galvanic cells - edwardsville school district 7 - edwardsville
TRANSCRIPT
Galvanic Cells
Chapter 17 Sections 1, 2 &4
Electrochemistry e- e-
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Electrochemistry- Standard Reduction Potentials e- e-
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Activity Series
Metals
Halogens
Li
Rb
K
Ba
Ca
Na
Mg
Al
Mn
Zn
Cr
Fe
Ni
Sn
Pb
H
Cu
Hg
Ag
Pt
Au
F
Cl
Br
I
Electrochemistry- Standard Reduction Potentials e- e-
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Electrochemistry- Standard Reduction Potentials e- e-
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Electrochemistry – Galvanic Cells
Electrons are transferred directly when reactants collide
No work is obtained – instead heat is released
How can we obtain work?
Separate the oxidizing and reducing agents
require the e- to go through a wire
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MnO41- + 8H+ + 5e- Mn2+ + 4H2O
Fe2+ Fe3+ + 1e-
Reduction - Mn
Oxidation - Fe
Oxidizing agent - MnO41-
Reducing Agent – Fe2+
Current flows for a second then stops
Something more needs to added.
p. 792
Electrochemistry – Galvanic Cells
It needs a salt bridge
Solutions must be connected so that ions can flow and keep the net
charge in each container at zero.
A salt bridge is a U-tube with electrolyte (pastey stuff) or porous disk.
Choose a substance that would be noninteractive (something made of
spectator ions if it’s a paste)
It allows ions to flow without mixing the solutions
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Electrochemistry – Galvanic Cells
A galvanic cell changes chemical energy to electrical energy.
Components
1. Two separate solutions (oxidizing & reducing agent)
2. A wire
3. A salt bridge
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REDUCING
AGENT
OXIDIZING
AGENT
ANODE
oxidation
X Y + e-
p. 793
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CATHODE
reduction
X + e- Y
Electrochemistry – Galvanic Cells
Be able to diagram a cell, label the parts and the flow.
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ZnSO4
REDUCING
AGENT
CuSO4
OXIDIZING
AGENT
ANODE
oxidation
Zn2+ Zn + e-
p. 793
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CATHODE
reduction
Cu + 2e- 2Cu2+
Zn Cu
Zn2+
SO4 2- Cu2+
SO4 2-
What is the
cathode?
Reduction
occurs at the
cathode.
Copper has a
greater
reduction
potential
Which way
do e- flow?
Anode to Cathode
What is
happening at
the salt bridge?
cations
anions
What are the
agents?
standard
reduction
potentials
see
the
atoms
Electrochemistry- Standard Reduction Potentials
Line Notation for the cell – so you don’t have to draw the cell
• Anode on the left; Cathode on the right
• Separate the half cells with a ||
• Separate the electrode from the solution with a |
ANODE CATHODE
Zn(s) | Zn 2+(aq) || Cu3+
(aq) | Cu(s)
Oxidation chamber || Reduction chamber
electrode | soln || soln | electrode
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Electrochemistry – Galvanic Cells
Cell Potential – the “pull” or driving force that makes electrons go from the reducing agent to the oxidizing agent
•The potential to lose e- The push from the element mostly likely to loose to the one that will gain
• E cell
•Also called electromotive force, emf, of the cell
•Unit = volt = 1 joule of work per columb of charge transferred
– V = 1 J / 1 C
•Measured with a voltmeter
– Would measure less than cell potential
– Because it doesn’t measure the frictional heating of the wire
– New voltmeters use a negligible amount of current so they are used
•Measured with a potentiometer
– Variable voltage device (powered from a cell circuit)
– Adjusted so no current flows in the cell circuit
– Then cell potential = voltage setting but opposite sign
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Electrochemistry – Galvanic Cells
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.
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Electrochemistry- Standard Reduction Potentials
If you don’t want the electrode to participate in the reaction pick a “chemically inert” metal or element.
Like Au or Pt. But what would be cheaper?
