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Chapter 20: Electron Transfer Reactions
You are expected to know how to assign oxidation states and
use them to determine which reactants are oxidized and which
are reduced in a redox reaction. You should also be familiar
with the terms oxidizing agent and reducing agent. If you are
not fully comfortable with these concepts (summarized below),
please review the relevant sections of Chapter 5 (Kotz).
Rules for Assigning Oxidation States
1. Pure elements have oxidation states of 0.
2. Ions have oxidation states that add up to the charge of the
ion (e.g. Na+ is +1, Mg2+ is +2, Al3+ is +3, oxidation states in
SO42- add up to -2)
3. Hydrogen has an oxidation state of +1 unless bonded to a
less electronegative atom (i.e. a metal or boron – in that
case, it has an oxidation state of -1) or in H2 (see rule 1).
4. Fluorine has an oxidation state of -1.
5. Oxygen has an oxidation state of -2 unless bonded to
fluorine or another oxygen.
6. Halogens other than fluorine have oxidation states of -1
unless bonded to oxygen or a more electronegative halogen.
7. Assign the rest of the atoms by process of elimination. The
oxidation states must add up to the total charge of the
molecule or ion.
A reactant that gains electrons (oxidation state becomes more
negative) is reduced. It helps another reactant be oxidized, so
it’s also a oxidizing agent.
A reactant that loses electrons (oxidation state becomes more
positive) is oxidized. It helps another reactant be reduced, so
it’s also a reducing agent.
For more complicated molecules, the oxidation state of each
atom can be found by drawing the Lewis structure and treating
every bond as ionic (unless it is between two atoms of the same
element). This is the “opposite” of finding formal charge – in
which every bond is treated as perfectly covalent.
e.g. Acetic acid is CH3CO2H. Calculate the average oxidation
state of each carbon atom then determine the actual
oxidation state of each carbon atom. (χC = 2.5; χH = 2.2)
Half-Reactions
Any redox reaction must, by definition, consist of both an
oxidation and a reduction. As such, we can break the redox
reaction into an oxidation half-reaction (electrons are ‘products’)
and a reduction half-reaction (electrons are ‘reactants’).
e.g. A redox reaction occurs between magnesium metal and
acid (H+):
Mg(s) + 2 H+(aq) → Mg2+(aq) + H2(g)
____ is reduced to ____, and the reduction half-reaction is:
____ is oxidized to ____, and the oxidation half-reaction is:
Adding two half-reactions gives the overall redox reaction.
The example above could have been balanced by inspection, but
more complex redox reactions are most easily balanced by
generating the two half-reactions, balancing them then adding
them such that the electrons cancel out.
We will focus on electrochemical reactions in aqueous solution
at standard state. Thus, temperature is 25 ˚C, acidic solutions
have [H3O+] = 1 M, and basic solutions have [OH-] = 1 M.
Sidenote: The pH of a standard state acidic solution is ____.
The pH of a standard state basic solution is ____.
To balance a redox reaction, follow the following steps:
1. Assign enough oxidation states to determine which element
is reduced and which is oxidized.
2. Separate the overall reaction into two half-reactions. (Don’t
worry about water, H+ or OH- yet.)
3. Balance the elements in each half-reaction.
• Balance all elements except H and O.
• In acidic solution:
o First, balance O by adding H2O.
o Then, balance H by adding H+.
• In basic solution:
o First, balance the half reactions as if in acid.
o Then, use OH- to cancel out the H+ (making one H2O
for every H+ + OH-), adding the same number of OHto each side of the equation.
4. Balance the half-reactions for charge by adding electrons.
5. Multiply each half-reaction by whatever factor is necessary
so that the electrons will cancel out.
6. Add the two half-reactions, simplifying as necessary.
7. Check that the final equation is balanced (including charge!)
Balance each of the half-reactions below, and indicate whether it
is an oxidation half-reaction or a reduction half-reaction.
I2(aq) →
I-(aq)
in acid
VO2+(aq) →
V3+(aq)
in acid
CrO2-(aq) →
CrO42-(aq)
in base
NiO2(aq)
in base
Ni(OH)2(aq) →
Use the half-reaction method to balance each of the redox
reactions below.
