OXIDATION REDUCTION Reactions and Electrochemical Cells Voltaic
Cells: Using Spontaneous Reactions to Generate Electrical Energy
Cell Potential: Output of a Voltaic Cell Free Energy and Electrical
Work Electrolytic Cells: Using Electrical Energy to Drive
Nonspontaneous Reactions
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OXIDATION STATES The oxidation state (oxidation number) is an
indicator of the degree of oxidation of an atom in a chemical
compound. The formal oxidation state is the hypothetical charge
that an atom would have if all bonds to atoms of different elements
were 100% ionic.
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ASSIGNING OXIDATION STATES (IUPAC) the oxidation state of a
free element (uncombined element) is zero for a simple (monatomic)
ion, the oxidation state is equal to the net charge on the ion
hydrogen has an oxidation state of 1 and oxygen has an oxidation
state of 2 when they are present in most compounds. (Exceptions
hydrides and peroxides) the algebraic sum of oxidation states of
all atoms in a neutral molecule must be zero, while in ions the
algebraic sum of the oxidation states of the constituent atoms must
be equal to the charge on the ion
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Key Points About Redox Reactions o Oxidation (electron loss)
always accompanies reduction (electron gain). o The oxidizing agent
is reduced, and the reducing agent is oxidized. o The number of
electrons gained by the oxidizing agent always equals the number
lost by the reducing agent.
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REDOX reactions - transfer of electrons between species. All
the redox reactions have two parts: OxidationReduction
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The Loss of Electrons is Oxidation. An element that loses
electrons is said to be oxidized. The species in which that element
is present in a reaction is called the reducing agent. The Gain of
Electrons is Reduction. An element that gains electrons is said to
be reduced. The species in which that element is present in a
reaction is called the oxidizing agent.
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A summary of terminology for oxidation-reduction (redox)
reactions.
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A SUMMARY OF REDOX TERMINOLOGY Zn loses electrons. Zn is the
reducing agent and becomes oxidized. The oxidation number of Zn
increases from 0 to +2. REDUCTION One reactant loses electrons.
Reducing agent is oxidized. Oxidation number increases. Hydrogen
ion gains electrons. Hydrogen ion is the oxidizing agent and
becomes reduced. The oxidation number of H decreases from +1 to 0.
Other reactant gains electrons. Oxidizing agent is reduced.
Oxidation number decreases. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2
(g)
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HALF-REACTION METHOD FOR BALANCING REDOX REACTIONS Summary:
This method divides the overall redox reaction into oxidation and
reduction half-reactions. o Each reaction is balanced for mass
(atoms) and charge. o One or both are multiplied by some integer to
make the number of electrons gained and lost equal. o The
half-reactions are then recombined to give the balanced redox
equation.
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Half-Reaction Method for Balancing Redox Reactions Advantages:
o The separation of half-reactions reflects actual physical
separations in electrochemical cells. o The half-reactions are
easier to balance especially if they involve acid or base. o It is
usually not necessary to assign oxidation numbers to those species
not undergoing change.
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HALF-REACTION METHOD FOR BALANCING REDOX REACTIONS 1.Divide the
equation into two half reactions. 2.Balance each half reaction -
balance all elements other than H and O - balance the number of
oxygen by adding H 2 O as needed - balance the number of hydrogen
by adding H + as needed - balance the charges by adding e - as
needed 3.Multiply each half-reactions by integers such as the
number of electrons lost equals the number of electrons gained.
4.Add the reactions cancelling species that appear on both sides.
5.Check
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Balancing Redox Reactions in Acidic Solution Mn +2 + BiO 3 - Bi
+3 + MnO 4 -
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SAMPLE PROBLEM: Balance the following equation: NaCN + H 2 O +
KMnO 4 NaCNO + MnO 2 + KOH CN - + MnO 4 - CNO - + MnO 2 (s) (basic
solution)
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Figure 21.3 General characteristics of voltaic and electrolytic
cells.