Carbon. Really? Does carbon conduct electricity?
Graphite (gr) does!
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Electrochemistry- Standard Reduction Potentials
How would I know how any metal compares against any other?
We need a standard – one that can be oxidized or reduced.
2H+ + 2e- H20 reduction
H20 2H+ + 2e- oxidation
H+ H2+ possible but not common
The standard’s E cell would be zero.
Standard Hydrogen Electrode or SHE half cell
H2 can’t be solid so use a Pt electrode
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Electrochemistry- Standard Reduction Potentials
Can measure total potential of the cell E cell = 0.76 V
Can’t measure the potential of the half reactions ( or half cells)
Setting the standard potential for the hydrogen half reaction to zero
Allows us to assign values to all other half reactions
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2H+ + 2e- H2 Zn Zn2+ + 2e-
p. 794
Electrochemistry- Standard Reduction Potentials
E cell = E 2H+ H2 + E Zn Zn2+
0.76 = 0.000V + x
E Zn Zn2+ = 0.76 V
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2H+ + 2e- H2 Zn Zn2+ + 2e-
Denotes “standard state”
not the same as STP
Standard conditions p. 246
Compounds
gases = 1 atm
soln = 1M
Liquid or solid = pure
Element
1 atm
25 C
p. 794
Oxidation Reduction Reactions
Oxidation States
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Electrochemistry- Standard Reduction Potentials e- e-
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Electrochemistry- Standard Reduction Potentials
E cell = E Zn Zn2+ + E Cu2+ Cu
1.10V = 0.76 V + E Cu2+ Cu
E Cu2+ Cu = 0.34 V
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Cu2+ + 2e- Cu Zn Zn2+ + 2e-
p. 795
Electrochemistry- Standard Reduction Potentials
The half reaction with the largest potential will run as a reduction (as written)
The other will be oxidation; so it will run in reverse
So…
F2 + 2e- 2F1- E = 2.87 V (reduction)
but
2F1- F2 + 2e- E = - 2.87 V (oxidation)
E cell = E cathode + E anode reverse the sign of the anode
E cell = E cathode - E anode
Multiplying the half reaction so that the e- are equal doesn’t
change the E
E is an intensive property
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Electrochemistry- Standard Reduction Potentials
About Standard Reduction Potentials
•A negative reduction potential means it will most likely oxidize
•If the E cell is greater than zero you will get current,
spontaneously.
•If the E cell is zero then the system has reached equilibrium.
The cell is “dead”.
•Values in the table are predicting perfect conditions. You probably won’t get this in the lab. Why?
– Voltmeter
•
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Electrochemistry- Standard Reduction Potentials e- e-
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This is the chart given with the AP Exam.
Half reactions are written as reductions
Electrochemistry- Standard Reduction Potentials e- e-
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The strongest oxidizers
(oxidizing agents) have the
most positive reduction
potentials.
The strongest reducers
(reducing agents) have the
most negative reduction
potentials.
Electrochemistry- Standard Reduction Potentials e- e-
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The greater the difference
between the two, the
greater the voltage of the
cell.
Electrochemistry- Standard Reduction Potentials e- e-
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Will Br2 oxidize H2O2?
yes
Will Cd reduce Ag+?
yes
Can Al oxidize Au?
What are common metals in nature?
What are common metal ions in nature?
What are common non metal ions in nature?
a. Consider a galvanic cell based on the reaction
Al3+ (aq)
+ Mg (s) Al (s) + Mg2+(aq)
Find the half reactions, balance the cell reaction and calculate E for the cell.
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Electrochemistry- Sample Exercise 17.1 Page 797
b. Consider a galvanic cell based on the reaction
MnO4 1-
(aq) + H+ (aq) + ClO3
1- (aq) ClO4
1- (aq) + Mn2+
(aq) + H2O (l)
Find the half reactions, balance the cell reaction and calculate E for the cell.