MnO4-(aq) + HSO3-(aq) → Mn2+(aq) + SO42-(aq)
in acid
Cr2O72-(aq) + Fe2+(aq) → Cr3+(aq) + Fe3+(aq)
in acid
PbO2(s) + Cl-(aq) → ClO-(aq) + Pb(OH)3-(aq)
in base
Fe(OH)2(s) + CrO42-(aq) → Fe2O3(s) + Cr(OH)4-(aq)
in base
Voltaic Cells and Half-Cells
In Chapter 19, we used Gibbs free energy (∆G) to predict
whether a reaction would be product- or reactant-favoured.
When discussing electrochemistry, we use cell potential (E) for
the same purpose. There is a direct relationship between cell
potential and Gibbs free energy:
∆G = - n F E
where n is the number of moles of electrons transferred per mole
of reaction, F is Faraday’s constant (96,485 C/mol or 96.485
kC/mol) and E is the cell potential in volts (1 V = 1 J/C).
Under standard conditions, this translates to:
∆G˚ = - n F E˚
Note that the cell potential will always have the opposite sign of
the Gibbs free energy:
If ∆G < 0 then E > 0 and the reaction is ___________-favoured.
If ∆G > 0 then E < 0 and the reaction is ___________-favoured.
In electrochemistry, a reaction which is product-favoured
releases electrochemical energy which can be used to do work.
The reaction is said to produce a voltage. Any electrochemical
cell using a product-favoured reaction is called a Voltaic cell.
A reaction which is reactant-favoured can be driven forward
by applying a greater opposing voltage. This process is called
electrolysis, and it occurs in electrolytic cells.
In the lab, we have seen that we can do redox reactions in one
container (e.g. reducing Cu2+ to Cu with zinc); however, this
does not allow us to harness any electrochemical energy. If we
want to obtain useful/measurable electrochemical energy, we
need to separate the two half-reactions into compartments called
half-cells. This prevents the zinc from reacting with the Cu2+
directly – instead the electrons must travel from one half-cell to
the other via connected electrodes. Meanwhile, a salt bridge
prevents charges from building up in either half-cell by allowing
inert ions to pass between the two half-cells.
Just as electrons are attracted to cations, they are also attracted
to a cathode. In an electrochemical cell, electrons are generated
by a(n) ___________________ reaction at the anode then travel
to the cathode where they are consumed in a(n)
___________________ reaction.
In addition to the Zn and Cu2+ necessary for the forward
reaction, this cell also contains the Cu and Zn2+ necessary for the
reverse reaction. This makes the reaction reversible – which is
necessary in order to accurately measure the cell potential. To
completely describe the electrochemical cell, we list the
reactants and products at both electrodes – with the oxidation
reaction at the left so that the electrons “flow” from left to right:
Zn(s) | Zn2+ (aq, 1 M) || Cu2+ (aq, 1 M) | Cu(s)
The vertical lines (|) represent phase boundaries (e.g. between
the solid Zn electrode and the Zn2+ solution). Double vertical
lines (||) indicate double boundaries – such as a salt bridge. If
the components of a redox reaction co-exist in solution without a
phase boundary, the oxidized and reduced forms are separated
with a comma. This is often the case when the electrical current
is introduced via an inert electrode (such as platinum metal).
e.g. Pt(s) | Fe2+ (1 M), Fe3+ (1 M) for the oxidation of Fe2+ to Fe3+
A handy mnemonic device to remind you to put the cathode
(reducing half-reaction) at the right is “Right Red Cat”.
e.g. For the cell shown on the previous page, confirm that the
cell potential is +1.10 V.
∆Gf˚(Cu2+) = 65.52 kJ/mol, ∆Gf˚(Zn2+) = -147.0 kJ/mol
Generally, any discrepancy between the calculated and
measured cell potentials is due to non-standard conditions –
most often concentrations that are not exactly 1 M.
There is another way to calculate the potential of an
electrochemical cell. We can add the electrochemical potentials
for each of the two half-cells (i.e. for the two half-reactions).