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Figure 21.5 A voltaic cell based on the zinc-copper
reaction.
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Notation for a Voltaic Cell Components of anode compartment
(oxidation half-cell) Components of cathode compartment (reduction
half-cell) Phase of lower oxidation state Phase of higher oxidation
state Phase of lower oxidation state Phase boundary between
half-cells Examples:Zn( s ) | Zn 2+ ( aq ) || Cu 2+ ( aq ) | Cu ( s
) Zn( s ) Zn 2+ ( aq ) + 2e - Cu 2+ ( aq ) + 2e - Cu( s ) graphite
| I - ( aq ) | I 2 ( s ) || H + ( aq ), MnO 4 - ( aq ) | Mn 2+ ( aq
) | graphite inert electrodes
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Figure 21.6 A voltaic cell using inactive electrodes.
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SAMPLE PROBLEM The oxidation-reduction reaction, Cr 2 O 7 2-
(aq) + 14H + (aq) + 6I - (aq) 2Cr 3+ (aq) + 3I 2 (s) + 7H 2 O(l) is
spontaneous. Draw a diagram of the cell and indicate the reactions
occuring at the electrodes (anode, cathode), the direction of
electron migration, the direction of ion migration, the signs of
the electrodes. Write the notation for the cell.
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Describing a Voltaic Cell with Diagram and Notation Draw a
diagram, show balanced equations, and write the notation for a
voltaic cell that consists of one half-cell with a Cr bar in a
Cr(NO 3 ) 3 solution, another half-cell with an Ag bar in an AgNO 3
solution, and a KNO 3 salt bridge. Measurement indicates that the
Cr electrode is negative relative to the Ag electrode. Identify the
oxidation and reduction reactions and write each half- reaction.
Associate the (-)(Cr) pole with the anode (oxidation) and the (+)
pole with the cathode (reduction).
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Determining an unknown E o half-cell with the standard
reference (hydrogen) electrode.
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E o cell = E o cathode - E o anode.
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Sample Problem 21.3Calculating an Unknown E o half-cell from E
o cell A voltaic cell houses the reaction between aqueous bromine
and zinc metal: Br 2 ( aq ) + Zn( s ) Zn 2+ ( aq ) + 2Br - ( aq ) E
o cell = 1.83 V Calculate E o bromine, given E o zinc = -0.76 V.
The reaction is spontaneous as written since the E o cell is (+).
Zinc is being oxidized and is the anode. Therefore, the E o bromine
can be found using E o cell = E o cathode - E o anode.
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Sample Problem 21.3Calculating an Unknown E o half-cell from E
o cell PLAN: SOLUTION: The reaction is spontaneous as written since
the E o cell is (+). Zinc is being oxidized and is the anode.
Therefore, the E o bromine can be found using E o cell = E o
cathode - E o anode. Anode: Zn( s ) Zn 2+ ( aq ) + 2e - E = +0.76 V
E o Zn as Zn 2+ ( aq ) + 2e - Zn( s ) is -0.76 V E o cell = E o
cathode - E o anode = 1.83 = E o bromine - (-0.76) E o bromine =
1.83 + (-0.76) = 1.07 V
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Li+/Li -3.05 Sn+4/Sn+2+0.13 Br2/Br+1.07
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Cd +2 /Cd -0.40 Cu +2 /Cu +0.34 Fe +2 /Fe -0.44 Ni +2 /Ni -0.25
Zn +2 /Zn -0.76
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By convention, electrode potentials are written as reductions.
