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Chapter 20:
Electrochemistry

Chemistry: The Molecular Nature

of Matter, 6E
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Electrochemistry

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Study of chemical reactions that can produce

electricity or use electricity to produce desired

product.
Study of interchange of chemical and electrical
energy
Electrochemical reaction always involves
oxidation-reduction reactions
Electron transfer reactions
Electrons transferred from one substance to
another

Also called redox reactions

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Oxidation-Reduction Reactions

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Oxidation and reduction reactions occur in

many chemical and biochemical systems
Combustion reactions
Photosynthesis

6CO2 + 6H2O C6H12O6 + 6O2

Mitochondrial Respiration
NADH NAD+
Methane monoxygenase

CH4 + NADH + H+ + O2 CH3OH + NAD+ + H2O

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Involves two processes:

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Oxidation Reduction Reactions

Oxidation loss of electrons from one reactant

Na

(s)

Na+ + e

Oxidation Half Reaction

Reduction gain of electrons from another

reactant

Cl2 (g) + 2 e 2 Cl Reduction Half Reaction

Net reaction
2 Na (s) + Cl2 (g) 2 Na+ + 2 Cl
Oxidation and reduction ALWAYS occur together
Can't have one without the other
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Oxidation-Reduction Reaction

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Substance that accepts the e's

Takes e's from another substance

is itself reduced
e acceptor
Cl2 (g) + 2 e 2 Cl

Reducing Agent

Substance that donates e's

Releases e's to another substance

is itself oxidized
e donor
Na (s) Na+ + e
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Oxidation Numbers

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Way to keep track of e's

Rules for Assigning Oxidation Numbers
1. Sum of all oxidation numbers of atoms in molecule
or polyatomic ion must equal charge on particle
2. Oxidation number:
1.
2.
3.
4.
5.

of
of
of
of
of

any free element is zero (0)

any simple, monoatomic ion is equal to charge on ion
fluorine in its compounds is 1
hydrogen in its compounds is +1
oxygen in its compounds is 2

3. If there is a conflict between 2 rules, apply the

rule with lower number and ignore conflicting rule.

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Define Oxidation-Reduction in Terms

of Oxidation Number
Oxidation
in oxidation number or more + oxidation
number
Leo (Loss of es)
Oil (oxidation is e loss)

Reduction

in oxidation number or more oxidation

number
Ger (Gain of es)
Rig (Reduction is e gain)
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Your Turn!

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What is the oxidation number of Cl in HClO4?

A. +1
B. +3
C. +5
D. +7

H is +1 and O is -2. There are 4 O atoms.

+1 +(4)(-2) + x = 0
x = +7
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Spontaneous Redox Reactions

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When a copper wire is placed in a solution of

silver nitrate
Silver metal spontaneously precipitates
Copper ion spontaneously forms as evidenced by
blue color of solution

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Electrochemistry

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Possible to separate oxidation and reduction

processes and cause to occur in two different
locations
Can use spontaneous redox reaction to produce
electricity
Can use electricity to make nonspontaneous redox
reactions happen
Biology does this by coupling nonspontaneous
redox reactions with spontaneous reactions that
provide the driving force.
Can harness electrical energy to do work
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Galvanic (or Voltaic) Cell

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Electrochemical cell in which spontaneous

redox reactions occur
Produce electricity spontaneously
Energy released by this spontaneous redox
reaction can be used to perform electrical
work
Transfer of electrons takes place through an
external pathway (wire) rather than directly
between reactants
Ex. Batteries used to power laptops, cell phones,
camera, etc.
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Spontaneous Redox Reactions

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Ex 1. Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

Ex 2. Cu2+(aq) + Zn(s) Cu(s) + Zn2+(aq)
Lets look at Ex.1 in detail
Imagine dividing up the reaction in Ex. 1 into
individual oxidation and reduction half reactions
Cu(s) Cu2+(aq) + 2e
2Ag+(aq) + 2e 2Ag(s)

Physically this can be accomplished by having a

strip of the given metal (called an electrode) in
a solution of the corresponding ion
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Example of Spontaneous Redox

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Silver metal placed in a solution of AgNO3

Copper metal placed in solution of Cu(NO3)2
Each compartment is called a half-cell

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Half-Cells

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When metal ion collides with If metal atom on surface of

electrode and gains
electrode loses electrons,
electrons, ion becomes
becomes oxidized

reduced

Left on their own, each individual cell quickly establishes an

equilibrium between metal and ions in solution.
M(s) ' Mn+(aq) + ne
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Electrode Names

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By combining 2 different half-cells we can

cause es to flow from 1 cell to the other
One half-cell undergoes oxidation = anode
Other half-cell undergoes reduction = cathode
Anode = electrode at which oxidation (e
loss) occurs
Cathode = electrode at which reduction (e
gain) occurs
An Ox and a Red Cat

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Learning Check:

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Identify cathode and anode in these acidic

solutions
CrO3 (s) + MnO2 (s) MnO4 (aq) + Cr3+ (aq)
CrO3 (s) + 6H+ + 3 e Cr3+ (aq) + 3H2O (redn)
MnO2(s) + 2H2O MnO4(aq) + 4H+ + 3e (oxidn)
Cathode = reduction = CrO3 Cr3+ (aq) half rxn
Anode = oxidation = MnO2(s) MnO4(aq) half rxn

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Learning Check:

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Identify the cathode and anode in these acidic

solutions
H2O2(aq) + CO2(g) H2C2O4(aq) + O2(g)
1H2O2 (aq) O2 (g) + 2H+ (aq) + 2e
2CO2 (g) + 2H+ + 2e H2C2O4 (aq)

(oxidn)
(redn)

Cathode = reduction =
2CO2 (g) H2C2O4 (aq)
Anode = oxidation = H2O2 (aq) O2 (g)
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half rxn
half rxn
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Your Turn!

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Identify the half-reaction that occurs at the anode for

the reaction:
4H+(aq) + 3S(s) + 4NO3-(aq) 3SO2(g) + 4NO(g) + 2H2O
A.
B.
C.
D.

3S(s) + 6H2O 3SO2(g) + 12H+(aq) + 12e12e- + 4NO3-(aq) + 16H+(aq) 4NO(g) + 8H2O

3S(s) + 6H2O 3SO2(g) + 12H+(aq) + 8e8e- + 4NO3-(aq) + 16H+(aq) 4NO(g) + 8H2O

Oxidation occurs at the anode and oxidation is the loss of

electrons. Answer C is not a balanced equation.
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How can we get electrons to migrate into

the other solution?