Note that it is impossible to directly measure the potential for a
half-cell since there would be no current and therefore no
voltage. Instead, cell potential is measured against an arbitrary
zero reference point, the standard hydrogen electrode (SHE):
2 H+(aq) + 2 e- → H2(g)
E˚ = 0 V
Like ∆G, ∆H, and ∆S, reversing the reaction reverses the sign of
E˚; therefore, we can also say that:
H2(g) → 2 H+(aq) + 2 eE˚ = 0 V
The SHE is an electrode in which hydrogen gas is bubbled over
a platinum electrode in the presence of 1 M H+(aq).
When the standard hydrogen electrode is used as the anode in an
electrochemical cell, the measured cell potential will be the
cathode’s standard reduction potential.
e.g. An electrochemical cell consists of a standard hydrogen
electrode (as the anode) and a copper cathode:
Cu2+(aq) + 2 e- → Cu(s)
E˚ = + 0.337 V
H2(g) → 2 H+(aq) + 2 eE˚ = 0 V
Cu2+(aq) + H2(g) → 2 H+(aq) + Cu(s)
E˚ = + 0.337 V
Similarly, we can use standard reduction potentials to calculate a
cell potential as long as we remember to reverse the reaction
at the anode. Note that it is never necessary to multiply
standard reduction potentials by any coefficient other than -1.
The number of electrons transferred is already accounted for –
the volt is a unit of energy-per-charge-transferred.
Calculate the potential for each of the cells below.
Cd(s) | Cd2+ (aq, 1 M) || Ni2+ (aq, 1 M) | Ni(s)
Sn(s) | Sn2+ (aq, 1 M) || Au3+ (aq, 1 M) | Au(s)
Note that both of these examples give positive cell potentials.
They could both be used as voltaic cells.
Also, note that tables of standard reduction potentials are
typically arranged in order of decreasing reduction potential.
This means that the reactants at the top of the table are the most
easily reduced, or the strongest ________________ agents.
Similarly, the products at the bottom of the table are the most
easily oxidized, or the strongest _________________ agents.
Use a table of standard reduction potentials to propose a voltaic
cell that we have not looked at so far.
Use a table of standard reduction potentials to propose an
electrolytic cell.
Name a stronger reducing agent than magnesium.
Which is a stronger oxidizing agent, bleach or peroxide?
The Nernst Equation
In the real world, cells rarely operate under standard conditions.
Even if we construct a cell under standard conditions, reactants
get consumed and their concentrations drop below 1 M. As
reactants are consumed, they no longer produce as many
electrons, and the cell potential decreases. This is why batteries
run out, and we have to either recharge or discard them. This
effect is quantified by the Nernst equation:
E = E° -
RT
lnQ
nF
where E is the actual cell potential (in V), E˚ is the standard cell
potential (in V), R is the ideal gas constant (in J·mol-1·K-1), T is
temperature (in K), n is number of moles of electrons transferred
per mole of reaction, F is Faraday’s constant (96,485 C/mol) and
Q is the reaction quotient (see chapters 16 and 18).
At 25 ˚C, the Nernst equation simplifies to:
E = E° -
0.0257 V
lnQ
n
Recognize that this version of the Nernst equation is only
applicable at 25 ˚C and, since you can always look up R and F,
really isn’t that much of a time-saver.
Looking back at the general form of the Nernst equation, we can
confirm that consuming reactants lowers cell potential. This is
because consuming reactants will increase Q (more products,
less reactants) and therefore increase the term subtracted from
E˚. We will see that it takes a large change in concentration to
significantly affect cell potential – which is a very good thing, or
batteries wouldn’t work long enough to be practical!
e.g. A cell is formed from a Zn2+/Zn half-cell in which [Zn2+] =
0.0500 M and a Cl2/Cl- half-cell in which [Cl-] = 0.0500 M
and PCl2 = 1.25 atm.
(a) Provide a cell notation for this cell.
(b) Calculate the E actual for this cell.
