a. introduction to electrochemistry
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A. Introduction to Electrochemistry
electrochemistry is the branch of chemistry that studies
electron transfer occurs in
Electrochemistry
eg)
electron transfer in chemical reactions
living systems
photosynthesis, cellular respiration
also occurs in
eg)
non-living systems
combustion, bleaching, metallurgy
is a
is a oxidation
reduction gain of electrons “GER”
loss of electrons “LEO”
eg) Mg(s) Mg2+(aq) + 2e
eg) Fe3+(aq) + 3e Fe(s)
2Cl(aq) Cl2(g) + 2e
Br2(l) + 2e 2Br(aq)
B. Redox Reactions
oxidation and reduction reactions occur together, hence the term
the reduction and oxidation reactions are called the
“adding” the half reactions together will give you the that takes place during the
redox reaction
redox
half reactions
net ionic equation
the e lost in the oxidation half reaction the e gained in the reduction half reaction
must equal
you may have to of the half reactions to balance the e
(ions not changing) are included!
multiply one or both
spectator ions NOT
Example 1Given the following reaction, write the half reactions and the net ionic equation.
Na(s) + LiCl(aq) Li(s) + NaCl(aq) 0 1+ 1– 0 1+ 1–
ox red Cl- is spectator
Ox:
Red:
Net:
Li+(aq) + 1e- Li(s)
Na(s) Na+(aq) + 1e-
Li+(aq) + Na(s) Li(s) + Na+
(aq)
Example 2Given the following reaction, write the half reactions and the net ionic equation.
Zn(s) + Au(NO3)3(aq) Au(s) + Zn(NO3)2(aq) 0 3+ 1– 0 2+ 1–
ox red NO3- is spectator
Ox:
Red:
Net:
Au3+(aq) + 3e- Au(s)
Zn(s) Zn2+(aq) + 2e-
2 Au3+(aq) + 3 Zn(s) 2 Au(s) + 3 Zn2+
(aq)
[ ]
[ ]
3
2
C. Spontaneous Redox Reactions
chemical reactions which occur on their own, without the input of , are called
not all reactions are spontaneous
the substance that is is called the ( ) (it causes the oxidation by taking e-)
the substance that is is called the ( ) (it causes the reduction by
giving up e-)
additional energy spontaneous
reduced oxidizing agent OA
oxidized reducing agent RA
in the table of redox half reactions (see pg 9 in Data Booklet), the is at the top left and the is at the bottom right
the rule states that a spontaneous reaction occurs if the agent is above the
agent in the table of redox half reactions
OA + spontaneous RA reaction
RA + non- spontaneousOA reaction
strongest oxidizing agent (SOA)strongest reducing agent (SRA)
redox spontaneity oxidizing
reducing
Try These:For each of the following combinations of substances, state whether the reaction would be spontaneous or non-spontaneous:
Cr3+(aq) with Ag(s)
I2(s) with K(s)
H2O2(l) with Au3+(aq)
Sn2+(aq) with Cu(s)
Fe2+(aq) with H2O (l)
non-spontaneous
non-spontaneous
non-spontaneous (both ways)
spontaneous
spontaneous
D. Predicting Redox Reactions
we will be predicting the strongest or most dominating reaction that occurs when substances are mixed (other reactions do take place because of atomic collisions!)
Steps:
1. List all species present as reactants
dissociate and
do not dissociate include ions if it is
always include
soluble ionic compounds acids
molecular compounds
H+(aq) acidic
H2O(l)
3. Identify the and using the table.
4. Write out the for the SOA and SRA.
5. Determine the
6. Determine
SOA SRA
half reactions
net ionic reaction
spontaneity
2. Identify each as or (***some can be both so memorize them… , , , )
OA RAFe2+ Cr2+ Sn2+ H2O(l)
Example 1Predict the most likely redox reaction when chromium is placed into aqueous zinc sulphate.
SOA (Red):
SRA (Ox):
Net:
Zn2+(aq) + 2e- Zn(s)
Cr(s) Cr2+(aq) + 2e-
Zn2+(aq) + Cr(s) Zn(s) + Cr2+
(aq)
Cr(s) Zn2+(aq) SO4
2-(aq) H2O(l)
RA OA OA with H2O(l) OA/RAS S
spont
Example 2Predict the most likely redox reaction when silver is placed into aqueous cadmium nitrate.
