A. Redox Reactions
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Transcript of A. Redox Reactions
A. Redox Reactions
electrochemistry is the branch of chemistry that studies
Electrochemistry
electron transfer in chemical reactions
is a oxidation
loss of electrons “LEO”
eg) Mg(s) Mg2+(aq) + 2e
2Cl(aq) Cl2(g) + 2e
is a reduction
gain of electrons “GER”
eg) Fe3+(aq) + 3e Fe(s)
Br2(l) + 2e 2Br(aq)
oxidation and reduction reactions occur together, hence the term redox
the reduction and oxidation reactions are called the
“adding” the half reactions together will give you the that takes place during the redox reaction
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
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-)
reduced oxidizing agent OA
oxidized reducing agentRA
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.
3 Zn(s) + 2 Au(NO3)3(aq) 2 Au(s) + 3 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
B. Spontaneous Redox Reactions chemical reactions which occur on their own,
without the input of , are called
not all reactions are spontaneous
additional energy spontaneou
s
in the table of redox half reactions (see pg 7 in Data Booklet), the is at the top left and the is at the bottom right
strongest oxidizing agent (SOA) strongest reducing agent (SRA)
the rule states that a spontaneous reaction occurs if the agent is above the agent in the table of redox half reactions
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
C. 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 compoundsacids
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 must be higher than SRA to be spontaneous)
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
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+ (aq) OA with H+ (aq), H2O(l)
OA/RASS
[ ]
5
MnO4-(aq) SO4
2-
(aq) OA OA/ RA
spont
D. 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 7 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
Put the oxidizing agents in order from strongest to weakest.
Put the reducing agents in order from strongest to weakest.
Ag+(aq), Cu2+(aq), Pb2+(aq), Zn2+(aq)
Zn(s), Pb(s), Cu(s), Ag(s)
Redox Table
+ e- Ag(s)
Cu2+(aq) + 2e-
Hg2+(aq) + 2e-
Zn2+(aq) + 2e-
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 reactionZn(s) + Cu2+(aq) Zn2+(aq) +
Cu(s) Hg(l) + Ag+(aq) no reaction
Ag+(aq)
Cu(s)Zn(s)
Hg(l)
Label each as OA or Ra
Example 2 (continued):
Put the oxidizing agents in order from weakest to strongest.
Put the reducing agents in order from weakest to strongest.
Zn2+(aq), Cu2+(aq), Ag+(aq), Hg2+(aq)
Hg(l), Ag(s), Cu(s), Zn(s)
Redox Table
Z2 (g) + 2e- + 2e- 2Y-
(aq)
+ 2e- 2W-
(aq)
X2 (g) + 2e-
SOA
SRA
Example 3:Generate a redox table given the following data: 2X-(aq) + Y2(g) spontaneous reaction
2Z-(aq) + Y2 (g) no reaction2Z-(aq) + W2 (g) spontaneous
reaction
Label each as OA or RA
Y2 (g)
2X-(aq)
2Z-(aq)W2 (g)
Example 3 (continued):
Put the oxidizing agents in order from strongest to weakest .
Put the reducing agents in order from strongest to weakest .
W2(g),
X-(aq),
Z2(g), Y2(g), X2(g)
Y-(aq), Z-(aq), W-(aq)
E. Oxidation Numbers (States) an 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 numberappears
molecular compound ion
charges
1. if the number then has occurred
decreases reduction
2. if the number then has occurred
increases oxidation
neutral molecule polyatomic 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 H12O6
C2 H5 O H+1
–2+1+5
+1
+12
figuring out oxidation numbers can help to identify whether a reaction is a or not
for it to be a redox reaction, there has to be an in oxidation number and a in oxidation number seen in the reaction
Ag(s) + NaNO3(aq) Na(s) + AgNO3(aq)
PbSO4(aq) + 2 KI(aq) PbI2(s) + K2SO4(aq)
eg)
redox reaction
0 +1-1 -1+10
Ag increases oxidized
Na decreases reduced…
redox!!!
nothing changes NOT a redox reaction!
bothincrease decrease
+2-2 -2+2-1-1+1 +1
electron transfer occurs in eg)
living systems photosynthesis, cellular respiration
also occurs in
eg)
non-living systems
combustion, bleaching, metallurgy
F. 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
G. Balancing Redox Reactions sometimes most reactants and products are
known but the complete reaction is not given…called a reaction skeleton
1. Half Reaction Method 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 :
+6 2 +32
4 = 1+3+6 8 = 2
(Cr is already balanced)
+3 e– +2 H2O(l)
+4 H+(aq) CrO42-(aq) CrO2
-(aq)
net charge = –1
net charge = –1
(+6) (+3)
Example 2:Balance the following half reaction:
+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
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 reactions H2O(l) H+(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
SO42-
(aq)
Net
2. Oxidation Number Method 1. Assign oxidation numbers.
2. Balance the substances that change in oxidation number.
3. Use a to join the reducing agent with its corresponding product (ignore the H+(aq) and H2O(l)) and a to join the oxidizing agent with its corresponding product. 4. On each line, write the in oxidation number # of atoms.