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Electrochemistry- Sample Exercise 17.1 Page 797
What if you have a substance that is not solid for the electrode?
Look at the example on page 799.
It is the same reaction as our Redox Titration Lab Fe2+ + MnO4 1-
In the example iron is solid but MnO4 1- / Mn2+ is a solution.
Fe2+ Fe
MnO4 1- Mn2+
What is the line notation?
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Electrochemistry- Standard Reduction Potentials
Fe ǀ Fe2+ ǁ H+, MnO4 1-, Mn2+ ǀ Pt
Include the inert
electrode
Don’t include
the water
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Electrochemistry- Try #29 Page 831
© 2009, Prentice-Hall, Inc.
Electrochemistry- Gold/Nickel Voltaic Cell e- e-
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Au Ni
Au3+(aq)
X-(aq) X-(aq)
Ni2+(aq)
K+(aq) NO3-(aq)
Au3+ + 3e- Au
E0 = +1.50 V Ni2+ + 2e- Ni E0 = -0.25 V
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Electrochemistry- Standard Reduction Potentials e- e-
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A cell will always run spontaneously in the direction that produces a positive cell
potential
Nickel is the anode E cell = E cathode - E anode flip the nickel
Current flows from anode to cathode
Au3+ + 3e- Au
E0 = 1.50 V
reduction: cathode
Ni Ni2+ + 2e-
E0 = 0.25 V
oxidation: anode
Au Ni
Au3+(aq)
X-(aq) X-(aq)
Ni2+(aq)
K+(aq) NO3-(aq)
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E0(V)
Au3+ + 3e- Au 1.50
Ni Ni2+ + 2e- 0.25
2 Au3+ + 3 Ni 2 Au + 3 Ni2+ 1.75
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Line Notation:
ANODE CATHODE
Ni(s)|Ni2+ (aq, 1 M)||Au3+ (aq, 1 M)|Au(s)
oxidation reduction
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© 2009, Prentice-Hall, Inc.
Electrochemistry- Aluminum/Nickel Cell e- e-
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Al Ni
Al3+(aq)
X-(aq) X-(aq)
Ni2+(aq)
K+(aq) NO3-(aq)
Al3+ + 3e- Al
E0 = -1.66 V Ni2+ + 2e- Ni E0 = -0.25 V
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E0(V)
Al Al3+ + 3e- 1.66
Ni2+ + 2e- Ni -0.25
2Al + 3Ni 2+ 3Ni + 2Al3+ 1.41
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Line Notation:
ANODE CATHODE
Al(s) | Al3+(aq, 1 M) || Ni2+
(aq, 1 M) | Ni (s)
oxidation reduction
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Sketch the Ag/Ag+ & Zn/Zn2+ galvanic cell. Show the direction of e- flow and the direction of ion migration. Calculate E0 for the cell. Give the line notation for the cell.
Electrochemistry- Extra Problem e- e-
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Zn Ag
Zn2+(aq)
X-(aq) X-(aq)
Ag+(aq)
K+(aq) NO3-(aq)
Zn2+ + 2e- Zn
E0 = - 0.76 V
Ag+ + e- Ag
E0 = + 0.80 V
e- ?
Sketch a system having a solid lead electrode and a lead(IV) oxide electrode immersed in sulfuric acid. Show the direction of e- flow and the direction of ion migration. Calculate E0 for the cell. Give the line notation for the cell.
Electrochemistry- Extra Problem e- e-
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Pb
Pb
O2
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H2SO4(aq)
PbSO4 + 2e- Pb + SO42-
E0 = - 0.35 V
PbO2 + 4H+ + SO4
2- + 2e- PbSO4 + 2H2O
E0 = + 1.69V
The Lead-Acid Car Battery
Electrochemistry- Standard Reduction Potentials e- e-
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Cu(s) + 2Ce 4+ (aq)
2Cu 2+ (aq)
+ Ce 3+ (aq)
Remember E is positive. Treat this like an equilibrium.