SOA (Red):
SRA (Ox):
Net:
Cd2+(aq) + 2e- Cd(s)
Ag(s) Ag+(aq) + e-
Cd2+(aq) + 2 Ag(s) Cd(s) + 2 Ag+
(aq)
Ag(s) Cd2+(aq) NO3
-(aq) H2O(l)
RA OA OA with H+ (aq) OA/RAS S
[ ]2
nonspont
Example 3Predict the most likely redox reaction when potassium permanganate is slowly poured into an acidic iron (II) sulphate solution.
SOA (Red):
SRA (Ox):
Net:
MnO4-(aq) + 8H+
(aq) + 5e- Mn2+(aq) + 4H2O(l)
Fe2+(aq) Fe3+
(aq) + e-
MnO4-(aq) +8H+
(aq) + 5Fe2+(aq) Mn2+
(aq) + 4H2O(l) + 5 Fe3+(aq)
K+(aq) H+
(aq) Fe2+(aq) H2O(l)
OA OA with H+ (aqOA with H+ (aq), H2O(l)
OA/RASS
[ ]5
MnO4-(aq) SO4
2-(aq)
OA OA/ RA
spont
E. Generating Redox Tables you can be given data for certain ions and elements
then be asked to generate a redox table like the one on pg 9 of you Data Booklet (a smaller version!)
you may have to generate a table from real or fictional elements and ions
the tables that we use are all written as half reactions
reduction
Example 1Generate a redox table given the following data:
Cu2+(aq) Zn2+
(aq) Pb2+(aq) Ag+
(aq)
Cu(s)
Zn(s)
Pb(s)
Ag(s)
indicates no reaction indicates a reaction
Redox Table
Ag+(aq) + e- Ag(s)
Cu2+(aq) + 2e- Cu(s)
Pb2+(aq) + 2e- Pb(s)
Zn2+(aq) + 2e- Zn(s)
SOA
SRA
Redox Table
Ag+(aq) + e- Ag(s)
Cu2+(aq) + 2e- Cu(s)
Hg2+(aq) + 2e- Hg(l)
Zn2+(aq) + 2e- Zn(s)
SOA
SRA
Example 2:Generate a redox table given the following data: Cu(s) + Ag+
(aq) Cu2+(aq) + Ag(s)
Zn2+(aq) + Ag(s) no reaction
Zn(s) + Cu2+(aq) Zn2+
(aq) + Cu(s)
Hg(l) + Ag+(aq) no reaction
Redox Table
Z2 + 2e- 2Z-(aq)
Y2 + 2e- 2Y-(aq)
W2 + 2e- 2W-(aq)
X2 + 2e- 2X-(aq)
SOA
SRA
Example 3:Generate a redox table given the following data: 2X-
(aq) + Y2 spontaneous reaction
2Z-(aq) + Y2 no reaction
2Z-(aq) + W2 spontaneous reaction
1. pg 572 Lab 13.A, 2. pg 573 # 11-143. pg 5 in workbook
F. Oxidation Numbers (States) an or is the charge an atom
to have when found in a or charged
can be used when you have a where there are no to determine if oxidation or reduction is occurring
how do you use a change in the number?
oxidation number stateappears
molecular compound ion charges
1. if the number then has occurred decreases reduction
2. if the number then has occurred increases oxidation
neutral moleculepolyatomic ion
Rules for Assigning Oxidation Numbers:
1. In a pure element, the oxidation number is .
2. In simple ions, the oxidation number is equal to the .
zero
ion charge
3. In most compounds containing hydrogen, the oxidation number for hydrogen is . (Exception is the metal hydrides eg) LiH where the oxidation number of hydrogen is ).