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Connect the two solutions with a metal wire that will

allow es to pass from one cell to another
Now es can flow, but the reaction still will not initiate.
Why?
Consider what would happen to solution if es did flow
from one cell to the other

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Electron Flow Between Half-Cells

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At copper end, 2 es are given up

Cu(s) Cu2+(aq) + 2e
Travel across with wire to the silver half-cell
2Ag+(aq) + 2e 2Ag(s)
What happens to the charge in each half-cell?
Each time the reaction occurs

Net charge at copper end by 2

Net charge at silver end by 2

Violates principle of electroneutrality

Cannot have solution with a net charge

We need to balance the charge in

order for the es to flow
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Metallic Conduction

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Two Types of Charge Conduction

External to the cell
Electrical charge is transported from one electrode
to the other by movement of electrons through the
wires
How metals conduct electricity

Electrolytic Conduction

Inside electrochemical cells

Electrical charge is carried through the liquid by
movement of ions
Transport of electrical charge by ions
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Salt Bridge

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Needed to complete circuit

Tube filled with solution of an electrolyte
Salt composed of ions not involved in cell reaction
KNO3 and KCl often used

Porous plugs at each end of tube

Prevent solution from pouring out

Salt bridge

Enable ions from salt bridge to migrate between

half-cells to neutralize charges in cell compartments
Anions always migrate toward anode
Cations always migrate toward cathode

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Electron Flow Between Half-Cells

ly

Now the spontaneous reaction occurs

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As es flow from one half-cell to other, salt bridge can

transfer corresponding charge back to 1st half-cell
Cu loses 2 es giving Cu2+
These 2 es travel through anode up through wire and across
to other half-cell.
Here 2 Ag+ ions in solution pick up the 2 es and form 2Ag(s)
which precipitate out onto electrode
Ions flow through salt bridge to
complete the circuit
Cations flow into Ag half-cell
Anions flow into Cu half-cell

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Summary of a Galvanic Cell

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Half-cells (compartments containing reactants

for each half-reaction)
Electrodes to conduct current through solution
Wire connecting two half-cells
Salt bridge to offset ion
movement
Resistance to electrical flow
Supporting electrolyte

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Your Turn!

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Which of the following species would be

appropriate for a salt bridge solution?
A. AgCl
B. C12H22O11 sucrose (sugar)
C. NaCl
D. C6H6

The solution needs to be an electrolyte. AgCl is

not soluble and the organic compounds are not
ionic.
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Cell Reaction

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Net overall reaction in the cell

To get add individual half-reactions, after make
sure that
number of es gained in reduction = number of es
lost in oxidation
(oxidation)
Cu(s) Cu2+(aq) + 2e
2Ag+(aq) + 2e 2Ag(s)
(reduction)
2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)

Use ion electron method to balance halfreactions (see Ch 6)

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Standard Cell Notation

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Drawing sketches of electrochemical cells can

be cumbersome
Simpler representation called standard cell
notation used instead
anode

cathode

Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)

Anode electrode where oxidation occurs is

placed on left
Cathode electrode where reduction occurs in
placed on right
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Salt Bridge

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Cell Notation

cathode

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Cu(s)|Cu2+(aq)||Ag+(aq)|Ag(s)

anode
electrode

anode
electrolyte

cathode
electrolyte

cathode
electrode

Single slash = boundary between phases (solid

electrode and aqueous solution of ions)
Double slash represents salt bridge
Separates cell reactions

In each half (half-cell)

Electrodes appear at outsides

Reaction electrolytes in inner section
Species in same state separated with ;
Concentrations shown in ( )
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Learning Check

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Write the standard cell notation for the

following electrochemical cells:
Fe (s) + Cd2+ (aq) Cd(s) + Fe2+(aq)
Anode = ox = Fe(s)
Cathode = red = Cd2+(aq)
Fe(s)|Fe2+(aq)||Cd2+(aq)|Cd(s)
Al(s) + Au3+(aq) Al3+(aq) + Au(s)
Anode = ox = Al(s)
Cathode = red = Au3+(aq)
Al(s)|Al3+(aq)||Au3+(aq)|Au(s)
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Learning Check

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Where there are no conductive metals involved

in a process, an inert electrode is used. C(gr)
or Pt(s) are often used. Write the Standard
Cell notation for the reactions
H2O2(aq) + CO2(g) H2C2O4(aq) + O2(g)
Anode = ox = H2O2(aq) | O2(g)
Cathode = an = CO2(g) | H2C2O4(aq)
C(gr)|H2O2(aq); H+(aq)|O2(g)||CO2(g)|H2C2O4(aq); H+(aq)|C(gr)
Pt(s)|H2O2(aq); H+(aq)|O2(g)||CO2(g)|H2C2O4(aq); H+(aq)|Pt(s)
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Your Turn!

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Write the standard cell notation (Pt electrodes) for the

following reaction:
2Mn3+(aq) + 2I-(aq) Mn2+(aq) + I2(s)
A. Pt(s)|Mn3+(aq); Mn2+(aq)||I-(aq)|I2(s)|Pt(s)
B. Pt(s)|I-(aq)|I2(s)||Mn3+(aq); Mn2+(aq)|Pt(s)
C. Mn3+(aq)|Pt(s); Mn2+(aq)||I-(aq)|I2(s)|Pt(s)
D. Pt(s)|Mn3+(aq); I-(aq)||Mn2+(aq)|I2(s)|Pt(s)
Oxidation reaction is on the right and reduction
reaction is on the left of the salt bridge (||).
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Measuring Cell Potential

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Instead of just a wire we can connect a

voltmeter between 2 half-cells
Voltmeter
Measures potential difference between 2 half-cells

Unit of Potential Volt (V)

Measure of amount of energy

(Joules, J) that can be
delivered per SI unit of
charge (coulomb, C)

1 V = 1 J/C

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Voltage

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Voltage is driving force for reactions

Greater voltage = greater driving force

Voltage or potential of galvanic cell varies with

amount of current flowing through the circuit
Voltage is always positive in direction of a
spontaneous reaction
Cell Potential (Ecell)
Maximum voltage that a given cell can generate
Depends on
Composition of electrodes
[Ions] in half-cells
Temperature
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Standard Cell Potential, Ecell

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To compare potentials for different cells, we

need a set of standard conditions
Ecell means at standard conditions
T = 25C = 298 K
All gases at P = 1 atm
All [Ions] = 1 M

Rarely larger than a few volts

Ex. 2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)