The pH meter is a familiar example of
electrochemistry under nonstandard
conditions. It is actually a voltmeter
attached to a glass electrode. The
bulb at the bottom of the glass
Silver wire
electrode is made of a very thin layer
of glass – so thin that a surface
AgCl (s)
potential difference is created when
the internal [H+] and external [H+]
KCl (saturated)
differ. Because the protons cannot
Porous membrane
actually cross the glass membrane,
this potential can be measured (by the
0.1 M HCl
voltmeter) and used to calculate the
external [H+] – which is readily
Glass membrane
converted to the pH of the solution.
Essentially, we are dealing with an
The Glass Electrode
electrochemical cell in which both
half-reactions involve H+/H2 and all of the cell potential is
coming from the concentration gradient. The glass electrode
also contains two Ag+/Ag reference electrodes which are not
included in our calculations. Thus, our half-cells are:
2 H+(aq) + 2 e- → H2(g)
H2(g) → 2 H+(aq) + 2 e-
and
To calculate the external (solution) [H+], we use the Nernst
equation:
RT
E = E° -
nF
lnQ
The standard state cell potential of an H+/H2 anode and H+/H2
cathode would be 0 V therefore all of the potential is coming
from the concentration gradient:
Eglass
=0electrode
RT
lnQ
nF
The reaction quotient is [H+]2inside / [H+]2outside and, as we can see
in the half-cell equations, 2 electrons are transferred, so n = 2:
Eglass
=0electrode
2
[H+ ]inside
RT
ln + 2
2F [H ]
outside
This reduces to:
Eglass
+
RT [H ]inside
ln
=electrode
F [H+ ]
outside
We know that ln(x) = 2.303 log10(x), R = 8.3145 J·mol-1·K-1,
F = 96485 C·mol-1 and under standard conditions T = 298.15 K.
This gives:
+
Eglass
electrode
= - 0.0592log
[H ]inside
[H+ ]outside
Switching from the natural logarithm allows us to calculate pH
values directly (since pH = -log[H+]):
Eglass
electrode
0.0592
Or:
Eglass
= - (log[H+ ]inside − log[H+ ]outside)
electrode
0.0592
= pHinside - pHoutside
Finally, we get:
pHoutside = pHinside
−
Eglass
electrode
0.0592
The reason we have to calibrate pH meters is that the two
reference electrodes aren’t 100% identical, and the electrons
don’t move 100% efficiently. Calibration lets us factor this in.
As you will be aware from lab, pH meters (glass electrodes)
suffer a few limitations – particularly at extreme pH values:
• In strongly acidic solutions (very low pH), pH meters tend
to read high, for reasons that are not yet understood.
• In solutions where [H+] is low and [Na+] is high, the pH
meter will begin to respond to Na+ as well as H+.
• The pH meter measures [H+] in the solution immediately
next to the electrode. In well-buffered solutions, it
responds to changes in [H+] very quickly. In poorly
buffered solutions, it responds to changes in [H+] slowly.
• A dry glass electrode needs to be soaked in water for hours
before it can be used to [H+] with any accuracy.
Cell Potentials and Equilibrium Constants
As we have seen, the Nernst equation allows us to relate cell
potential to the reaction quotient under nonstandard conditions:
E = E° -
RT
lnQ
nF
This equation is very similar to an equation we used to relate
Gibbs free energy to reaction quotient:
∆G = ∆G° + RT lnQ
This should come as no surprise given that cell potential is really
Gibbs free energy “in disguise”. (∆G = - nFE).
If we consider the system at equilibrium (E = 0, Q = K), we can
therefore relate cell potential to the equilibrium constant under
standard conditions:
0 = E° -
RT
lnK
nF
or
E° =
RT
lnK
nF
This is a common way to measure equilibrium constants –
measure the cell potential under standard (non-equilibrium)
conditions then use the above formula to calculate K.
e.g. The cell potential for the reaction below under standard
conditions is +0.11 V. Calculate the equilibrium constant
at 25 ˚C.
