Transcript Balancing Redox Using the Ion

Oxidation-Reduction
Reactions
Oxidation-Reduction Reactions
“LEO says GER”
L
E lose electrons, oxidize
O
says
G
E gain electrons, reduce
R
What is a Redox Reaction?
• Redox is a term for reactions in which the
acceptance of an electron (reduction) by a
material is matched with the donation of an
electron (oxidation).
• Possible Redox Reactions:
– Synthesis Reactions
– Decomposition Reactions
– Single Replacement Reactions
How Do We Know if a
Reaction is a Redox Reaction?
• Check oxidation number of each element in
a reaction to determine if there is both
oxidation and reduction.
• Oxidation number: the charge, or apparent
charge, an atom has when bonded to another
atom.
• There are rules for determining oxidation
numbers!
Redox Example
• Synthesis reaction:
2Ca (s) + O2 (g)  2CaO (s)
•
•
•
•
Ca loses electrons; it is oxidized.
O gains electrons; it is reduced.
Ca is reducer or reducing agent.
O is oxidizer or oxidizing agent.
Another Redox Example
• Single Replacement reaction:
Mg(s) + 2HCl(ag)  MgCl2(aq) + H2(g)
• Mg loses electrons = oxidized
• H gains electron = reduced
Oxidation Number Rules
• Any atom of an element has an oxidation number
of zero.
• Any monoatomic metallic ion has an oxidation
number equal to its charge. (Transition metals have
multiple oxidation numbers.)
• Oxygen has an oxidation number of -2.
– Oxygen exceptions:
• With fluorine, oxygen is +2
• In peroxide, oxygen is -1
Oxidation Number Rules, Cont.
• Hydrogen has an oxidation number of +1.
• Exception: In a metal hydride, hydrogen is -1.
• Nonmetals may have multiple oxidation
numbers, both positive and negative.
• The sum of the oxidation numbers in a
compound equals zero.
Assigning Oxidation Numbers
• Assign oxidation numbers to all the atoms in
each of the following:
HNO3
CuCl2
O2
H2O2
MgSO4
CO2
Na2C2O4
Cr2O3
PbSO4
PbO2
(NH4)2Ce(SO4)3
Ag
Balancing Redox Reactions
Using the Ion Electron
Method
Acidic or Neutral Equations
• Write an ionic equation in which all
soluble species are dissociated.
HNO3(aq) + H2S(g) 
NO (g) + S8(s) + H2O(l)
Becomes
H+(aq) + NO3-(aq) + H2S(g) 
NO (g) + S8(s) + H2O(l)
Acidic or Neutral Equations
• Write a Skeletal Equation (one that
includes only substances involved in
redox reaction).
NO3- + H2S  NO + S8
Acidic or Neutral Equations
Write equations for the half reactions.
Balance the mass in each except for
oxygen and hydrogen.
Red: NO3
NO
Ox: 8H2S

S8
Acidic or Neutral Equations
• Add oxygen in the form of water to
the oxygen-deficient side.
Red: NO3-  2H2O + NO
Ox: 8H2S

S8
Acidic or Neutral Equations
• Add hydrogen in the form of H+ to
the hydrogen-deficient side
Red: NO3- + 4H+  2H2O + NO
Ox: 8H2S
 16H+ + S8
Acidic or Neutral Equations
Add electrons as e- to the more positive side.
Red: 3e- +NO3- + 4H+
 2H2O +
Ox: 8H2S  16H+ + S8 + 16e-
NO
Acidic or Neutral equations
Find L.C.M. of electrons and multiply as
needed.
Red: (3e- +NO3- + 4H+  2H2O + NO)16
Ox: 3(8H2S  16H+ + S8 + 16e- )
Acidic or Neutral equations
Add the half cells. Cancel out items common
to both half cells.
Red: 48e- +16NO3- + 64H+  32H2O + 16NO
Ox: 24H2S  48H+ + 3S8 + 48e16H++ 16NO3- + 24H2S16NO + 32H2O + 3S8
Acidic or Neutral equations
If possible, place these coefficients in the
original equation (primarily done if equation
has neutral species).
16HNO3(aq) + 24H2S(g) 
16NO (g) + 3S8(s) + 32H2O(l)
Basic Equations
• Write an ionic equation in which all
soluble species are dissociated.
• What dissociates? Strong electrolytes
only!
NH3(g) + O2(g)  NO(g) + H2O(l)
Remains the same (nothing dissociates).
Basic Equations
• Write a Skeletal Equation (one that
includes only substances involved in
redox reaction).
NH3(g) + O2(g)  NO(g) + H2O(l)
Basic Equations
Write equations for the half reactions.
Balance the mass for each element
except oxygen and hydrogen.
Ox: NH3