If you need batteries with higher voltages,

arrange several cells in series
Ex. car batteries

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Electrical Potential

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Competition between two half-cells

Every half-cell has tendency to gain electrons

and proceed as reduction half-reaction
Reduction Potential, Ered
Relative ease of gaining electrons

Standard Reduction Potential, Ered

Reduction potential measured under standard

conditions
All [solutes] = 1M
All gases at P = 1 atm
T = 25C
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Standard Cell Potential, Ecell

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When two half-cells are connected

The one with larger (more +) Ered acquires electrons

Undergoes reduction

The one with lower (more ) Ered donates electrons

Undergoes oxidation

Measure cell potential, Ecell

Represents magnitude of difference between Ered of

one half-cell and Ered of other half-cell
Is always taken as positive number for spontaneous
redox reactions

cell

Of substance
red reduced

Jespersen/Brady/Hyslop

red

Of substance
oxidized

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Ex: Calculating Ecell

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Look at reaction in zinc-copper cell

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s)

Zinc is oxidized
Copper is reduced
Half reduction reactions
Zn2+ (aq) + 2 e Zn (s)
Cu2+ (aq) + 2 e Cu (s)
Reaction for Cu2+ must have greater
tendency to proceed than Zn2+
as Cu2+ is the one reduced

Ecell = ECu2+ EZn2+

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Reference Electrode

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No way to measure reduction potential of isolated

half-reaction
Can only measure difference in potential between 2 half-cells

To assign Ered to half reactions

Must arbitrarily choose a reference electrode (half-reaction)

Set its potential to exactly 0.00V

Standard Hydrogen Electrode

H2 gas at 1 atm bubbled over

Pt electrode coated with finely divided Pt
to provide large surface area
T = 25C
1.00 M [H+]

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Standard Hydrogen Electrode

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2H+(aq, 1.00M) +2e ' H2(g, 1 atm) EH+ = 0.00 V

Double arrow indicates only that reaction is
reversible
Not that there is true equilibrium
Whether oxidation or reduction occurs depends on
what half-reaction this is paired with

How to list in cell notation

Pt(s), H2(g)|H+(aq)||Cu2+(aq)|Cu (s)

Zn(s)|Zn2+(aq)|| H+(aq)|Pt(s), H2(g)

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Ex. Galvanic cell of Cu/Cu2+ with Hydrogen

Electrode

Need to know which half-cell undergoes reduction and

which undergoes oxidation
When measuring potential in galvanic cell
Cathode (reduction) carries + charge
Anode (oxidation carries charge

Use voltmeter to determine potential

Shows that Cu carries + charge (cathode)
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Ex. Galvanic cell of Cu/Cu2+ with

Hydrogen Electrode
Cu2+(aq) + 2e ' Cu(s)
(cathode)
(anode)
H2(g) ' 2H+(aq) + 2e
Cu2+(aq) + H2(g) ' Cu(s) + 2H+(aq) (cell reaction)

Ecell = Ereduced Eoxidized

Ecell = Ecathode Eanode
Ecell = ECu2+ EH+
0.34 V = ECu2+ 0V
ECu2+ = 0.34 V
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Ex. Galvanic cell of Zn/Zn2+ with

Hydrogen Electrode

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In this case the hydrogen electrode is positive so it

is the cathode
Zn(s) ' Zn2+(aq) + 2e
2H+(aq) + 2e ' H2(g)

(anode)
(cathode)

Zn(s) + 2H+(aq) ' Zn2+(aq) + H2(g) (cell reaction)

Ecell = Ereduced Eoxidized

Ecell = Ecathode Eanode
Ecell = EH+ EZn2+
0.76 V = 0.00V EZn2+
EZn2+ = 0.76 V
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Standard Reduction Potentials (Ered)

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See Table 20.1 in the Jespersen text

All tabulated as reduction potentials

oxidized form + electrons reduced form

Can be ions, elements or compounds
Arranged from top to bottom by Ered

High values of Ered (Ered > 0) means

substance is easily reduced
More positive value of Ered = more likely to
undergo reduction

F2(g, 1atm) + 2 e 2 F(aq, 1M) E = 2.87 V

Most easily reduced
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Standard Reduction Potentials (Ered)

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Low value of Ered (Ered < 0) means substance

is easily oxidized
More negative value of Ered = less likely to
undergo reduction

Li+ (aq, 1M) + e Li(s) E = 3.05 V

More likely the reverse reaction occurs
E = +3.05 V
Li(s) Li+ (aq, 1M) + e

When reaction reversed, sign of E reversed

All substances are compared to H+, which has a
Ered of 0.00 V.
Sign of Ered is sign of electrode when attached to
H+/H2
Jespersen/Brady/Hyslop

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Using Ecell to Calculate Ered

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Ex. Go back to galvanic cell with

2Ag+(aq) + Cu(s) 2Ag(s) + Cu2+(aq)
Ecell = 0.46V
Cu(s) undergoes oxidation
ECu = + 0.34 V

Ag+(aq) undergoes reduction

What is Ered for the half-reaction of Ag+?

Ecell = E Ag+ ECu

+0.46 V = EAg+ (+0.34 V)
EAg+ = 0.46 V + 0.34 V = +0.80 V
Jespersen/Brady/Hyslop

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Can we Use Ered to Predict

Spontaneous Reactions?

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Yes!!
For substances in two half-reactions
Substance with more positive Ered
always occurs as written
Reduction
Other half-reaction with less positive Ered
always forces to run in reverse
Oxidation
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Predicting Reaction Spontaneity

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When have list of Ered as given in Table 20.1

Half-reaction with more positive potentials
(higher up in table) and half-reaction occurs as
written
Other half-reaction (lower in table) is reversed and
occurs as oxidation

For spontaneous reaction

Reactants found in left side on higher half-reaction

and on right side of lower half-reaction

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Predicting Reaction Spontaneity

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Ex. What spontaneous reaction occurs when Cr

and Au are added to a solution of Au3+ and
Cr3+ ?
E = 0.74 V
Cr3+(aq) + 3e Cr(s)
Au3+(aq) + 3e Au(s)
E = +1.50 V
Au is more + so reduction
Cr (oxidations) so half reaction is reversed
Au3+(aq) + 3e Au(s)
Cr(s) Cr3+(aq) + 3e
Au3+(aq) + Cr(s) Au(s) + Cr3+(aq)
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Is this Answer Reasonable?