+1
–1
5. The sum of oxidation numbers of all atoms in a substance must equal the of the substance. ( for compounds and of the polyatomic ion)
eg) sum of MgO = sum of PO43- =
4. In most compounds containing oxygen, the oxidation number for oxygen is . (Exception is the peroxides eg) H2O2, Na2O2 where the oxidation number of oxygen
is ) –1
–2
net chargeZero the charge
0 –3
Example What is the oxidation number (state) for the element identified in each of the following substances:
a) N in N2O N2 O
individual oxidation numbers
sum of oxidation numbers
–2
–2 = 0+2
+1
b) N in NO3-
N O3–
–2
–6 = –1+5
+5
c) C in C2H5OH –2+1
= 0–4
–2
d) C in C6H12O6 –2
–12 = 00
0C6 H12 O6
C2 H5 O H+1
–2 +1+5
+1
+12
G. Balancing Redox Reactions sometimes most reactants and products are known but
the complete reaction is not given…called a reaction
we can use the oxidation numbers to balance the equations either in or conditions:
skeleton
acidic basic
1. In an Acidic Solution
1. Assign
2. Balance the that changes in oxidation number.
3. Add to balance the change in oxidation number.
4. Balance O using
5. Balance H using
6. Check that the half-reaction is balanced with respect to
oxidation numbers (ON).
element
e– total
H2O(l).
H+(aq).
net charge.
(ON subscript coefficient)
Example 1:Balance the following half reaction assuming acidic conditions:
+6 2 +3 2
4 = 1+3+6 8 = 2
(Cr is already balanced)
+3 e– + 2 H2O(l) +4 H+(aq) CrO4
2-(aq) CrO2
-(aq)
net charge = –1 net charge = –1
(+6) (+3)
Example 2:Balance the following half reaction assuming acidic conditions:
+1 2 0
+1 4 = 0
+6 e– + 4 H2O(l) +6 H+(aq) HClO2(aq) Cl2(g)
net charge = 0 net charge = 0
(+6) (0)+3
+3
2
1. In a Basic Solution
1. Assign
2. Balance the that changes in oxidation number.
4. Balance equation charge
5. Balance H using
6. Check that the half-reaction is balanced with respect to
oxidation numbers.
element
OH–(aq).
H2O(l).
oxygen.
3. Add to balance the change in oxidation number.
e– total(ON subscript coefficient)
Example 1:Balance the following half reaction assuming basic conditions:
+4 2 +6 2
8 = 2+6+4 6 = 2
(S is already balanced)
+ 2 e– + H2O(l) +2 OH–(aq) SO3
2-(aq) SO4
2-(aq)
# oxygen = 2 + 3 = 5 # oxygen = 4 + 1 = 5
(+4) (+6)
3. Putting it All Together
now we are going to combine coming up with our own half reactions with figuring out the net redox reaction
Steps
1. Assign
2. Separate the partial net equation into two (omit any , , or ).
3. Balance each half-reaction.
4. of the equations so e lost = e gained.
5. Add the equations to produce a balanced
6. Check to see if all elements and charges are balanced.
oxidation numbers.
half reactionsH2O(l) H+
(aq) OH-(aq)
net ionic equation
Simplify.
Multiply one or both
Example 1:Balance the following using oxidation numbers, assuming acidic conditions:
+42
+48 = 2
+3 e– + 2 H2O(l) +4 H+(aq) CrO4
2-(aq) CrO2
-(aq)
(+6) (+3)
+6
+6CrO4
2-(aq) + SO3
2-(aq) CrO2
-(aq) + SO4
2-(aq)
2
6 = 2
22
84 = 1 = 2+3 +6
+6+3
+ 2 e– + H2O(l) + 2 H+(aq) SO3
2-(aq) SO4
2-(aq)
(+4) (+6)
Red
Ox [ ]
[ ]2
3
8 H+(aq) + 2 CrO4
2-(aq) + 3 H2O(l) + 3 SO3
2-(aq) 2 CrO2
-(aq) + 4 H2O(l) + 3 SO4
2-(aq) + 6 H+
(aq)
2 H+(aq) + 2 CrO4
2-(aq) + 3 SO3
2-(aq) 2 CrO2
-(aq) + H2O(l) + 3 SO4
2-(aq)
Net
Example 2:Balance the following using oxidation numbers, assuming basic conditions:
+32
6 = 2
+4 e– + 6 OH–(aq) +3 H2O(l) TeO3
2-(aq) Te(s)
(+4) (0)
+4
+4TeO3
2-(aq) + Cr3+
(aq) Te(s) + Cr2O72-
(aq) 2
14 = 2+12
+60
+ 6 e– + 14 OH– (aq) + 7 H2O(l) Cr3+ (aq) Cr2O7
2-(aq)
(+6) (+12)
Red
Ox [ ]
[ ]3
2
9 H2O(l) + 3 TeO32-
(aq) + 28 OH-(aq) + 4 Cr3+
(aq) 3 Te(s) + 18 OH-(aq) + 2 Cr2O7
2-(aq) + 14 H2O(l)
3 TeO32-
(aq) + 10 OH-(aq) + 4 Cr3+
(aq) 3 Te(s) + 2 Cr2O72-
(aq) + 5 H2O(l) Net
2
4. Disproportionation
disproportionation occurs when one element is both oxidized and reduced in a reaction
eg)2 H2O2(aq) 2 H2O(l) + O2(g)
-1 -2 0
Cl2(g) + 2 OH-(aq) ClO-
(aq) + Cl-(aq) + H2O(l)
0 +1 -1
H. Redox Stoichiometry
stoichiometry can be used to predict or analyze a quantity of a chemical involved in a chemical reaction
in the past we have used balanced chemical equations to do stoich calculations
1. Calculations
we can now apply these same calculations to balanced redox equations
Example 1What is the mass of zinc is produced when 100 g of chromium is placed into aqueous zinc sulphate.