Increase [Ce 4+ ], which direction is favored?
Forward. Increase the driving force on e-, increase the E Increase [Cu 2+ ] or [Ce 3+], which direction is favored?
Reverse. Decrease the driving force on e-, decrease the E
Sample Exercise 17.5 page 803
2Al + 3 Mn2+ 2Al3+ + 3 Mn
a. [Al3+] = 2.0M; [Mn2+] = 1.0M
b. [Al3+] = 1.0M; [Mn2+] = 3.0M
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Electrochemistry- Concentration on E
Al3+ decrease E < 0.48V
Mn2+ increase E > 0.48V
Cell Potentials depend on concentration
Can construct cells with same components but different concentrations
Ag ǀ 0.10M Ag2+ ǁ 1.0 M Ag2+ ǀ Ag
E cell =0.80V – 0.80 V = 0
However since the concentration is not equal,
the half cell potentials are not 0.80V.
This is a concentration cell. Voltages are typically small
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Electrochemistry- Concentration on E
The Nernst Equation
Derived from the dependence of free energy on concentration
*** Will learn about this after thermodynamics***
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Electrochemistry- Concentration on E
n E = E cell -
0.0591 log (Q)
moles of e-
Like Keq but not necessarily at equilibrium
Remember (l) and (s) don’t go into the Keq
Why not? It’s all about concentration – solids and liquids have a definite volume
•The M for solids and liquids is 1 M= mol/L
•If you change moles for solids and liquids then the volume changes proportionally
Quotient = Keq
[P]x
[R]y
The Nernst Equation
Derived from the dependence of free energy on concentration
*** Will learn about this after thermodynamics***
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Electrochemistry- Concentration on E
n E = E cell -
0.0591 log (Q)
moles of e-
Trying to get the max voltage?
•Choose the half reactions that give max volts – be practical Li or F won’t work
•Make the reactant concentration greater than 1 & the product concentration less than 1
That way [P]x / [R]y is 0.XX and then log 0.XX is a negative
number and you’re adding to the E cell
Quotient = Keq
[P]x
[R]y
Handy ideas about logs
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Electrochemistry- Sample Exercise 17.7 page 806
= log 1 – log x = -log x log 1
x log x 2 = 2 log x
log 100 = 2
Sig. Figs for Logarithms For any log, the number to the left of the decimal point is called the characteristic, and the number to the right of the decimal point is called the mantissa. The characteristic only locates the decimal point of the number, so it is usually not included when determining the number of significant figures. The mantissa has as many significant figures as the number whose log was found. Example 1: log 5.43 x 1010 = 10.735 The number has 3 significant figures, but its log ends up with 5 significant figures, since the mantissa has 3 and the characteristic has 2.
log xy = log x + log y
log x y = y log x log y√ x = log x1/y = (1/y ) log x
Describe the cell based on the following half reactions:
VO2+
(aq) + 2H1+ (aq) + e- VO 2+
(aq) + H2O (l)
Zn2+ (aq) + 2e- Zn (s)
T = 25 C
[VO2+] = 2.0 M
[H1+] = 0.50M
[VO 2+ ] = 1.0 x 10-2 M
[Zn2+ ] = 1.0 x 10-1 M
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Electrochemistry- Sample Exercise 17.7 page 806
Describe the cell based on the following half reactions:
VO2+
(aq) + 2H1+ (aq) + e- VO 2+
(aq) + H2O (l)
Zn2+ (aq) + 2e- Zn (s)
Where T= 25 C
[VO2+] = 2.0 M
[H1+] = 0.50M
[VO 2+ ] = 1.0 x 10-2 M
[Zn2+ ] = 1.0 x 10-1 M
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Electrochemistry- Sample Exercise 17.7 page 806
Solution
1. Balance First
2. Calculate the standard cell potential
3. Use the Nernst Equation
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Electrochemistry- Try #59 Page 832
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Electrochemistry- Try #57 Page 832