NO
Red: O2

H2O
Basic Equations
• To balance oxygen, add oxygen in the form
of water to the oxygen-deficient side.
Ox:
Red: O2
H2O + NH3  NO

2H2O
Basic Equations
• To balance hydrogen, add hydrogen in the
form of H+ to the hydrogen-deficient side.
Ox: H2O + NH3  NO + 5H+
Red: O2 + 4H+

2H2O
Basic Equations
• For each H+ added, add an OH- to both sides.
Ox: 5OH- + H2O + NH3  NO + 5H++ 5OHRed: O2 + 4H+ + 4OH-  4OH- + 2H2O
Basic Equations
• Combine H+ and OH- to form water, and
cancel out water molecules on both sides of
half reactions.
Ox: 5OH- + H2O + NH3  NO + 5H2O
Red: O2 + 4H2O  4OH- + 2H2O
Ox: 5OH- + NH3  NO + 4H2O
Red: O2 + 2H2O  4OH-
Basic Equations
Add electrons as e- to the more positive side.
Ox:
5OH- + NH3  NO + 4H2O + 5e-
Red: 4e- + O2 + 2H2O  4OH-
Basic Equations
Find L.C.M. of electrons and multiply as
needed.
Ox: (5OH- + NH3  NO + 4H2O + 5e-)4
Red: 5(4e- + O2 + 2H2O  4OH-)
Basic Equations
Add the half cells. Cancel out items common to
both half cells.
Ox: 20 OH- + 4 NH3  4NO + 16H2O + 20eRed: 20 e- + 5 O2 + 10 H2O  20 OH4NH3 + 5 O2  4NO + 6H2O
Basic Equations
If possible, place these coefficients in the
original equation (primarily done if
equation has neutral species).
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(l)
Galvanic (Voltaic)
Cells
• The energy released in a spontaneous redox
reaction is used to perform electrical work.
• Galvanic cells are devices in which chemical
energy is changed to electrical energy.
• Electricity = movement of charged particles
• Electrochemical cells = Galvanic cell = battery
• Galvanic cells are spontaneous.
Galvanic Cells
• If a strip of Zn is placed in a solution of CuSO4,
Cu is deposited on the Zn and the Zn dissolves
by forming Zn2+.
• Zn is spontaneously oxidized to Zn2+ by Cu2+.
• The Cu2+ is spontaneously reduced to Cu0 by Zn.
• The entire process is spontaneous.
Galvanic Cell Description
• Galvanic cells consist of
– Two solid metals for the electrodes
– Anode (oxidation): Zn(s)  Zn2+(aq) + 2e- AN OX
– Cathode (reduction): Cu2+(aq) + 2e-  Cu(s) RED CAT
– Electrolytic solution in each compartment
– Salt bridge (electrolyte + jello-like material): connects
both compartments so ions flow to keep net charge in
each compartment zero.
– Wire connecting each electrode
Salt Bridge or Porous Disk
Galvanic Cell Description
• As oxidation occurs, Zn is converted to Zn2+
and 2e-. The electrons flow towards the
anode where they are used in reduction
reaction.
• Flow of electrons = electricity = work
• Zn electrode loses mass and Cu electrode
gains mass.
• Galvanic cell eventually “dies.”
“Rules” of Galvanic Cells
At the anode (oxidation) electrons are
products.
2. At the cathode (reduction) electrons are
reactants.
3. Electrons cannot swim.
1.
Example Packet p 7
• The following redox reaction is spontaneous:
• Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq) +
3I2(s) + 7H2O(l)
• A voltaic cell is created using potassium
dichromate and sulfuric acid in one beaker and a
solution of KI in another. A salt bridge connects the
two beakers and platinum is used for the electrodes.
Indicate the reactions at the anode and cathode, the
direction of e- flow, and the signs at the electrodes.
Example Packet p 8
• The two half-reactions in a voltaic cell are
Zn(s) → Zn2+(aq) + 2eClO3-(aq) + 6H+(aq) + 6e- → Cl-(aq) + 3H2O(l)
a) Which reaction occurs at anode, which at cathode?
b) Which electrode is consumed in the cell reaction?
c) Which electrode is positive?
Cell Potential
• The flow of e-’s from anode to cathode is
spontaneous.
• e-’s flow from anode to cathode because the
cathode has a lower electrical potential
energy than the anode.
• Potential difference = difference in electrical
potential. Measured in volts.
Cell Electromotive Force
• One volt is the potential difference required to
impart one joule of energy to a charge of one
coulomb:
1J
1V 
1C
