at a current of 12.0 A.
practice:1
Slide 133 / 144
How many minutes will it take to plate out 6.36 g of Cu metal
from a solution of Cu2+ using a current of 12 amps in an electrolytic cell?
practice:2
Slide 134 / 144
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
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 metal 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 min
D
225 min
E
13,480 min
Slide 137 / 144
· Batteries · Hydrogen fuel cells · Corrosion · Corrosion prevention · Biology
Slide 138 / 144
Slide 139 / 144
Slide 140 / 144
Slide 141 / 144
Slide 142 / 144
Slide 143 / 144
electron transport in Mitochondria ......
Slide 144 / 144