SOA (Red):
SRA (Ox):
Net:
Zn2+(aq) + 2e- Zn(s)
Cr(s) Cr2+(aq) + 2e-
Cr(s) + Zn2+(aq) Zn(s + Cr2+
(aq) )
Cr(s) Zn2+(aq) SO4
2-(aq) H2O(l)
RA OA OA with H2O(l) OA/RASRA SOA
m = 100 gM = 52.00 g/moln = m M = 100 g 52.00 g/mol = 1.923… mol
Cr(s) + Zn2+(aq) Zn(s) + Cr2+
(aq)
m = ? M = 65.39 g/mol n = 1.923… mol x 1/1 = 1.923… mol
m = nM = (1.923…mol)(65.39 g/mol) = 125.75 g = 126 g
Example 2What volume of 1.50 mol/L potassium permanganate is needed to react with 500 mL of 2.25 mol/L acidic iron (II) sulphate solution?
SOA (Red):
SRA (Ox):
Net:
MnO4-(aq) + 8H+
(aq) + 5e- Mn2+(aq) + 4H2O(l)
Fe2+(aq) Fe3+
(aq) + e-
MnO4-(aq) +8H+
(aq) + 5Fe2+(aq) Mn2+
(aq)+ 4H2O(l) + 5Fe3+(aq)
K+(aq) MnO4
-(aq) H+
(aq) H2O(l) Fe2+(aq) SO4
2-(aq)
OA with H+(aq) OA/RA
SRASOAOA OA OA/RA OA with H+
(aq) OA with H2O(l)
[ ]5
v = ? c = 1.50 mol/L n = 1.125 mol x 1/5 = 0.225 mol v = n Cv = 0.225 mol 1.50 mol/L = 0.150 L
MnO4-(aq) +8H+
(aq) + 5Fe2+(aq) Mn2+
(aq)+ 4H2O(l) + 5 Fe3+(aq)
v = 0.500 L c = 2.25 mol/L n = cv = (2.25 mol/L)(.500L) = 1.125 mol
a titration is a lab process used to determine the of a substance needed to react completely with another substance
this volume can then be used to calculate an unknown using stoichiometry
2. Titrations
one reagent ( - ) is slowly added to another ( - ) until an abrupt change ( ) occurs, usually in colour
volume
concentration
titrant OAsample RAendpoint
in redox titrations, two common oxidizing agents are used because of their and :
1.
2.
colour strengthpermanganate ions (MnO4
-(aq)) – purple
dichromate ions (Cr2O72-
(aq)) – orange
eg)MnO4-(aq) + 8 H+
(aq) + 5 e- Mn2+(aq) + 4 H2O(l)
purple colourlesscolourless
as long as the sample (RA) in the flask is reacting with the the sample will be
permanganate ions (dichromate ions)colourless (orange)
when the reaction is complete, any unreacted permanganate ions will turn the sample (pink) (with dichromate, sample goes from orange to green)
purple
the volume of titrant (OA) needed to reach the endpoint is called the equivalence point
the of the titrant must be accurately known
permanganate solution as it oxidizes distilled water
concentration
decomposes
the concentration of the permanganate solution must be calculated against a (a solution of known concentration) before it can be used in a titration itself
this is done just prior to the titration
primary standard
ExampleFind the concentration of (standardize) the KMnO4(aq) solution
by titrating 10.00 mL of 0.500 mol/L acidified tin (II) chloride with the KMnO4(aq).