• Electromotive force (emf): force required to
push electrons through external circuit.
• Cell potential (Ecell) = emf of a cell.
• Ecell : standard emf (standard cell potential) at
standard conditions (1M solutions at 25 C)
Standard Reduction (Half-Cell)
Potentials
• We have convenient tables of standard
reduction potentials in aqueous solutions.
• Note: oxidation will NOT be on table, just
use reduction value.
• Standard reduction potentials, Ered are
measured relative to the standard hydrogen
electrode (SHE).
•SHE is the cathode. It consists of a Pt electrode
in a tube placed in 1 M H+ solution. H2 is
bubbled through the tube.
•For the SHE, we assign
2H+(aq, 1M) + 2e-  H2(g, 1 atm)
•E of zero.
Calculating Standard Reduction
Potentials
• emf calculated from standard reduction potentials:
Ecell  Ered cathode  Ered anode
• Consider Zn(s)  Zn2+(aq) + 2e-.
• Ecell = Ered(cathode) - Ered(anode)
• Ecell = 0 V - Ered = -0.76 V.
• Standard reduction potentials must be written as
reduction reactions:
• Zn2+(aq) + 2e-  Zn(s), Ered = -0.76 V.
Standard Reduction Potentials
• Reactions with Ered > 0 are spontaneous
reductions relative to the SHE.
• Since Ered = -0.76 V, reduction of Zn2+ in the
presence of the SHE is not spontaneous but the
oxidation of Zn with the SHE is spontaneous.
• Changing the stoichiometric coefficient does not
affect Ered.
• Therefore,
• 2Zn2+(aq) + 4e-  2Zn(s), Ered = -0.76 V.
Calculating E for a Cell
• Recall Ecell  Ered cathode  Ered anode
Combining two half-reactions often requires two
manipulations:
1. Half-reaction with largest potential will run as
written (reduction) and other half-reaction will
run in reverse. Change sign of anode reaction
and add it to reduction reaction.
2. Balance half-reactions but DO NOT change E!
Example Packet p 10
• We have a Zn-Cu2+ voltaic cell:
• Zn(s) + Cu2+(aq, 1 M) → Zn2+(aq, 1 M) + Cu(s)
• Ecell = 1.10 V
• Given that the standard reduction potential of zinc
ion is -0.76 V, calculate the Ered for the reduction of
copper (II) ion to solid copper.
Example Packet p 11
• Using the standard reduction potentials listed
in Table 201., calculate the standard emf of a
cell with the following reaction:
• Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq)
+ 3I2(s) + 7H2O(l)
Example Packet p 12
• A voltaic cell is based on the following two
standard half reactions:
• Cd2+(aq) + 2e- → Cd(s)
• Sn2+(aq) + 2e- → Sn(s)
• Determine the reactions that occur at the
cathode and the anode.
• Determine the cell potential.
Standard Reduction Potentials
• Reactions with Ered < 0 are spontaneous
oxidations relative to the SHE.
• The larger the difference between Ered values,
the larger Ecell.
• In a Galvanic cell (spontaneous) Ered(cathode)
is more positive than Ered(anode).
• A negative E indicates a nonspontaneous
process.
Example Packet p 14
• Using the table of standard reduction
potentials, determine whether the following
reactions are spontaneous under standard
conditions.
• a. Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g)
• b. Cl2(g) + 2I-(aq) → 2Cl-(aq) + I2(s)
Oxidizing and Reducing Agents
• The more positive Ered the stronger the oxidizing
agent on the left.
• The more negative Ered the stronger the reducing
agent on the right.
• A species on the higher to the left of the table of
standard reduction potentials will spontaneously
oxidize a species that is lower to the right in the
table.
• That is, F2 will oxidize H2 or Li; Ni2+ will oxidize
Al(s).
Example Packet p 13
• Using Table 20.1, rank the following
ions in order of increasing strength as
oxidizing agents.
• Nitrate, silver ion, dichromate
Line Notation for a Galvanic Cell
• Need handy notation for describing cells.
• Anode on left, cathode on right separated by
double vertical lines for salt bridge.
• Phase difference in compartments shown by
single vertical line.
• Ex: Mg(s)Mg2+(aq)  Al3+(aq)Al(s)