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Use reduction potentials to predict Ecell for the

reaction.
E = 0.74 V
Cr3+(aq) + 3e Cr(s)
Au3+(aq) + 3e Au(s)
E = +1.50 V
Net Reaction =
Au3+(aq) + Cr(s) Au(s) + Cr3+(aq)
Ecell = Ereduced Eoxidized
Ecell = EAu3+ ECr3+
Ecell = +1.50 V (0.74 V)
Ecell = +2.24 V
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Calculating Cell Potentials, Ecell

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Predict the reaction that will occur when Mg

and Cd are added to a solution containing
Mg2+ and Cd2+ ion.
E = 0.40 V
Cd2+(aq) + 2e Cd(s)
E = 2.37 V
Mg2+(aq) + 2e Mg(s)
Since Cd2+ higher in Table (more positive), it
is reduced and Mg is oxidized
Cd2+(aq) + 2e Cd(s) (reduction)
Mg(s) Mg2+(aq) + 2e (oxidation)
Cd2+(aq) + Mg(s) Cd(s) + Mg2+(aq)
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Calculating Cell Potentials, Ecell

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What would be the cell reaction and cell

potential for the galvanic cell employing the
following half-reactions?
E = 0.28 V
Co2+(aq) + 2e Co(s)
Au3+(aq) + 3e Au(s)
E = +1.50 V
Au3+ has more positive Ered so
reduction
2[ Au3+(aq) + 3e Au(s) ]
2+(aq) + 2e
Co
(
s
)

Co
oxidation
]
3[
2Au3+(aq) + 3 Co(s) 2Au(s) + 3Co2+(aq) net
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Calculating Cell Potentials, Ecell

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Ecell = Ereduced Eoxidized

Ecell = EAu3+ ECo2+
Ecell = +1.50 V (0.28 V)
Ecell = +1.78 V
Note: Although multiplying half-reaction by factors
to make electrons cancel,
Do NOT multiply Ered by these factors
Ered are intensive quantities, do not depend on
amount
Units = V
Same number of joules available for each coulomb of
charge regardless of total number of electrons shown
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Predicting if Reaction is Spontaneous

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Given any redox reaction

Calculate Ecell for reaction as written
If Ecell > 0 V (positive), then reaction is
spontaneous
Galvanic cell = spontaneous
Ecell always positive

If Ecell < 0 V (negative), then reaction is

nonspontaneous
Electrolytic cell = nonspontaneous
Ecell always negative
Jespersen/Brady/Hyslop

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Using Ecell to Predict Spontaneity

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Determine whether the following reaction is

spontaneous as written
6I(aq) + BrO3(aq) + 6H+(aq) 3I2(s) + Br(aq) + 6H2O
Ered = + 0.54 V
I2(s) + 2e ' 2I(aq)
BrO3(aq) + 6H+(aq) + 6e ' Br(aq) + 6H2O
Ered = + 1.44 V
BrO3(aq) + 6H+(aq) + 6e ' Br(aq) + 6H2O redn
3 x [2I(aq) ' I2(s) + 2e ]
oxidn
Ecell = EBrO3 EI2
Ecell = 1.44 V (+0.54 V)
Ecell = +0.90 V
spontaneous
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Using Ecell to Predict Spontaneity

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Determine whether the following reaction is

spontaneous as written
Au(s) + Al3+(aq) Au3+(aq) + Al(s)
Al3+(aq) + 3e Al(s)
Au3+(aq) + 3e Au(s)

Au(s) Au3+(aq) + 3e
Al3+(aq) + 3e Al(s)

E = 1.66 V
E = +1.50 V

oxidation
reduction

Ecell = EAl3+ EAu3+

Ecell = 1.66 V (+1.50 V)
Ecell = 3.16 V
nonspontaneous
Jespersen/Brady/Hyslop

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Your Turn!

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Is the following reaction spontaneous and what is its

standard cell potential?
2H2O2(aq) + S(s) SO2(g) + 2H2O
H2O2(aq) + 2H+(aq) + 2e- 2H2O Eo = +1.77 V
SO2(g) + 4H+(aq) +4e- S(s) + 2H2O Eo = +.45 V
A. spontaneous; +2.22V
B. spontaneous; +1.32V
C. spontaneous; +3.09V
D. non-spontaneous; -1.32V
Eocell = (+1.77 0.45) V = +1.32 V
spontaneous since Eocell > 0
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Are Ecell and G Related?

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Yes, both let us predict spontaneity of reaction

For chemical reaction
G = maximum work system can do

In electrical system

Maximum work = nFEcell

Where
n = number of moles of electrons transferred
F = Faradays constant = number of Coulombs of
charge equivalent to 1 mole of e
1 F = 96,486 C
Ecell = potential of cell in Volts
Jespersen/Brady/Hyslop

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coulombs joules

=J

mol e
coulomb

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Max Work = mol e

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Relationship between Ecell and G

Ecell

Equating the 2 expressions for maximum work gives

G = nFEcell

If using standard potentials, you can calculate

standard free energy changes

G = nFEcell
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Calculating Cell Potentials, Ecell

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Ex. Calculate G for the following reaction

3+
3+
Au (aq) + Al(s) Au(s) + Al (aq)
Ecell = +3.16 V
Solving
Std conditions so T = 25C = 298 K

96 ,485C
= (3 mol e )
1 mol e

1J
(3.16V )

1C 1V

G = 914,678 J/mol = -915 kJ/mol

Jespersen/Brady/Hyslop

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Applications of Electrochemistry

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Determination of Equilibrium Constant, K

If we combine
G = RT lnK C
with
G = nFEcell
We get
nFEcell = RT lnK C
Rearranging gives

Jespersen/Brady/Hyslop

E cell
o

RT
=
ln K C
nF

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Determine KC from Ecell

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Ex. Calculate KC for the redox reaction

6I(aq) + BrO3(aq) + 6H+(aq) 3I2(s) + Br(aq) + 6H2O

Ecell = +0.90 V

Determine n

BrO3(aq) + 6H+(aq) + 6e ' Br(aq) + 6H2O redn

6I(aq) ' 3I2(s) + 6e
oxidn

So n = 6
T = 298 K
R = 8.3145 J/(molK)
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Next Rearrange eqn to get lnKc

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Determine KC from Ecell

E cell
o

RT
=
ln K C
nF

Plug values into equation

ln K C

E cell nF
=
RT
o

0.90 J / C 6 96,485C / mol

ln K C =
8.3145J / mol K 298K
lnKc = 210.3

K C = e 210.3 = 2.1 10 90

Ecell = +0.90 V, G = 521 kJ/mol, Kc = 2.1 1090

Reaction spontaneous; goes to completion
Jespersen/Brady/Hyslop

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Your Turn!