When pairing two half-cells, you must reverse one reduction
half-cell to produce an oxidation half-cell. Reverse the sign of
the potential. The reduction half-cell potential and the oxidation
half-cell potential are added to obtain the E o cell. When writing
a spontaneous redox reaction, the left side (reactants) must
contain the stronger oxidizing and reducing agents. Example:Zn( s )
+ Cu 2+ ( aq ) Zn 2+ ( aq ) + Cu( s ) stronger reducing agent
weaker oxidizing agent stronger oxidizing agent weaker reducing
agent Writing Spontaneous Redox Reactions
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Sample Problem 21.4 Writing Spontaneous Redox Reactions
PROBLEM:(a) Combine the following three half-reactions into three
balanced equations (A, B, and C) for spontaneous reactions, and
calculate E o cell for each. E o = 0.96 V(1) NO 3 - ( aq ) + 4H + (
aq ) + 3e - NO( g ) + 2H 2 O( l ) E o = -0.23 V(2) N 2 ( g ) + 5H +
( aq ) + 4e - N 2 H 5 + ( aq ) E o = 1.23 V(3) MnO 2 ( s ) + 4H + (
aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l )
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Relative Reactivities (Activities) of Metals 1. Metals that can
displace H 2 from acid. 2. Metals that cannot displace H 2 from
acid. 3. Metals that can displace H 2 from water. 4. Metals that
can displace other metals from solution.
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The activity series of the metals.
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Free Energy and Electrical Work G -E cell -E cell = -w max
charge charge = nF n = mol e - F = Faraday constant F = 1 V = 1 F =
G = w max = charge x (-E cell ) G = -nFE cell In the standard state
- G o = -nFE o cell G o = - RT ln K E o cell = - ( ) ln K 96,485 C
mol e - 9.65 x 10 4 J Vmol e - J C RT n Fn F
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The interrelationship of G 0, E 0 cell, and K. < 0 0 > 0
0 < 0 > 1 1 < 1 G o = -RT lnK
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The Effect of Concentration on Cell Potential G = G o + RT ln Q
-nF E cell = -nF E cell + RT ln Q E cell = E o cell -ln Q RT nF
When Q [product], ln Q E o cell. When Q >1 and thus [reactant]
0, so E cell < E o cell. When Q = 1 and thus [reactant] =
[product], ln Q = 0, so E cell = E o cell. E cell = E o cell - log
Q 0.0592 n
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Figure 21.12 A concentration cell based on the Cu/Cu 2+
half-reaction.
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Figure 21.13 The laboratory measurement of pH.
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Alkaline battery. Figure 21.14
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Silver button battery. Figure 21.15
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Lead-acid battery. Figure 21.16
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Nickel-metal hydride battery. Figure 21.17
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Lithium-ion battery. Figure 21.18
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Figure 21.19 Hydrogen fuel cell.
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Figure 21.20 The corrosion of iron.
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Figure 21.21 The effect of metal-metal contact on the corrosion
of iron. Faster corrosion
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Figure 21.22 The use of sacrificial anodes to prevent iron
corrosion.
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ELECTROLYTIC CELL
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ELECTROLYSIS OF MOLTEN SALTS
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Predicting the Electrolysis Products of Aqueous Ionic Solutions
What products form during electrolysis of aqueous solutions of the
following salts: (a) KBr; Compare the potentials of the reacting
ions with those of water, remembering to consider the 0.4 to 0.6 V
overvoltage. The reduction half-reaction with the less negative
potential, and the oxidation half- reaction with the less positive
potential will occur at their respective electrodes. E o = -2.93
V(a) K + ( aq ) + e - K( s ) E o = -0.42 V2H 2 O( l ) + 2e - H 2 (
g ) + 2OH - ( aq ) The overvoltage would make the water reduction
-0.8 to -1.0 V, but the reduction of K + is still a higher
potential so H 2 ( g ) is produced at the cathode. The overvoltage
would give the water half-cell more potential than Br -, so the Br
- will be oxidized. Br 2 ( l ) forms at the anode. E o = 1.07 V2Br
- ( aq ) Br 2 ( l ) + 2e - 2H 2 O( l ) O 2 ( g ) + 4H + ( aq ) + 4e
- E o = 0.82 V
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A summary diagram for the stoichiometry of electrolysis.
