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2014
Chemistry Electrolytic Cells and Electroplating
Adv
ance
d Pl
acem
ent Presenter Copy
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H Li Na K Rb
Cs Fr
He
Ne Ar
Kr
Xe
Rn
Ce
Th
Pr
Pa
Nd U
Pm Np
Sm Pu
Eu
Am
Gd
Cm
Tb Bk
Dy Cf
Ho
Es
Er
Fm
Tm Md
Yb
No
Lu Lr
1 3 11 19 37 55 87
2 10 18 36 54 86
Be
Mg
Ca Sr
Ba
Ra
B Al
Ga In Tl
C Si
Ge
Sn
Pb
N P As
Sb Bi
O S Se Te Po
F Cl
Br I At
Sc Y La Ac
Ti Zr Hf
Rf
V Nb Ta Db
Cr
Mo W Sg
Mn Tc Re
Bh
Fe Ru
Os
Hs
Co
Rh Ir Mt
Ni
Pd Pt §
Cu
Ag
Au §
Zn Cd
Hg §
1.00
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4 12 20 38 56 88
5 13 31 49 81
6 14 32 50 82
7 15 33 51 83
8 16 34 52 84
9 17 35 53 85
21 39 57 89
58 90
*Lan
than
ide
Ser
ies:
†Act
inid
e S
erie
s:
* †
59 91
60 92
61 93
62 94
63 95
64 96
65 97
66 98
67 99
68 100
69 101
70 102
71 103
22 40 72 104
23 41 73 105
24 42 74 106
25 43 75 107
26 44 76 108
27 45 77 109
28 46 78 110
29 47 79 111
30 48 80 112
Perio
dic
Tabl
e of
the
Elem
ents
§Not
yet
nam
ed
ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS
Throughout the test the following symbols have the definitions specified unless otherwise noted.
L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)
ATOMIC STRUCTURE
E = hν
c = λν
E = energy ν = frequency
λ = wavelength
Planck’s constant, h = 6.626 × 10−34 J s Speed of light, c = 2.998 × 108 m s−1
Avogadro’s number = 6.022 × 1023 mol−1 Electron charge, e = −1.602 × 10−19 coulomb
EQUILIBRIUM
Kc = [C] [D]
[A] [B]
c d
a b, where a A + b B c C + d D
Kp = C
A B
( ) ( )
( ) ( )
c dD
a b
P P
P P
Ka = [H ][A ][HA]
+ -
Kb = [OH ][HB ][B]
- +
Kw = [H+][OH−] = 1.0 × 10−14 at 25°C
= Ka × Kb
pH = − log[H+] , pOH = − log[OH−]
14 = pH + pOH
pH = pKa + log [A ][HA]
-
pKa = − logKa , pKb = − logKb
Equilibrium Constants
Kc (molar concentrations)
Kp (gas pressures)
Ka (weak acid)
Kb (weak base)
Kw (water)
KINETICS
ln[A] t − ln[A]0 = − kt
[ ] [ ]0A A1 1
t
- = kt
t½ = 0.693k
k = rate constant
t = time t½ = half-life
ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS
Throughout the test the following symbols have the definitions specified unless otherwise noted.
L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)
ATOMIC STRUCTURE
E = hν
c = λν
E = energy ν = frequency
λ = wavelength
Planck’s constant, h = 6.626 × 10−34 J s Speed of light, c = 2.998 × 108 m s−1
Avogadro’s number = 6.022 × 1023 mol−1 Electron charge, e = −1.602 × 10−19 coulomb
EQUILIBRIUM
Kc = [C] [D]
[A] [B]
c d
a b, where a A + b B c C + d D
Kp = C
A B
( ) ( )
( ) ( )
c dD
a b
P P
P P
Ka = [H ][A ][HA]
+ -
Kb = [OH ][HB ][B]
- +
Kw = [H+][OH−] = 1.0 × 10−14 at 25°C
= Ka × Kb
pH = − log[H+] , pOH = − log[OH−]
14 = pH + pOH
pH = pKa + log [A ][HA]
-
pKa = − logKa , pKb = − logKb
Equilibrium Constants
Kc (molar concentrations)
Kp (gas pressures)
Ka (weak acid)
Kb (weak base)
Kw (water)
KINETICS
ln[A] t − ln[A]0 = − kt
[ ] [ ]0A A1 1
t
- = kt
t½ = 0.693k
k = rate constant
t = time t½ = half-life
GASES, LIQUIDS, AND SOLUTIONS
PV = nRT
PA = Ptotal × XA, where XA = moles A
total moles
Ptotal = PA + PB + PC + . . .