Trial 1 2 3 4
Final Volume (mL)
18.40 35.30 17.30 34.10
Initial Volume (mL)
1.00 18.40 0.60 17.30
Volume of (mL)
Endpoint Colour
pink light pink light pink light pink
titrant17.40 16.90 16.70 16.80
endpoint average is calculated by using 3 volumes within 0.20 mL
Endpoint average = 16.90 mL + 16.70 mL + 16.80 mL3
= 16.80 mL
SOA (Red):
SRA (Ox):
Net:
MnO4-(aq) + 8H+
(aq) + 5e- Mn2+(aq) + 4H2O(l)
Sn2+(aq) Sn4+
(aq) + 2 e-
2MnO4-(aq) + 16H+
(aq) + 5Sn2+(aq) 2Mn2+
(aq)+ 8H2O(l) + 5Sn4+(aq)
K+(aq) MnO4
-(aq) H+
(aq) H2O(l) Sn2+(aq) Cl-
(aq)
OA with H+(aq) OA/RA
SRASOAOA OA OA/RA RA
[ ]5
determine net ionic redox equation
Analysis:
[ ] 2
use net redox equation to calculate KMnO4(aq) concentration
2MnO4-(aq) + 16H+
(aq) + 5Sn2+(aq) 2Mn2+
(aq)+ 8H2O(l) +5Sn4+(aq)
v = 0.01680 L C = ? n = 0.00500 mol x 2/5 = 0.00200 mol C = n vC = 0.00200 mol 0.01680 L = 0.119 mol/L
v = 0. 01000 L c = 0.500 mol/L n = cv = (0.500 mol/L)(0.01000 L) = 0.00500 mol
I. Electrochemical Cells1. Voltaic/Galvanic Cells
are devices that convert energy into energy
in redox reactions, e- are transferred from the to the
the transfer of e- can occur through a separating the two substances in containers called
electric cells chemicalelectrical
oxidized substance reduced substance
conducting wire
half cells
a is an arrangement where are joined so that the and can move between them
are made of good conducting materials so e- can flow…can be the of the solution or inert such as
the is a solution that contains ions and will transmit ions (charged particles)
voltaic or galvanic cell two half cells ionse-
electrodes metal
carbon
electrolyte
the electrode where occurs is called the
if the anode is a metal, it mass as the cell operates
the anode is labelled as since it is the electrode where the electrons originate
oxidationanode
loses
the move to the since this electrode
negative
anions anode loses electrons (leaving a net positive charge in the electrode)
the electrode where occurs is called the
if the cathode is a metal, it mass as the cell operates
reductioncathode
gains
the cathode is labelled as since
the move to the since this electrode
positive the anode is labelled negative
cations cathodegains electrons (leaving a net negative charge in the electrode)
electrons flow from the to the
ions must be able to to their attracting electrode (either through the or a ) otherwise a buildup of charge will occur opposing the movement of e-
the flow of ions through the solution and e- through the wire maintains overall electrical neutrality
anode (LEOA)cathode (GERC)
moveporous cup salt bridge
through a connecting wire
2. Standard Reduction Potentials
are the ability of a half cell to
these potentials are measured using a
each half reaction listed in the Data Booklet has an E value measured in assigned to it
reduction potentials attract e-
voltmeter
volts
all values in the table are arbitrarily assigned based on a standard
the half reaction has been set as the standard and has an E value of
hydrogen cell 0.00 V
3. Predicting Voltage of a Voltaic Cell the standard cell potential is determined by
the for the two half reactions
the on the E value for the half reaction must be
if you multiply an equation to balance e-, you multiply the E value (voltage is independent of number of e- transferred)
(Enet) adding
sign oxidationreversed
DO NOT
a E net is a reaction positive spontaneous
a E net is a reaction negative nonspontaneous
E values
Example Calculate the E net for the reaction of Zn(s) with CuSO4(aq).