anode
cathode
Description of a Galvanic Cell
• Complete description includes 4 items:
• The cell potential (always positive) and the
overall balanced cell reaction.
• The direction of e- flow given by halfreactions to obtain positive cell potential.
• Designation of anode and cathode.
• Notation with electrode/ions present in each
compartment.
Zumdahl Example pp 832-833
• Describe completely the galvanic cell based
on the following half-reactions under
standard conditions.
Ag+ + e- → Ag
Fe3+ + e- → Fe2+
E = 0.80 V
E = 0.77 V
∆G and Cell Potential
• We can show that G = -nFE
• G is the change in free-energy, n is the number of
moles of electrons transferred, F is Faraday’s
constant, and E is the emf of the cell.
• We define
1F = 96,500 C/mol = 96,500 J/V mol
• Since n and F are positive, if G < 0 then E  0 for
a spontaneous reaction.
Example Packet p 15
• Use the standard reduction potentials to calculate
the standard free-energy change for the following
reactions. Use
• G = -nFE and 1F = 96,500 J/V mol
• 4Ag(s) + O2(g) + 4H+(aq) → 4Ag+(aq) + 2H2O(l)
• What are the values of E and ∆G for:
• 2Ag(s) + 1/2O2(g) + 2H+(aq) → 2Ag+(aq) + H2O(l)
Nernst Equation
• A Galvanic cell is functional until E = 0 at which
point equilibrium has been reached.
• The point at which E = 0 is determined by the
concentrations of the species involved in the redox
reaction.
• The Nernst equation relates emf to concentration
using
G  G  RT ln Q
•
and noting that  nFE   nFE   RT ln Q
Workable Nernst Equation
• This rearranges to give the Nernst equation:
RT
E  E 
ln Q
nF
• The Nernst equation can be simplified by
collecting all the constants together using a
temperature of 298 K and the base 10 logarithm:
E = E - (0.0592 V/ n) log Q
Remember that n is number of moles of electrons.
Example Packet p 17 (top)
• Calculate the emf at 298 K generated by the
cell involving the following reaction:
• Cr2O72-(aq) + 14H+(aq) + 6I-(aq) → 2Cr3+(aq)
+ 3I2(s) + 7H2O(l)
• When [Cr2O72-] = 2.0 M, [H+] = 1.0 M , and
[Cr3+] = 1.0 x 10-5 M, and [I-] = 1.0 M.
Example Packet p 17 (bottom)
• If the voltage of the Zn-H+ cell is 0.45 V at
298 K when [Zn2+] = 1.0 M and PH2 = 1.0
atm, what is the concentration of hydrogen
ion?
• Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)
Concentration Cells
Nernst equation can be used to generate a cell that has
an emf based solely on difference in concentration.
• One compartment will consist of a concentrated
solution, while the other has a dilute solution.
• Example: 1.00 M Ni2+(aq) and 1.00 10-3 M
Ni2+(aq).
• The cell tends to equalize the concentrations of
Ni2+(aq) in each compartment.
• The concentrated solution has to reduce the amount
of Ni2+(aq) [to Ni(s)], so must be the cathode.
Concentration Cells
A concentration cell will function until the
concentration of ions in each cell is equal.
Cell emf and Equilibrium
• A system is at equilibrium when G = 0.
• From the Nernst equation, at equilibrium and
298 K (E = 0.00 V and Q = Keq):
0.0592
0  E 
ln K eq
n
nE
log K eq 
0.0592
So, if we know cell emf, we can calculate
K.
Example Packet p 18
• A voltaic cell is constructed with two hydrogen
electrodes. Electrode 1 has P = 1.00 atm and an
unknown [H+]. Electrode 2 is a standard hydrogen
electrode (P = 1.00 atm, [H+] = 1.00 M.) At 298 K
the measured cell voltage is 0.211 V, and the
electrical current is observed to flow from Electrode
1 through the external circuit to Electrode 2.
• Use E = E - (0.0592 V/ n) log Q; Ecell = 0.00 V
• a) Calculate [H+] for the solution at Electrode 1.
• b) What is the pH?
Example Packet p 19
• Using standard electrode reduction potentials ,
calculate the equilibrium constant for the
oxidation of iron (II) ion by oxygen in acidic
solution:
O2(g) + 4H+(aq) + 4Fe2+(aq) → 4Fe3+(aq) + 2H2O(l)
Batteries
• A battery is a selfcontained
electrochemical power
source with one or