Faradays Law of Electrolysis: The amount of substances that
undergoes oxidation or reduction at each electrode during
electrolysis is proportional to the amount of electricity that
passes through the cell 1.0 Faraday = 96,500 C 1 ampere = 1
C/s
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CORROSION PROTECTION 1.Plating the metal with a thin layer of
less easily oxidized 2.Connecting a metal to a sacrificial anode (a
more active metal preferentially oxidized 3.Metal oxides to form
naturally on the surface 4.Galvanizing (coating with zinc)
5.Protective coating, such as paint
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Latimer Diagram Frost Diagram DIAGRAMMATIC REPRESENTATIONS OF
REDUCTION POTENTIAL DATA Pourbaix Diagram Table of redox potentials
are organized by the voltage of the process rather than by the
chemical species involved. Difficult to use if one wants to
understand the complex redox chemistry of the elements
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Reminder: Thermodynamic data will predict which reactions ought
to occur, but cannot determine whether they happen at an observable
rate or not. Most of the redox reactions of inorganic compounds are
rapid reactions, but there are many times when thermodynamics
predicts more than one possible product, and where the actual
product is selected by the rate of reaction.
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Written with the most oxidized species on the left and the most
reduced species on the right. Oxidation number decrease from left
to right and the E 0 values are written above the line joining the
species involved in the couple. LATIMER DIAGRAMS
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Look at the Latimer diagram of nitrogen in acidic solution a bc
de fgh
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FROST DIAGRAM N 2
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FROST DIAGRAMS - INTERPRETATION The lowest point represent the
most thermodynamically stable form of the element A move downward
the plot represent a thermodynamically favorable process A species
at the to right of the diagram is oxidizing. Strengths relative to
their position. The slope of the line drawn between two points
divided by the number of electrons transferred is the E o for the
half reaction. A positive slope means the reduction process is
positive. Any state represented on a convex point is
thermodynamically unstable with respect to disproportionation Any
state represented on a concave point is thermodynamically stable
with respect to disproportionation,
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the lowest lying species corresponds to the most stable
oxidation state of the element What do we really get from the Frost
diagram?
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The oxidizing agent - couple with more positive slope - more
positive E The reducing agent - couple with less positive slope If
the line has ive slope- higher lying species reducing agent If the
line has +ive slope higher lying species oxidizing agent Oxidizing
agent? Reducing agent?
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Identify all the species that are unstable with respect to
disproportionation
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DISPROPORTIONATION. ANOTHER EXAMPLE
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Comproportionation reaction
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Comproportionation is spontaneous if the intermediate species
lies below the straight line joining the two reactant species.
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NE 0 Identify species that are unstable to disproportionation
and comproportionation
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Disproportionation
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Comproportionation In acidic solution Mn and MnO2 Mn 2+ Rate of
the reaction hindered insolubility? In basic solution MnO2 and
Mn(OH)2 Mn2O3
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Consider the Latimer Diagrams for Pb and Tin in acid solution.
A)Calculate values of nE for each oxidation state of lead and tin,
and construct a comparative Frost diagram for these two elements in
acid solution. B) Which of the six species you have plotted is the
strongest oxidizing agent? Justify your answer. C) Which of the six
species you have plotted is the strongest reducing agent? Justify
your answer. D) What product(s) would form if PbO 2 was mixed with
Sn 2+ in acidic solution? Write a balanced equation.
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POURBAIX DIAGRAMS Graphical representations of thermodynamic
and electrochemical equilibria between metal and water, indicating
thermodynamically stable phases as a function of electrode
potential and pH. -predicts the spontaneous direction of reactions.
-estimates the composition of corrosion products. -predicts
environmental changes that will prevent or reduce corrosion
attack.
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SAMPLE PROBLEM: Consider the Pourbaix diagram for manganese.
(a)What form(s) will manganese take in lake and stream water, where
the pH = 6-8 and E o = 0.6-0.7 V?