SOA (Red):
SRA (Ox):
Net:
Cu2+(aq) + 2e- Cu(s)
Zn(s) Zn2+(aq) + 2e-
Zn(s) + Cu2+(aq) Cu(s + Zn2+
(aq)
Zn(s) Cu2+(aq) SO4
2-(aq) H2O(l)
RA OA OA with H2O(l) OA/RAS S
E = +0.34 V
E = +0.76 V
Enet = +1.10 V
4. Shorthand Notation
line separates
double line represents the or and separates the two
ORZn(s) / Zn2+(aq) // Cu2+
(aq) / Cu(s)
Zn(s) / Zn2+(aq) // Cr2O7
2-(aq) , H
+(aq) , Cr3+
(aq)/ C(s)
(/) phases
(//) porous cup salt bridgehalf reactions
comma separates (,) chemical species in the same phase
***anode // cathode
5. Drawing Cells when drawing a cell from the shorthand notation, you have
to be able to label the
you also have to show and label the and including and
anode, cathode, positive terminal, negative terminal, electrolytes, direction of e- flow, directions of cation and anion flow
reduction half reaction, oxidation half reaction net reaction
E values, Enet spontaneity
ExampleDraw and fully label the following electrochemical cell: Al(s)/ Al3+
(aq) // Ni2+(aq) / Ni(s)
Ni(s)Al(s)
Ni2+ (electrolyte)
Al3+ (electrolyte)
e- V
anions
cations
cathode positive terminal
anodenegative terminal
SOA (Red):
SRA (Ox):
Net:
Ni2+(aq) + 2e- Ni(s)
Al(s) Al3+(aq) + 3e-
2 Al(s) + 3 Ni2+(aq) 3 Ni(s + 2 Al3+
(aq) )
Al(s) Al3+(aq) Ni2+
(aq) H2O(l)
RA OA OA OA/RAS S
Ni(s)
RA
[ ]
[ ]
3
2
E = –0.26 V
E = +1.66 V
Enet = +1.40 V
spontaneous: yes
J. Commercial Cells are made by connecting two or more voltaic
cells in
the of the battery is the of the
batteries series (one after the other)
voltageindividual cells
sum
there are many types of batteries:
a) Dry Cell common batteries of clocks, remote
controls, noisy kids toys etc.
Cathode (Red): 2 MnO2(s) + H2O(l) + 2e- Mn2O3(aq) + 2 OH-(aq) E= +0.79 V
Anode (Oxid): Zn(s) Zn2+(aq) + 2e- E = +0.76 V
Net: 2 MnO2(s)+ H2O(l) + Zn(s) Mn2O3(aq) + 2 OH-(aq) + Zn2+
(aq) Enet = +1.55 V
the produced causes irreversible side reactions to occur making recharging impossible
OH-
1.5 V and 9 V
b) Nickel-Cadmium
one type of battery
Cat (Red): 2 NiO(OH)(s) + 2 H2O(l) + 2e- 2 Ni(OH)2(s) + 2 OH-(aq) E= +0.49 V
An (Oxid): Cd(s) + 2 OH-(aq) 2 Cd(OH)2(s) + 2e- E = +0.76 V
Net: 2 NiO(OH)(s)+ 2 H2O(l) + Cd(s) 2 Ni(OH)2(s)+ 2 Cd(OH)2(s) Enet = +1.25 V
rechargeable
c) Lead Storage Battery
where serves as the anode, and serves as the cathode
both electrodes dip into an electrolyte solution of
are connected in series
typical car battery lead lead coated with lead dioxide
sulfuric acid
six cells
Cat (Red): PbO2(s)+ HSO4-(aq) + 3 H+
(aq) + 2e- PbSO4(s)+ 2 H2O(l) E= +1.68 V
An (Oxid): Pb(s) + HSO4-(aq) PbSO4(s) + H+
(aq) + 2e- E = +0.36 V
Net: Pb(s)+ PbO2(s) + 2 H+(aq) + 2 HSO4
-(aq) 2 PbSO4(s)+ 2 H2O(l) Enet = +2.04 V
d) Fuel Cells
cells where
the energy from this reaction can be used to
one type is the
is pumped in at the while is pumped in at the (which both have a lot of surface area)
reactants are continuously supplied
run machines
hydrogen-oxygen fuel cell
hydrogen gas anodeoxygen gas cathode
pressure is used to push the H2 through a platinum
catalyst which splits the H2 into 2H+ and 2e-
Cathode (Red): O2(g) + 4 H+(aq) + 4e- 2 H2O(l) E= V
Anode (Oxid): 2 H2(g) 4 H+(aq) + 4e- E = V
Net: O2(g) + 2 H2(g) 2 H2O(l) Enet = +1.23 V
the 2e- move through an external circuit towards the cathode generating electrical energy