more voltaic cell.
• When the cells are
connected in series,
greater emfs can be
achieved.
Lead-Acid Battery
• A 12 V car battery consists
of 6 cathode/anode pairs
each producing 2 V.
• Cathode: PbO2; Anode: Pb
• Wood or glass-fiber spacers
used to prevent electrodes
form touching.
• Battery condition monitored
by measuring density of
acid solution. 3-5 yr life.
Alkaline Battery
emf = 1.55 V
at room temp.
• Anode: Zn cap (Zn powder mixed in a gel)
• Cathode: MnO2, NH4Cl and C paste; reduction
of MnO2
• Graphite rod in center is inert cathode.
• Alkaline battery: NH4Cl replaced with KOH.
Fuel Cells
• Direct production of electricity from fuels.
• Galvanic cell with continuously-supplied
reactants.
• H2-O2 fuel cell primary source of
electricity on Apollo moon flights.
• By-products are steam and hydroxide.
• Fuel cells being used in cars.
Corrosion
• Oxidation of metals
• Oxidation of most metals by oxygen is
spontaneous.
• Metals such as Cu, Ag, Au, and Pt called
noble metals- resistant to oxidation.
• Most metals develop a layer a thin oxide
coating that protects inner atoms.
Corrosion of Iron
• Corrosion of iron is electrochemical process.
• Since Ered(Fe2+) < Ered(O2) iron can be
oxidized by oxygen.
• Water acts as salt bridge (steel does not rust
in dry air).
• Fe2+ initially formed can be further oxidized
to Fe3+ which forms rust, Fe2O3.xH2O(s).
•Corrosion can be prevented by coating iron with
paint or another metal.
•Galvanized iron coated with a thin layer of zinc.
Example Packet p 22
• Predict the nature of the corrosion
that would take place if an iron
gutter were nailed to a house using
aluminum nails.
Electrolysis
• Nonspontaneous reactions require an external
current in order to force the reaction to
proceed.
• Electrolysis reactions are nonspontaneous.
• In voltaic and electrolytic cells:
– reduction at cathode; oxidation at anode.
– However, in electrolytic cells, electrons are
forced to flow from the anode to cathode.
In electrolytic cells the anode is positive and the
cathode is negative. In galvanic cells the anode
is negative and the cathode is positive.
Example Packet p 24
• Electrolysis of AgF(aq) in an acidic
solution leads to the formation of silver
metal and oxygen gas.
• a. Write the half-reaction that occurs at
each electrode.
• b. Calculate the minimum emf for this
process under standard conditions.
Electroplating
• Ni plates on the inert electrode.
• Electroplating is important in protecting
objects from corrosion.
Quantitative Aspects of
Electrolyis
• How much material can we obtain with electrolysis?
• Consider Cu2+(aq) + 2e-  Cu(s).
– 2 mol of electrons will plate 1 mol of Cu.
– The charge of 1 mol of electrons is 96,500 C (1
F).
– Since Q = It, the amount of Cu can be calculated
from the current (I) and time (t) taken to plate.
Example Packet p 25
Calculate the number of grams of aluminum
produced in 1.00 hr by the electrolysis of molten
AlCl3 if the electrical current is 10.0 amperes
(A).
Use Q = I t, where Q = charge (Coulomb = C);
I = current (A); and t = time (sec)
Electrical Work
• Free-energy is a measure of the maximum
amount of useful work that can be obtained
from a system.

G

w
.
max
• We know
G   nFE .
 wmax   nFE
• If work is negative, then work is performed
by the system and E is positive.
Electrical Work
• emf can be thought about as a measure of the
driving force for a redox process.
• In an electrolytic cell an external source of
energy is needed to force the reaction to
proceed.
• To drive the nonspontaneous reaction the
external emf must be greater than Ecell.
• From physics: work measured in watts (W):
• 1 W = 1 J/s.
Last Example Packet p 26
• Calculate the number of kilowatt-hours
of electricity required to produce 1.0 x
103 kg of aluminum by the electrolysis
of Al3+ if the applied emf is 4.50 V.
• Remember: 1 J = 1 C-V
•
1 mol e-’s = 96,500 C
•
1 W = 1 J/s