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Determine the equilibrium constant, KC , for a

reaction that has a standard cell potential of
+2.55 V and transfers four electrons in the
reaction process.
A. infinite
B. 1.00
C. 1.6 x 10397
D. 3.2 x 10172

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Your Turn! - Solution

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2.55 J / C 4 96, 485C / mol

=
8.3145 J / mol K 298K
= 397.2

ln K C
ln K C

K C = 3.2 x 10

172

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Relationship between Ecell and Ecell

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Ecell = cell potential at standard conditions

All gases at 1 atm pressure
All ion concentrations = 1.00M
25C

When concentrations, pressures or temperature

changes, Ecell changes
Just as G changes

Ex. In operating battery

Potential gradually as reactants are used up and

cell reaction approaches equilibrium
When reaches equilibrium, Ecell = 0
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Relationship between Ecell and Ecell

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To determine effect of [ion] on Ecell look at

how G affected by [ion]
G = G + RT lnQ
Since G = nFEcell and G = nFEcell
Then
nFEcell = nFEcell + RT lnQ
Dividing both sides by nF gives

E cell = E cell
o

RT

ln Q
nF

Nernst equation

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Using the Nernst Eqn

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Need to construct mass action expression for

redox reaction in either
2+
[ ] in M for solutions
P in atm for gases

Q = ln

[Zn ]PH 2
+ 2

[H ]

Ex. For the reaction,

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

E cell = E cell
o

RT [Zn ]PH 2

ln
+ 2
nF
[H ]

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Ex. Using the Nernst Eqn

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Calculate the cell potential for a galvanic cell

employing the following half-reactions if
[Fe3+]=0.0100 M, [Fe2+]=0.250 M,
[Co2+]=0.050 M,
Fe3+(aq) + e Fe2+(aq) Ered = +0.77 V
Co2+(aq) + 2e Co(s)
Ered = 0.28 V
Solution
Fe3+ has more + Ered , so reduction
2 [ Fe3+(aq) + e Fe2+(aq) ]
reduction
Co(s) Co2+(aq) + 2e
oxidation
2Fe3+(aq) + Co(s) 2Fe2+(aq) + Co2+(aq) net rxn
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Ex. Using the Nernst Eqn

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n = total number of e transferred = 2

2+ 2
2+
RT
[
Fe
]
[
Co
]
o
E cell = E cell
ln
2F
[Fe3 + ]2
2Fe3+(aq) + Co(s) 2Fe2+(aq) + Co2+(aq)
Ecell = EFe ECo
Ecell = 0.77 V (0.28V) = +1.05 V
Plug in R, T, n and concentrations
3+

2+

E cell

8.3145J /(mol K ) 298K [0.250]2 [0.050]

= 1.05 V
ln
2 96,485C / mol
[0.0100]2

E cell

(0.03125)
= 1.05 V (0.0128 V) ln
(1.0 10 4 )

Ecell = +1.05 V 0.0738 V = 0.98 V

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Ex. Using the Nernst Eqn

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In order to determine H+ concentration, a

galvanic cell was constructed by connecting a
Cu electrode dipped in 0.50 M CuSO4 to a H2
electrode at gas pressure of 1 atm dipped in
the H+ solution of unknown concentration. A
value of 0.49 V is recorded for Ecell at 298K.
What is the pH of hydrogen half-cell?
Step 1: First write balanced redox reaction
and determine Ecell.
H2 (g) + Cu2+ (aq) 2H+ (aq) + Cu (s)
Ecell = ECu2+ EH+ = 0.34 V 0.00 V = 0.34 V
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Ex. Determination of pH from Ecell

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Step 2: Determine the form of the Nernst

equation for this reaction
n = 2 as 2 electrons are transferred
o

E cell = E cell

RT

2F

ln

[H + ]2

[Cu

2+

Step 3: Solve for [H+]

8.314 J /(mol K ) 298K

[H + ]2
0.49V = 0.34V
ln
2 (96 ,500C / mol )
(0.050)

[H + ]2
0.49 V 0.34 V = 0.01284 V ln
(0.050)
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Step 3: Solve for [H+]

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Ex. Determination of pH from Ecell

Solve for ln([H+]2/[Cu2+]PH2)
[H + ]2
ln

(0.050)

= 11.68

Take antilog

[H + ]2
= e 11.68 = 8.44 10 6
(0.050)

Solve for [H+]

[H+]2 = (8.44 x 106) x (0.050)

[H+ ] = 4.22 10 7 = 6.5 10 4 M

pH = log[H+] = 3.19
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Your Turn!

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What is the pOH of the following reaction when the

pressure of O2 is 1 atm, the temperature is 25oC, the
concentration of Al3+ is 1.00M, and the cell potential
reads 1.341V?
3O2(g) + 6H2O + 4Al(s) 4Al3+(aq) + 12OH-(aq)
O2(g) + 2H2O + 4e- 4OH-(aq) Eo = +0.401V
Al3+(aq) Al(s) + 3eEo = -1.66V
A. 12.99
B. 7.00
C. 1.01
D. Cannot be determined from the data given
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Your Turn! - Solution

8.3145 J / (mol K ) 298K

1.341V = 1.281V
ln[OH ]12
12 (96, 485C / mol )
[OH ] = 9.667 x 10 2
pOH = 1.01

Jespersen/Brady/Hyslop

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Electrolytic Cell

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Nonspontaneous
Use electrical energy to force nonspontaneous
reaction to occur
Electrodes switch relative to galvanic cells
What was cathode becomes anode where oxidation occurs
What was anode becomes cathode where reduction occurs

Type of reaction that occurs when you recharge

batteries
Substance undergoing electrolysis must be molten or
in solution so ions can move freely and conduction can
occur
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Molten NaCl cell (801C)

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Example Electrolysis Cell

Inert electrodes placed into
molten liquid and connected
to electrical source
Cathode: Na+() + e Na()
Anode: 2Cl() Cl2 (g) + 2e

cathode

anode

Multiply first half-cell by 2

to balance electrons

2Na+() + 2e 2Na()
2Na+() + 2Cl() 2Na() + Cl2(g) (Net)
Since nonspontaneous
electrolysis
2Na+() + 2Cl() 2Na() + Cl2(g)
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Electrolytic vs. Galvanic Cells

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Electrolysis means application of electricity