n = mM
K = °C + 273
D = m
V
KE per molecule = 12
mv2
Molarity, M = moles of solute per liter of solution
A = abc
1 1
pressurevolumetemperaturenumber of molesmassmolar massdensitykinetic energyvelocityabsorbancemolarabsorptivitypath lengthconcentration
Gas constant, 8.314 J mol K
0.08206
PVTnm
DKE
Aabc
R
Ã
- -
=============
==
M
1 1
1 1
L atm mol K
62.36 L torr mol K 1 atm 760 mm Hg
760 torr
STP 0.00 C and 1.000 atm
- -
- -====
THERMOCHEMISTRY/ ELECTROCHEMISTRY
products reactants
products reactants
products reactants
ln
ff
ff
q mc T
S S S
H H H
G G G
G H T S
RT K
n F E
qI
t
D
D
D D D
D D D
D D D
=
= -Â Â
= -Â Â
= -Â Â
= -
= -
= -
�
heatmassspecific heat capacitytemperature
standard entropy
standard enthalpy
standard free energynumber of moles
standard reduction potentialcurrent (amperes)charge (coulombs)t
qmcT
S
H
Gn
EIqt
============ ime (seconds)
Faraday’s constant , 96,485 coulombs per moleof electrons1 joule
1volt1coulomb
F =
=
Electrochemistry Electrolytic Cells and Electroplating
What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with electrochemical processes:
E°cell; cell potential; reduction or oxidation; anode; cathode; salt bridge; electron flow; voltage; electromotive force; galvanic/voltaic; electrode; battery; current; amps; time; grams (mass); plate/deposit; electroplating; identity
of metal; coulombs of charge
ELECTROCHEMICAL CELLS
ELECTROCHEMICAL TERMS
Electrochemistry the study of the interchange of chemical and electrical energy
OIL RIG Oxidation Is Loss, Reduction Is Gain (of electrons)
LEO the lion says GER Lose Electrons in Oxidation; Gain Electrons in Reduction Oxidation the loss of electrons, increase in charge
Reduction the gain of electrons, reduction of charge
Oxidation number the assigned charge on an atom
Electrochemical Cells: A Comparison
Galvanic (voltaic) cells
spontaneous oxidation-reduction reaction
Is separated into 2 half-cells
Electrodes made from metals (inert Pt or C if ion to ion or gas)
Battery − its cell potential drives the reaction and thus the e−
Electrolytic cells non-spontaneous oxidation-reduction reaction
Usually occurs in a single container
Usually inert electrodes
Battery charger − requires an external energy source to drive the reaction and e−
Copyright © 2014 National Math + Science Initiative, Dallas, Texas. All Rights Reserved. Visit us online at www.nms.org Page 1
Electrochemistry – Electrolytic Cells and Electroplating
Electrolysis and Non-spontaneous Cells
The Electrolytic Cell: What is what and what to know
An electrolytic cell is an electrochemical cell in which the reaction occurs only after adding electrical energy; thus it is not thermodynamically favorable – it requires a driving force
Electrolytic reactions often occur in a single container; not 2 separate ½ cells like Voltaic cells
The calculated E° for an electrolytic cell is negative (i.e. that is the amount of potential required to drive the reaction).
This driving force causes the electrons to travel from the positive electrode to the negative electrode, the exact opposite of what you would expect. Remember: EPA – Electrolytic Positive Anode
Anode − the electrode where oxidation occurs. Cathode − the electrode where reduction occurs.
Electron Flow − From Anode To CAThode
§ Electrolytic reactions typically occur aqueous solutions (or in molten liquids)
§ If there is no water present, you have a pure molten ionic compound thus § the cation will be reduced § the anion will be oxidized
§ If water is present, you have an aqueous solution of the ionic compound, thus § You must decide which species is being oxidized and reduced; the ions or the water!
§ No alkali or alkaline earth metal can be reduced in an aqueous solution - water is more easily reduced. § Polyatomic ions are typically NOT oxidized in an aqueous solution - water is more easily oxidized.
§ When it comes to water, be familiar with the following
§ REDUCTION OF WATER: 2 H2O() + 2 e− → H2(g) + 2 OH− E° = −0.83 V
§ OXIDATION OF WATER:
2 H2O()→ O2(g) + 4 H+ + 4 e− E° = −1.23 V
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Electrochemistry – Electrolytic Cells and Electroplating
Electrolysis: What to Do and How to Do it! An electric current is applied to a 1.0 M KI solution. The possible reduction half reactions are listed in the table.
Reduction Half Reactions E° (V) 2 H2O() + 2 e− → H2(g) + 2 OH−
−0.83 V
O2(g) + 4 H+ + 4 e− → 2 H2O() +1.23 V
I2(g) + 2 e− → 2 I− +0.53 V
K+ + e− → K(s) −2.92 V
(A) Write the balanced half-reaction for the reaction that takes place at the anode. (B) Write the balanced overall reaction for the reaction that takes place.