Ni(s) + Sn 2+(aq)
Ni 2+(aq) + Sn(s)
Commercial Voltaic Cells and Storage Batteries
Thus far, we have looked at voltaic cells schematically. We can
construct cells of the types shown – and they work – but they
wouldn’t be particularly portable or practical. More familiar
voltaic cells come in batteries. Technically, the term battery
refers to a series of voltaic cells connected in series so that their
voltages add up.
e.g. A 12 V car battery is really a series of six 2 V cells.
A 9 V radio battery is really a series of six 1.5 V cells.
A standard 1.5 V “battery” is just one 1.5 V cell.
We can divide commercial voltaic cells into two categories:
• primary batteries, which can only be used once
• secondary batteries, which can be recharged
The category of the cell depends on whether or not the redox
reactions are reversible.
Note that, in most commercial voltaic cells, pastes of inorganic
salts are used. This allows compact storage of the salts. The
water in the paste dissolves some salt to make a saturated
solution in which the ion concentrations do not significantly
decrease until near the end of the cell’s usefulness. The solution
also allows for ion migration (necessary for flow of charge).
Dry Batteries (Zinc-Carbon Cell, or Leclenché Cell)
“Flashlight batteries” are primary
batteries that are single voltaic cells.
The anode is zinc with an insulating
wrapper, and the cathode is a carbon
rod at which manganese is reduced.
Anode (oxidation; E˚ = + 0.76 V)
Zn → Zn2+ + 2eCathode (reduction; E˚ = + 0.74 V)
2 MnO2 + 2 NH4+ + 2 e- →
Mn2O3 + 2 NH3 + 2 H2O
This gives a total voltage of 1.5 V for a fresh dry cell. The
ammonia forms co-ordination complexes with zinc cations,
[Zn(NH3)4]2+. If this gas is generated too quickly, though, it
forms an insulating layer around the cathode and the battery
needs to “rest” before it can be used again.
Alkaline Batteries
“Flashlight batteries” have a limited shelf-life because the acidic
NH4Cl corrodes the zinc can. They also “die” quickly relative to
other types of batteries (as the voltage drops quickly with use).
A common alternative is the alkaline battery. It uses similar
redox chemistry, but in an alkaline (basic) environment. This
minimizes corrosion of the zinc can and prevents the build-upof-ammonia problem. Alkaline batteries also tend use purer
starting materials and have better construction. As such, they
are worth their slightly higher cost and, today, most common
batteries are alkaline. They are also 1.5 V cells, operating under
nonstandard conditions. ([OH-] > 1 M.)
(+)
Cathode(+): paste containing
MnO2, graphite, and water
Outer steel jacket
Plastic sleeve
Anode(-): Paste containing
powdered zinc, KOH, and water
Inner steel jacket
Brass collector
(-)
Cell base
Anode (oxidation; E˚ = + 1.25 V)
Zn + 2 OH- → ZnO + H2O + 2eCathode (reduction; E˚ = + 0.15 V)
2 MnO2 + H2O + 2 e- → Mn2O3 + 2 OH-
Lithium Batteries
Like alkaline batteries, many lithium
batteries reduce manganese at the
cathode and have a basic environment.
The difference is that lithium is
oxidized at the anode (instead of zinc).
Otherwise, the design is similar.
Anode (oxidation; E˚ = + 3.04 V)
Li → Li+ + eCathode (reduction; E˚ = + 0.15 V)
2 MnO2 + H2O + 2 e- → Mn2O3 + 2 OHThere are also a variety of other lithium batteries using different
reductions at the cathode.
Mercury Batteries
Mercury batteries are relatively expensive and deliver 1.34 V
(less than the Mn-based cells). They are, however, very small
and maintain a fairly constant voltage throughout their lifetime.
As such, they are often used for hearing aids and other devices
requiring excellent reliability. They have a zinc-mercury
amalgam as the anode, and a graphite-HgO paste as the cathode.
Anode (oxidation; E˚ = + 1.25 V)
Zn + 2 OH- → ZnO + H2O + 2eCathode (reduction; E˚ = + 0.09 V)
HgO + H2O + 2 e- → Hg + 2 OH-
Zinc-Air Batteries
Similar to the mercury batteries in both construction and
application, zinc-air batteries have the interesting property that
the species reduced at the cathode is oxygen from the air. As
such, they have air holes (which must be uncovered before use).