Electrolytic means that a particular reaction is not
spontaneous

Electrolytic Cell

Galvanic Cell

Cathode is positive (reduction)

Anode is negative (oxidation)
Non-spontaneous
Requires a battery

Cathode is positive (reduction)

Anode is negative (oxidation)
Spontaneous
Is a Battery

In both galvanic and electrolytic cells, positive ions

move to cathode and negative ions move to anode
Attracted there in electrolytic cell by charges on electrodes
Diffuse there in galvanic cell to balance developing charges
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Electrical Conduction

Only occurs because reactions take place at surface of

electrode
Consider molten NaCl electrolytic cell

When charged anode placed in molten NaCl

Positive charge attracts Cl ions
positive charge on electrode pulls e away from Cl to form Cl
atoms that combine to form Cl2
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Electrical Conduction

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Cl2 is neutral so drifts away from charged electrode

Quickly replaced by other Cl or anions in solution
Other negative ions move toward anode causing
charge migration
Similarly positive ions migrate to cathode

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More complex
May have competing
reactions

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Electrolysis Reactions in Aq. Solution

Must consider possible

oxidation and reduction of
solute
Must also consider redox
reactions of water

Ex. K2SO4 in H2O

Expected products
K and S2O82
Actual products
H2 and O2 gases

Jespersen/Brady/Hyslop

cathode

anode

SO42

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Electrolysis of Aqueous K2SO4

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Why this difference in products?

Must examine reduction potential data

Cathode

K+ (aq) + e K(s)
E = 2.92 V
2H2O + 2e H2(g) + 2OH(aq)
E = 0.83 V
H2O much easier to reduce than K+

Anode

S2O82(aq) + 2e 2SO42
E = +2.01 V
O2(g) + 4H+(aq) + 4e 2H2OE = +1.23 V
S2O82 easier to reduce than O2
So H2O easier to oxidize than SO42
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 Alternatively calculate Ecell

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1st for K+ and S2O82 reaction

Then for H2O to H2 and O2

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Electrolysis of Aqueous K2SO4

K+ (aq) + e K(s)

2SO42 S2O82(aq) + 2e

Ecell = EK+ ES2O82 = 2.92 V (+2.01 V)

Ecell = 4.93 V

2H2O + 2e H2(g) + 2OH(aq)

2H2O O2(g) + 4H+(aq) + 4e

Ecell = EH2O EO2= 0.83 V (+1.23 V)

E cell = = 2.06 V

So water reaction costs much less energy

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Electrolysis of Aqueous K2SO4

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Net reaction
2 [ 2H2O + 2e H2(g) + 2OH(aq)]
2H2O O2(g) + 4H+(aq) + 4e
6H2O O2(g) + 2H2(g) + 4OH(aq) + 4H+(aq)
2
4H2O
Net = 2H2O H2(g) + O2(g)
So what is the role of K2SO4?

If not present, no electrolysis occurs

K2SO4 is the electrolyte
K+ and SO42 carry charges through solution
Without it no current can pass through solution
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Using Reduction Potentials to Predict

Electrolysis Products

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Suppose we carry out the electrolysis of CuBr2

in water. What products do we expect to
form?
Possible cathode reactions
Cu2+(aq) + 2e Cu(s)
E = +0.34 V
2H2O + 2e H2(g) + 2OH(aq)
E = 0.83 V

ECu2+ is more + so Cu2+ more likely to be

reduced at cathode

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Possible anode reactions

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Using Reduction Potentials to Predict

Electrolysis Products
Oxidation of Br or H2O

E = +1.07 V
Br2(aq) + 2e 2Br(aq)
O2(g) + 4H+(aq) + 4e 2H2OE = +1.23 V

EO2 more + so

O2 easier to reduce and Br easier to oxidize

Expected reaction

Cu2+(aq) + 2e Cu(s)
(cathode)
2Br(aq) Br2(aq) + 2e
(anode)
Cu2+(aq) + 2Br(aq) Cu(s) + Br2(aq) (net)
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Electrolysis of CuBr2 in H2)

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Blue color is Cu2+

Black bits on electrode
are Cu metal

Yellow-orange
color is Br2

Anode

Jespersen/Brady/Hyslop

Cathode

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Learning Check

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What are the expected products for the

electrolysis of an aqueous solution of Na2S with
two platinum electrodes connected to an
external electrical source?
Possible cathode reactions
Reduction of Na+ or H2O
Na+(aq) + e Na(s)
2H2O + 2e H2(g) + 2OH(aq)

E = 2.71 V
E = 0.83 V

H2O easier to reduce

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Possible anode reactions

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Using Reduction Potentials to Predict

Electrolysis Products
Oxidation of S2 or H2O

E = 0.48 V
S(s) + 2e S2(aq)
O2(g) + 4H+(aq) + 4e 2H2OE = +1.23 V

EO2 more + so

O2 easier to reduce and S2 easier to oxidize

Expected reaction

2H2O + 2e H2(g) + 2OH(aq) (cathode)

S2(aq) S(s) + 2e
(anode)
2H2O + S2(aq) H2(g) + S(s) + 2OH(aq) (net)
Jespersen/Brady/Hyslop

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Your Turn!

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Which of the following reactions is most likely

to occur at the anode?
A. Cl2 + 2eB. I2 + 2eC. Br2 + 2eD. Li+ + e-

Cl I Br Li

Eo
Eo
Eo
Eo

=
=
=
=

+1.36V
+0.54
+1.07
-3.05V

Lithium is the easiest to oxidize

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Michael Faraday

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Determined that amount of chemical change

occurring during electrolysis is directly
proportional to amount of electrical charge
passed through cell
Ampere (A)
SI unit of electrical current

Coulomb (C)

SI unit of charge
1 coulomb = 1 ampere 1 second
1 C = 1A s

Faraday (F)

1F = 96,485 C/mol e
Jespersen/Brady/Hyslop

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q = It = nF

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Faradays Equation

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Kinetics of Electrolysis

q = charge (coulombs, C)
I = current (Amperes, A, or C/s)
t = time (s)
n = moles of electrons transferred in process
F= Faradays constant (96,485 C/mol)

Units tell us how these quantities are related

This equation addresses kinetic aspects of
electrochemical cells
coulombs (C ) = current ( A ) time (s )

= mole e (n ) Faradays (F)

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Electroplating

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In Faradays equation, use number of moles

of electrons transferred, ne
Because we cant see electrons, we gauge
this by the amount of metal deposited or lost
Using the half-reaction and stoichiometry, we
can relate the metal to the number of moles
of electrons

molesmetal coefficien te
massmetal

= ne
MMmetal coefficien tm
Jespersen/Brady/Hyslop

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Calculations Related to Electrolysis

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In the electrolysis of a solution containing Ni2+

(aq), metallic Ni (s) deposits on the cathode.
Using a current of 0.150 A for 12.2 min, what
mass of nickel will form?
Faradays
(mole e)

Charge (C)

1mol reactant

n mol e

Mol reactant
(or product)
MM

As=C

Current
& time

Jespersen/Brady/Hyslop

Mass (g) reactant

(or product)

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Solution

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1. Calculate charge passing through cell in

12.2 min

Coulombs (C) = current (A) time (s)

C = 0.150 A 12.2 min 60 s/min
C = 109.8 C

2. Calculate moles of electrons

1mol e
mol e = 109.8C
= 1.138 10 3 mol e
96,485C

Jespersen/Brady/Hyslop

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Solution (cont.)