(C) Which reaction takes place at the cathode?
First realize the solution contains 3 important species: K+, I−, and H2O. From those 3 species, you must decide which is being oxidized and which is being reduced. Where to start – Water can be both oxidized and reduced in an aqueous solution. The I− cannot be further reduced (it has a 1– charge) The K+ cannot be further oxidixed (it has a 1+ charge) For OXIDATION: either I− or H2O will be OXIDIZED… Write the oxidation half-‐reactions – Remember, the more positive oxidation potential will be oxidized. 2 H2O → O2 + 4 H+ + 4 e− E° = −1.23 V 2 I− → I2 + 2 e− E° = −0.53 V (will be oxidized) Must decide which is being reduced, K+ or H2O… The more positive reduction potential gets to be reduced; plus a great rule of thumb to remember is NO alkali metal will be reduced in aqueous solution − water will be. 2 H2O+ 2 e− → H2 + 2 OH− E° = −0.83 V (will be reduced) K+ + e− → K E° = −2.92 V Put BOTH half-reactions TOGETHER: 2 I− → I2 + 2 e− E° = −0.53 V (oxidation) (occurs at the anode) 2 H2O+ 2 e− → H2 + 2 OH− E° = −0.83 V (reduction) (occurs at the cathode) Balanced Overall Reaction: 2 I− + 2 H2O → H2 + 2 OH− + I2 E° = −1.36 V
(D) What would you observe occurring at the cathode when current is applied to this solution?
Gas bubbles form from the production of hydrogen gas.
(E) Calculate the ΔG° for this reaction.
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Electrochemistry – Electrolytic Cells and Electroplating
J(2)(96500)( 1.36) 262000
molkJ262
mol
G n E
G
G
Δ ° = − ℑ °
Δ ° = − − =
Δ ° =
(F) Is this reaction thermodynamically favorable or non-thermodynamically favorable? Justify your answer.
Since ΔG° is positive (and E°cell is negative) the reaction is non-thermodynamically favorable.
The Electrical Energy of Electrolysis When running an electrical current through a solution to cause electrolysis, you can measure that current and determine how much of what substance is going to be produced – think stoichiometry too! They will ask you “how many grams of a metal can be plated” or “how long it will take to plate a given mass” § The amount of electrical charge flowing through an electrochemical cell is measured in coulombs (c). The rate at
which the charge flows (per second) is called the current and is measured in amperes, or amps symbolized by I for inductance. By definition,
amp(I ) = Coulomb(c)
sec(t)
§ A Faraday is the amount of charge associated with 1 mol of electrons. A Faraday has the value of 96,500 C.
Volt 1
Joule (J )Coulomb(c)
Amp 1
Coulomb(c)sec(t)
Faraday 96,500
Coulomb(c)mol of e−
Balanced REDOX Equation
mol of e−
mol of substance
# of Coulombs = IT or c = IT
time (t) MUST BE IN SECONDS!!!! § Using these units you can measure how much substance is going to be produced, how many coulombs or how many
amps were required, how long it takes, etc.…
§ A shortcut is the formula… (Molar Mass) = grams platedItnℑ
Calculate the mass of copper metal produced during the passage of 2.50 amps of current through a solution of copper(II) sulfate for 50.0 minutes.
(Molar Mass) = grams plated
(2.50)(50 60) (63.55) = grams plated = 2.47 Cu(2)(96500)
Itn
g
ℑ×
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Electrochemistry – Electrolytic Cells and Electroplating
Electrochemistry Cheat Sheet
E°cell; reduction and oxidizing agent; cell potential; reduction or oxidation; anode; cathode; salt bridge; electron flow; voltage; electromotive force; galvanic/voltaic; electrode; battery; current; amps; time; grams (mass); plate/deposit; electroplating; identity of metal; coulombs of charge
Electrolytic Cell Relationships
The Mnemonics apply to Electrolytic cells too: ANOX; REDCAT; FATCAT… Electrolytic Positive Anode
“the more positive reduction potential gets to be reduced” 96,500 Coulombs = 1 mole of electrons
“the more positive oxidation potential gets to be oxidized”
# electrons in balanced equations = # moles of electrons transferred
Reduction of H2O 2 H2O() + 2 e− → H2(g) + 2 OH− E° = −0.83 V
All time measurements must be in sec for electroplating/electrolysis problems
Oxidation of H2O 2 H2O → O2 + 4 H+ + 4 e− E° = −1.23 V (Molar Mass) = grams platedIt
nℑ
No alkali or alkaline earth metal can be reduced in an aqueous solution - water is more easily reduced.
Polyatomic ions are typically NOT oxidized in an aqueous solution - water is more easily oxidized.