the O2 is also pushed through the platinum catalyst forming
two oxygen atoms
the H+ ions and oxygen atoms combine to form water
+1.23
0.00
Hydrogen-oxygen Fuel Cell
need a source of hydrogen…reformers are used to convert CH4 or CH3OH into and
unfortunately, is a
about 24-32% efficient where gas-powered car is about 20% efficient
H2 CO2
CO2 greenhouse gas
K. Electrolytic Cells1. The Basics
in an electrical energy is used to force a chemical reaction to occur (opposite
of a voltaic cell)
commonly used to
these reactions have a Enet
electrolytic cell,nonspontaneous
negative
electroplate metals (eg. gold, silver, bronze, chromium etc), recharge batteries,
useful gases (eg. H2, O2, Cl2 etc) and split compounds into
the electrolytic cell is hooked up to a instead of load or external circuit) so the flow of e-
is
the of the electrolytic cell is connected to the of the battery and therefore is
battery (power supply
“pushed” by an outside force
cathodeanode negative
the of the electrolytic cell is connected to the of the battery and therefore is
anodecathode positive
all other conventions are the as voltaic cells same
Voltaic Cells Electrolytic Cells chemical to electrical energy electrical to chemical energy
usually contains porous cup or salt bridge
does not contain a porous cup or salt bridge
e– flow from anode to cathode oxidation at anode reduction at cathode cations migrate to cathode anions migrate to anode
Enet is positive (spont) Enet is negative (nonspont) has a voltmeter or external load has a power supply
cathode + anode – cathode – anode +
some processes are used in industry to produce gases, for example:
1. the for producing …aluminum oxide is electrolyzed using carbon electrodes …liquid aluminum is collected
2. a for producing …a saturated sodium chloride solution is electrolyzed …chlorine gas is formed and collected at the anodes
http://images.google.ca/imgres?imgurl=http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter18/Text_Images/FG18_18.JPG&imgrefurl=http://wps.prenhall.com/wps/media/objects/602/616516/Chapter_18.html&h=756&w=1600&sz=183&tbnid=4cFJrFlK4noQXM:&tbnh=70&tbnw=150&hl=en&start=11&prev=/images%3Fq%3Dhall-heroult%2Bcell%26svnum%3D10%26hl%3Den%26lr%3D
http://www.cheresources.com/chloralk.shtml
Hall-Heroult cell aluminum
chlor-alkali plant chlorine gas
Example 1An electric current is passed through a solution of nickel (II) nitrate using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.
Cathode SOA(Red):
Anode SRA(Ox):
Net:
Ni2+(aq) + 2e- Ni(s)
2 H2O(l) O2(g) + 4 H+(aq) + 4e-
2 Ni2+(aq) + 2 H2O(l) 2 Ni(s)+ O2(g) + 4 H+
(aq)
Ni2+(aq) NO3
-(aq) H2O(l)
OA OA with H+(aq) OA/RA
SRASOA
E = -0.26 V
E = -1.23 V
Enet = -1.49 V
[ ]2
Example 2An electric current is passed through a solution of potassium iodide using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.
Cathode SOA(Red):
Anode SRA(Ox):
Net:
2 H2O(l) + 2 e- H2(g) + 2 OH- (aq)
2 I-(aq) I2(s) + 2e-
2 H2O(l) + 2 I-(aq) H2(g) + 2 OH-
(aq) + I2(s)
K+(aq) I-
(aq) H2O(l)
OA RA OA/RASRA SOA
E = -0.83 V
E = -0.54 V
Enet = -1.37 V
Example 3An electric current is passed through a solution of copper(II) sulphate using a carbon electrode and a metal electrode. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.