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3. Calculate amount of Ni in g
1.138 10

mol e

1mol Ni

2mol e

58.69 g Ni

1mol Ni

= 0.033 g Ni
Could also do the entire calculation in one
step:

1mol e
1C

0.150 A 12.2 min 60s / min

96 ,485C A s
1mol Ni 58.69 g Ni

= 0.033g Ni

1mol Ni
2mol e
Jespersen/Brady/Hyslop

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Calculations Related to Electrolysis

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How long must a current of 0.800 A flow to

form 2.50 g of silver metal in an electroplating
experiment? The cathode reaction is Ag+(aq) +
e Ag(s)
1mol reactant

n mol e

Mol reactant
(or product)
MM

Faradays
(mole e)

Charge (C)

As=C

Mass (g) reactant

(or product)

Jespersen/Brady/Hyslop

Chemistry: The Molecular Nature of Matter, 6E

Current
& time
96

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Solution

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1. Calculate mole of electrons required

1mol Ag 1mol e

mol e = 2.500 g Ag

107.9 g Ag 1mol Ag
= 0.02317 mol e
2. Calculate quantity of charge

0.02317mol e

Jespersen/Brady/Hyslop

96,485 C
1mol e

= 2236 C

Chemistry: The Molecular Nature of Matter, 6E

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3. Calculate time required

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Solution

Charge (C ) 2236 C 1A s 1 min

time =
=

= 46.6 min
current (A ) 0.800 A
1C
60s

Could also do the entire calculation in one

step:

1mol Ag
1mol e
2.500 g Ag

107.9 g Ag 1mol Ag
96 , 485 C
1s /C
1 min

= 46 .6 min

0.800 A
60 s
1mol e
Jespersen/Brady/Hyslop

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Your Turn!

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60.3 minutes are required to deposit 1.00 g of

gold on the cathode from a solution of 1.00M
Au2+ solution. How many amps are used in the
electroplating process?
A. 0.250 A
B. 0.125 A
C. 15.0 A
D. 0.0625 A

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Your Turn! - Solution

1mol Au
2mol e
1.00 g Au

196.97 g Au 1mol Au
96, 485 C
1
1min

=i

65.3min 60s
1mol e
i = 0.250 amps

Jespersen/Brady/Hyslop

Chemistry: The Molecular Nature of Matter, 6E

100

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Batteries
Galvanic cells in action
Electroplating
Electrolysis in action

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Applications of Electrochemistry

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Batteries

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Galvanic cells that generate portable electrical

energy
Usually a collection of several linked in series to
get higher voltages
Ex. Car batteries,
usually 6 cells
each capable of
producing 2 V
so Net = 12 V

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Two Major Classes of Batteries

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Primary Cell
Secondary Cell
Non-rechargeable
Rechargeable
Ex. Alkaline dry cell Pb storage Battery

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Zn/MnO2 battery 1.5 V

Uses basic or alkaline electrolyte
Not rechargeable
Compared to dry cell

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Alkaline Battery

Longer shelf life

Delivers higher currents
Less expensive

Reactions:

Anode
Zn (s) + 2OH (aq) ZnO (s) + H2O + 2e
Cathode
2MnO2 (s) + H2O + 2e Mn2O3 (s) + 2OH
Net
Zn (s) + 2MnO2 (s) ZnO (s) + Mn2O3 (s)
Jespersen/Brady/Hyslop

(aq)

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Nickel Cadmium storage cell

Rechargeable

ly

Nicad Battery 1.4 V

High energy density

Release energy quickly
Rapidly recharged

Used in portable power tools,

CD players, electric cars
Reactions:

Anode: Cd(s) + 2OH(aq) Cd(OH)2(s) + 2e

Cathode: NiO2(s) + 2H2O + 2e Ni(OH)2(s) + 2OH
Net:
Cd(s) + NiO2(s) Cd(OH)2(s) + Ni(OH)2(s)

(aq)

Disadvantage = Cd toxicity, so disposal a problem

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Important Properties of Batteries

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Properties that modern battery designers consider for

applications such as
Cell phones, laptops, digital cameras, cordless tools and
pacemakers

Shelf-life

How long do batteries hold their charge when not in use?

Rate of energy output

High currents

Energy density

Ratio of available energy to battery volume

Specific energy

Ratio of available energy to weight

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Ni-MH Batteries

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Nickel Metal Hydride

Rechargeable
1.35V
Some metals and alloys can
absorb H2 gas and effectively
store it, release to redox reaction
Advantages:

50% more power/volume than NiCad

Useful longer

Reactions:

Anode: MH(s) + OH(aq) M(s) + H2O + e

Cathode: NiO(OH)(s) + H2O + e Ni(OH)2(s) + OH(aq)
Net:
MH(s) + NiO(OH)(s) Ni(OH)2(s) + M(s)
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Lithium Ion Batteries

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Rechargeable high specific energy (due to

low mass)
High energy density (due to very () E0red)
Dont actually involve true oxidation and
reduction
Li+ ions can slip between layers of atoms in
solids such as graphite or LiCoO2
Process called intercalation

Li+ ion battery is based on transport of Li+

ions
Jespersen/Brady/Hyslop

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108

Charge cycle

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Lithium Ion

Discharge cycle

Uncharged state

No Li+ ions in graphite

When charged

Li atoms leave LiCoO2

Travel through electrolyte
to graphite (C6)

Charging rxn:
LiCoO2 + C6 Li1xCoO2 + LixC6
When cell discharges to provide electrical power

Li+ ions move back through electrolyte to cobalt oxide which

electrons move through external circuit from C6 electrode to
CoO2 electrode
Let y = amount of Li+ transferred