Anode (oxidation; E˚ = + 1.25 V)
Zn + 2 OH- → ZnO + H2O + 2eCathode (reduction; E˚ = + 0.40 V)
½O2 + H2O + 2 e- → 2 OH-
Lead Acid Batteries
Unlike the examples we have looked
at thus far, lead acid batteries are
rechargeable (i.e. secondary batteries).
Generally, several lead acid cells (2 V
each) are connected in series to give a
higher-voltage battery.
Both the
cathode and anode involve reactions
between a lead source and hydrogen
sulfate (from the sulfuric acid used to
make the cell).
Anode (oxidation; E˚ = + 0.35 V)
Pb + HSO4- → PbSO4 + H+ + 2eCathode (reduction; E˚ = + 1.69 V)
PbO2 + HSO4- + 3 H+ + 2 e- → PbSO4 + 2 H2O
When the battery is recharged, a voltage is applied which makes
the two reactions run in reverse. As such, the PbSO4 deposited
on both electrodes is converted back into Pb, PbO2 and HSO4-.
Nickel-Cadmium Batteries (aka. Ni-cad batteries)
Ni-cad batteries are also rechargeable. They are more expensive
than lead acid batteries, but they have the advantages of being
smaller, longer lived and more easily recharged.
Anode (oxidation; E˚ = + 0.49 V)
Cd + 2 OH- → Cd(OH)2 + 2eCathode (reduction; E˚ = + 0.82 V)
NiO(OH) + H2O + e- → Ni(OH)2 + OHBecause they are not operating under standard conditions, the
nominal voltage of a Ni-cad battery is 1.2 V (not 1.3 V).
Electrolytic Cells
Recharging a battery is just one application
of electrolysis (the process by which a
reversible reactant-favoured redox reaction
is forced to go forward by applying a
voltage.) Electrolysis can also be used to
produce metals from solutions of their
cations – either as pure samples or in a
process called electroplating (adding a thin
layer of metal to the outside of an existing
object, see picture at right1).
Many steel parts are electroplated for strength:
• Car bumpers can be coated with first nickel then chromium.
• Bolts are often coated with either zinc or cadmium.
• Light fixtures can be coated with nickel then either chrome
or brass. (This is more for appearance than strength.)
A third example of electrolysis is the commercial production of
chlorine gas and sodium hydroxide from salt-water:
2 NaCl + 2 H2O → 2 NaOH + Cl2 + H2
What species is being oxidized?
What species is being reduced?
We generally measure the efficiency of an electrolytic cell by
comparing the number of moles of electrons consumed with the
number of moles of products made (factoring in stoichiometry
as necessary). Measuring the current gives us information from
which we can calculate the number of moles of electrons made:
1
http://en.wikipedia.org/wiki/Electroplating, last visited on April 5, 2006.
I=q
t
where I is current (in Amperes), q is charge (in Coulombs) and t
is the time for which the current was applied (in seconds). We
also know that the charge of one mole of electrons is 96 485
Coulombs per mole (Faraday’s constant, F).
Therefore:
I=
ne F
t
or
ne = It
F
(a) If you electrolyze a solution of Ni2+(aq) to form Ni(s) using a
current of 0.15 amp for 10 minutes, what mass of solid
nickel should be produced?
(b) If the electrolysis described in part (a) produced 25 mg of
nickel, what was the percent efficiency of this process?
Important Concepts from Chapter 20
• Assigning oxidation states, redox reactions, etc.
• Using half-reactions to balance redox reactions (in acid and in
base)
• Voltaic cells vs. electrolytic cells
• Cell potential (and how it relates to Gibbs free energy)
• Components of a voltaic cell
o electrodes (cathode and anode)
o salt bridge
• Cell notation
• Cells under nonstandard conditions (the Nernst equation)
• Relationship between E˚ and equilibrium constant
• Practical considerations for batteries, pH meters, etc.
• Applications of electrolytic cells
• Determining efficiency of electrolytic cells