( )( )sec( )
Coulomb qamp It
=
E°cell = −; not thermodynamically favored; ΔG = (+); K<1
Volt = 1Joule (J )
Coulomb(c)
ΔG° = −nℑΕ° Amp = Coulomb(c)
sec(t)
ΔG° = −RT lnK Faraday (ℑ) = 96,500
Coulomb(c)mol of e−
Inert electrodes are typically used in gas and ion to ion galvanic cells and in electrolytic cells
Connections
Thermo and Equilibrium Stoichiometry
Potential Pitfalls
Watch signs on voltages!! BE SURE units cancel out in your calculations.
Balancing overall reactions − make sure # of electrons is the same in both half reactions. Units on E°are volts
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Electrochemistry – Electrolytic Cells and Electroplating
NMSI SUPER PROBLEM An electric current is applied to two separate solutions for 30 minutes, under the same conditions using inert electrodes. Observations are noted in the table below.
Solution A – 1.0 M K2SO4 Solution B – 1.0 M CuSO4
Anode: gas bubbles Anode: gas bubbles
Cathode: gas bubbles Cathode: dark flakes formed on the electrode
In both reactions, water is oxidized according to the following oxidation half-reaction.
2 H2O ()→ O2(g) + 4 H+ + 4 e− E° = −1.23 V
(a) Write the balanced equation for the half-reaction that occurs at the cathode in
(i) Solution A 2 H2O + 2 e− → H2 + 2 OH− E° = −0.83 V Either K+ or H2O will be reduced. K+ + e− → K E° = −2.92 V 2 H2O() + 2 e− → H2(g) + 2 OH− E° = −0.83 V Since H2O has a more positive reduction potential it is more easily reduced than H2O. H2O is always reduced over an alkali metal
1 point for a correctly indicating the reduction of H2O over K+ 1 point for the correct reduction half-reaction.
(ii) Solution B
Cu2+ + 2 e− → Cu E° = +0.34 V Either Cu2+ or H2O will be reduced. Cu2+ + 2 e− → Cu E° = +0.34 V 2 H2O() + 2 e− → H2(g) + 2 OH− E° = −0.83 V Since Cu2+ has a more positive reduction potential it is more easily reduced than H2O.
1 point for a correctly balanced reaction showing the reduction of Cu2+ over H2O 1 point for the correct reduction half-reaction.
(b) For Solution A, is the reaction thermodynamically favorable or not thermodynamically favorable? Justify your
answer.
Non-thermodynamically favorable, since the reaction requires electric current the E° is negative (−) for this reaction, therefore ΔG is positive (+).
1 point for a correctly identifying Cu metal
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Electrochemistry – Electrolytic Cells and Electroplating
(c) In the electrolysis of the K2SO4 solution, identify the gas produced and describe a test that can be used to identify
the gas at the
(i) anode
At the anode, the oxidation of H2O releases O2 gas An identifying test for O2 gas is trapping the gas in a test tube and inserting a glowing wooden splint. In the presence of O2 gas a glowing wooden splint will re-ignite.
1 point for a correctly identifying O2 gas 1 point for a correctly identifying the O2 test.
(ii) cathode
At the cathode, the reduction of H2O releases H2 gas An identifying test for H2 gas is trapping the gas in a test tube and inserting a burning wooden splint. In the presence of H2 gas a popping sound will be observed.
1 point for a correctly identifying H2 gas 1 point for a correctly identifying the H2 test.
(d) Describe in the box below, what observations, if any, would be noted if a couple of drops of phenolphthalein
indicator were added around the cathode of both solutions. Phenolphthalein indicator is colorless in acidic solutions and turns pink in basic solutions. Justify your observations.
Solution A – 1.0 M K2SO4 Solution B – 1.0 M CuSO4
In solution A, since H2O is reduced the OH− ion is produced. This will create an alkaline solution that will turn pink in the presence of phenolphthalein. In solution B, Cu2+ is reduced to Cu. The phenolphthalein will have no observed effect on this solution since it is not becoming significantly more basic.
1 point for a correctly indicating the solution will turn pink with justification. 1 point for a indicating the solution will not change with justification.
(e) The dark flakes formed on the electrode in the electrolysis of Solution B were collected and dried. The mass of
these flakes was determined to be1.019 grams. (i) Identify the flakes.
The flakes are Cu metal, as Cu2+ is reduced to Cu. 1 point for a correctly identifying Cu
metal
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Electrochemistry – Electrolytic Cells and Electroplating
(ii) Calculate the amount of current that was passed through Solution B.
Itnℑ
(Molar Mass) = g
I = gnℑ(Molar Mass)t
= (1.019)(2)(96500)(63.55)(30)(60)
= 1.72amps
1 point for the correct number of Coulombs 1 point for the correct number of amps of current (both points are earned using the equation shown)
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