Discharge: Li1xCoO2 + LixC6 Li1x+yCoO2 + LixyC6

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Fuel Cells

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Galvanic cells in which reactants are continuously

supplied with reagents
Able to operate as long as supply of reactants is
maintained
Attractive power source for long-term generation of
electricity
Major advantages:

Clean burning
No electrode loss
Easily replenished
High operational temperature
Highly efficient
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110

Early design
Cathode: O2(g) + 2H2O + 4e 4OH(aq)
Anode:
H2(g) + 2OH(aq) 2H2O + 2e
Net:
2H2(g) + O2(g) 2H2O
Electrolyte

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Hydrogen-Oxygen Fuel Cell

Hot (~200C) concentrated

KOH in center compartment
In contact with 2 porous
electrodes containing Pt
catalyst that facilitates
reactions

Advances led to lower

operating temperatures and
use of H+ transfer membranes
Jespersen/Brady/Hyslop

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Application of Electrolysis

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Electroplating

Put fork and bar of silver in AgCN bath

Set potential so that cell runs electrolytically
Ag dissolves from Ag electrode
Ag deposited on metal utensil
cathode

Jespersen/Brady/Hyslop

anode

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Bar of Silver

112

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Uses of Electrolysis

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Separating metals in aqueous solution

If you have several metals in solution each with
different reduction potentials, you can often
separate by electrolysis
Start at very low voltage and gradually turn up
Ag+ + e Ag
E = 0.80 V
Cu2+ + 2e Cu E = 0.34 V
Zn2+ + 2e Zn
E = 0.76 V
Order of oxidizing ability: Ag+ > Cu2+ > Zn2+
So Ag plates out first, then Cu, then Zn as
voltage
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Uses of Electrolysis

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Production of Aluminum for bauxite ore

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Treat with NaOH (aq) to separate Al from other metals as

amphoteric oxide AlO2
Acidify with CO2 to form alumina, Al2O3 nH2O
React with cryolyte, AlF63 to form Al2OF6
Electrolysis of Al2OF6 + AlF63 to form Al(s)

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Electrorefining of
Metals

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Application of Electrolysis
Purifying metals such as
Cu

Ultrapure Cu sheets =
cathodes
Lower between slabs of
impure Cu = anodes
Aqueous CuSO4 electrolyte
solution
4 weeks for anodes to
dissolve and pure Cu to
plate onto cathodes
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Refining Copper

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 Steel mostly iron

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Corrosion of Iron

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Fe2O3 coating NOT impervious to further corrosion

Physical stress points leave strain in metal

More susceptible to corrosion

Creates anodic and cathodic regions
Electrochemical process
Anode region: Fe Fe2+ + 2e
Electrons flow through steel to cathode region
Cathode region: O2 + 2H2O + 4e 4OH

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Prevention of Corrosion

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How to conserve our natural resources

Primaryapply coating
Paint
Metal plating

Use chromium and tin to plate steel

Galvanizingusing zinc to coat steel
Zinc more active metal so oxidized first
Zn(s) = sacrificial coating
Alloyingmixture of metals
Stainless steel contains chromium and nickel

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 Cathodic Protection

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Prevention of Corrosion

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Method often used to protect steel in buried fuel

tanks and pipelines
Attach an active metal, such as magnesium, to
pipeline or tank to be protected
Since Mg better reducing agent that Fe
Mg supplies es
Prevents Fe from being
oxidized
Magnesium must be
replaced periodically

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Your Turn!

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Which metal would provide the best cathodic

protection for an iron tank?
A. Cr
B. Pb
C. Cu
D. Sn

Cr is a better reducing agent than the other

choices.
Jespersen/Brady/Hyslop

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Extra Practice

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Learning Check

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Calculate Ecell. Which are spontaneous?

Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s)
[0.799 0.337]V = 0.462V

Cr2O72(aq) + MnO2 (s) MnO4(aq) + Cr(s)

[1.33 1.695]V = 0.365V

Cu2+/Cu
0.337V
1.33V
Ag+/Ag
0.799V
0.126V
Pb2+/Pb
Jespersen/Brady/Hyslop

Cr2O72/Cr

MnO4/MnO2

1.695V

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Learning Check

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Calculate G0 in kJ.. Which reactions are spontaneous

under standard conditions?
Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)
G = 2 mol x 96,485 C/mol x 0.462 J/C
G = 89.1 kJ

Cr2O72(aq) + MnO2 (s) MnO4(aq) + Cr(s)

G = 6 mol x 96,485 C/mol x 0.365 J/C
G = 423 kJ

Cu2+/Cu
Ag+/Ag
Pb2+/Pb

0.337V
0.799V
-0.126V

Jespersen/Brady/Hyslop

Cr2O72/Cr
MnO4/MnO2

1.33V
1.695V

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Learning Check:

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Calculate K for the following cells:

Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s) E= 0.462 V
nF E o
ln K =
RT

2 96 , 485 C/mol 0.462 J/C

= 35.967
8.314 J/ (mol K ) 298.15 K

K = e35.967 = 4.18 1015

Cr2O72(aq) + MnO2 (s) MnO4(aq) + Cr(s)
E= 0.365 V
nFE o 12 96 , 485 C/mol 0.365 J/C
ln K =
=
= 170.486
RT
8.314 J/ (mol K ) 298.15 K

K = e170.486 = 9.10 10-75

Jespersen/Brady/Hyslop

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1.662V

Zn2+/Zn

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Al3+/Al

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Learning Check: Calculate Ecell

0.763V

Al|Al3+ (aq, 0.5M)||Zn2+(aq, 0.2M)|Zn

3Zn2+(aq) + 2Al(s) 3Zn(s) + 2Al3+(aq)
E = [0.763 (1.662)]V = 0.899V

0.02569V [0.5]2
E = 0.899V
ln
6
[0.2]3

E = 0.899 V 0.015 V = 0.884 V

Al|Al3+ (aq, 0.5M)||Zn2+ (aq, 1M)|Zn

0.02569V [0.5]2
E = 0.899V
ln
6
[1.0]3
E = 0.899
V + 0.006
V = 0.905 V
Jespersen/Brady/Hyslop
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Learning Check:

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What current must be supplied to deposit

3.00 g Au from a solution of AuCl3 in 200.0 s?
22.0 A
How much time (in s) does it take to plate
10.2 g of Ag+ using a 0.1mA power source?
9 107 s

Jespersen/Brady/Hyslop

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