Cathode SOA(Red):
Anode SRA(Ox):
Net:
Cu2+(aq) + 2 e- Cu(s)
2 H2O(l) + 2 Cu2+(aq) 2 Cu(s) + O2(g) + 4 H+
(aq)
Cu2+(aq) SO4
2-(aq) H2O(l)
OA RA OA/RASRASOA
E = +0.34 V
Enet = -0.89 V
2 H2O(l) O2(g) + 4 H+(aq) + 4e- E = -1.23 V
[ ]2
*** Note: is an exception to the rule… chlorine
when water and chlorine are competing as reducing agents, water is the stronger RA but is chosen because the transfer of e- from H2O to O2 is
more difficult …called overvoltage
chlorine
2. Quantitative Study of Electrolysis
quantitative analysis (stoich) provides information on necessary quantities, current and/or time for electrolytic reactions
one e- carries of charge
the unit for charge is the
this means that one of e- carry of charge
is called the (see Data Booklet pg 4)
(q) Coulomb (C)
1.60 x 10-19 C
mole 9.65 x 104 C
9.65 x 104 C/mol Faraday constant
ne- = q
F
where: ne- = number of moles of electrons (mol)q = charge in Coulombs (C)F = Faraday constant
= 9.65 x 104 C/mol
q = It
I = current in C/s or Amperes (A)t = time in seconds (s)
ne- = It
F
the above equations can be combined into one equation:
we can use these equations in stoichiometric calculations for current, time, mass, moles of a substance and moles e-
Example 1
Determine the number of moles of electrons supplied by a dry cell supplying a current of 0.100 A to a radio for 50.0 minutes.
I = 0.100 A (C/s)t = 50.0 min 60 s/min = 3000 sF = 9.65 x 104
C/mol
ne- = It F = (0.100 C/s)(3000 s) 9.65 x 104
C/mol
= 0.00311 mol
Example 2
An electrochemical cell caused a 0.0720 mol of e- to flow through a wire during a 3.00 hour period. Calculate the average current.
ne- = 0.0720 molt = 3.00 h 3600 s/h = 10 800 sF = 9.65 x 104
C/mol
ne- = It F0.0720 mol = I(10 800 s)
9.65 x 104 C/mol
I = 0.643 A
Example 3
If a 20.0 A current flows through an electrolytic cell containing molten aluminum oxide for 1.00 hours, what mass of Al(l) will be deposited at the cathode?
ne- = It F
= (20.0 A)(1.00 h 3600 s/h) 9.65 x 104
C/mol = 0.746…mol
n = 0.746…mol 1/3 = 0.248…molM = 26.98 g/molm = nM = (0.248…mol)(26.98g/mol) = 6.71 g
Al3+(l) + 3 e- Al(l)
L. Society and Technological Connections1. Rust and Corrosion
the metal is oxidized causing the loss of
corrosion can be viewed as the process of returning metals to their natural state (ore)
structural integrity
commonly, the oxide coating will scale off leaving new metal exposed an susceptible to corrosion
salt will by acting as a speed up the oxidation salt bridge
most metals develop a thin oxide coating which then protects their internal atoms against further oxidation
Fe(s)
H2O droplet
O2(g)
anode
cathode
Fe(OH)2(s)
rust
Cathode SOA(Red):
Anode SRA(Ox):
Net:
O2(g) + 2H2O(l) + 4e- 4 OH-(aq)
O2(g) + 2H2O(l) + 2Fe(s) 4 OH-(aq) + 2 Fe2+
(aq)
Fe(s) Fe2+(aq) + 2e- [ ]2
O2(g) + 2H2O(l) + 2Fe(s) 2 Fe(OH)2(s)
2. Prevention of Corrosion
other metals (eg. Zn, Cr, Sn) can be onto metals that you don’t want to corrode (eg. steel (Fe))
to protect metal from oxidation
this coating is of a metal that is a than the metal that is to be protected…the
coating metal will react instead and is called the
applying a coat of paint
plated
stronger reducing agent
sacrificial anode
3. Cathodic Protection
an is connected by a to the pipeline
used to protect steel (iron) in buried fuel tanks and pipelines
because Mg is a than the iron in the steel, the Mg supplies the and the iron becomes the cathode and is protected
active metal (eg. Mg) wire
stronger reducing agente- for reduction
is attached to the hull of ships to perform the same function titanium
4. Alloying
stainless steel contains chromium and nickel, changing steels reduction potential to one characteristic of
(basically unreactive)
alloying pure metals changes their reduction potential
noble metals like gold
5. Electroplating plating is the process of
by metal ions in solution
an object can be plated by making it the in an containing ions of the plating metal
depositing the neutral metal on the cathode reducing
cathodeelectrolytic cell
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