5. the RA and/or OA to balance the change in oxidation number.
6. the H2O(l) and the H+(aq).
line
line
change
Multiply
Balance
Example 1:Balance the following reaction using the oxidation number method.
__ H+(aq) + __MnO4-(aq) + __ SO3
2-(aq) __ MnO2(aq) + __ SO42-(aq) + __
H2O(l)
2 2 3 32 1+7 +4 +4 +6
= 3 1 atom =
= +2 1 atom =
3 2 =6
+2 3 =+6
Example 2:Balance the following reaction using the oxidation number method.
__ H2O(l) + __N2O4(g) + _ Br(aq) _ NO2 (aq) + _ BrO3
(aq) + __ H+(aq)
3 3 1 12 6+4 -1 +3 +5
= 1 2 atoms =
= +6 1 atom =
2 3 =6
+6
6
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/RA
SRA SOA
m = 100 gM = 52.00 g/mol
n = 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) + 5Fe3+
(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 (Cr2O7
2-(aq)) –
orange
as long as the sample (RA) in the flask is reacting with the the sample will be
permanganate ions (dichromate ions) colourless (green)
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
concentration
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
KMnO4(aq).
17.40
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 mL
3 = 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 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
chemicalelectrica
l
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 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 electrons (leaving a net charge in the electrode)
negative
anions anodeloses positive
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 anode is labelled negative
the move to the since this electrode electrons (leaving a net charge in the electrode)
positive
cations cathode
negativeaccepts
electrons flow from the to the through a connecting wire 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
anode (LEOA)cathode (GERC)
moveporous cup
salt bridge
electrical neutrality
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 cupsalt bridge half
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 cathode, anode, positive terminal, negative terminal, electrolytes, direction of e flow, and directions of cation and anion flow
you also have to show and label the reduction half reaction, oxidation half reaction and net reaction including E values, E net and 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 VAn (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 reactants are
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)
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 electrolytic cell, energy
is used to force a chemical reaction to occur (opposite of a voltaic cell)
commonly used to (eg. gold, silver, bronze, chromium etc),
these reactions have a Enet
nonspontaneous
negative
electroplate metals
(eg. H2, O2, Cl2 etc)
recharge batteries, useful gases
and split compounds into
electrical
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 or 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
Voltaic Cells Electrolytic Cells chemical to electrical
energy electrical to chemical
energy
usually contains porous cup or salt bridge
does not (usually) 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 Al
chlor-alkali plantchlorine 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/RA
SRA 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 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 our rules… chlorine (ions)
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
chloride ions
Example 4An electric current is passed through a solution of sodium chloride. Predict the anode and cathode reactions, overall reaction, and minimum voltage required.
Cathode SOA(Red):
Anode SRA(Ox):
Net:2 H2O(l) + 2 Cl-(aq) H2(g) + Cl2(g) + 2 OH- (aq)
Na+(aq) Cl-
(aq) H2O(l) OA RA OA/RA
SRA SOA
Enet = -2.19 V
2 Cl-(aq) Cl2(g) + 2e- E = -1.36
V
2 H2O(l) +2 e- H2(g) + 2 OH-
(aq)
E = -0.83 V
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 3)
(q) Coulomb (C)
1.60 x 10-19 C
mole 9.65 x 104
C
9.65 x 104
C/mol Faraday constant (F)
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 1An electrochemical cell caused a 0.0720 mol of e- to flow through a wire. Calculate the charge.
ne- = 0.0720 molF = 9.65 x 104
C/mol
ne- = q F0.0720 mol = q
9.65 x 104
C/mol q = 6948 C
= 6.95 103 C
Example 2Determine 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 3If 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. 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)
anodecathode
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)
M. Prevention of Corrosion
other metals (eg. Zn, Cr, Sn) can be onto metals that you don’t want to corrode (eg. steel (Fe))
applying a coating of 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
paint
plated
stronger reducing agent
sacrificial anode
Fe
Zn coating
this method is also calledcathodic protection
has been in use before the science of electrochemistry was developed
Sir Humphrey Davy first used cathodic protection on British naval ships in 1824!
because the attached metal is a than the iron in the steel, the more active metal supplies the and therefore the steel (iron) becomes the cathode and is protected
can be used to protect any metal but steel (iron) is most commonly protected
we use steel (iron) for many structures such as buried fuel tanks, septic tanks, pipelines, hulls of ships, bridge supports etc
to protect these structures by cathodic protection, an is connected by a to the structure
active metal (eg. Mg, Zn, Al)wire
stronger reducing agent
e- for reduction
Fe (tank)
more active metal eg) Mg, Zn, Al
stainless steel contains chromium and nickel, changing steels reduction potential to one characteristic of (basically unreactive)
another protection method is alloying pure metals, which changes their reduction
potential
noble metals like gold
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
electroplating
Electrolytic Cell