Introduction to Inorganic Chemistry/Redox Stability and Redox Reactions

Chapter 4: Redox Stability and Redox ReactionsEdit

Oxidation states of vanadium in acidic solution. From left to right the oxidation state goes from +5 to +2.

In redox reactions, one species is reduced (gains electrons) and another species is oxidized (loses electrons). Identifying the oxidation states of all species is an important first step for determining which species are oxidized, reduced, or unchanged. In terms of everyday life, redox reactions occur all of the time. One example of redox is with glycolysis, a process that occurs in all living things and involves the conversion of food into a usable energy1. As with a number of other reactions, oxygen does not undergo redox in glycolysis and is more of a bystander. Another example of redox is fire or combustion, such as in a car engine. In a car engine, carbon in the fuel is oxidized to carbon dioxide and carbon monoxide, while the oxygen present is reduced to form water upon combination with hydrogen atoms in the fuel1.

Oxidation-reduction reactions are important to understanding inorganic chemistry for several reasons:

• Transition metals can have multiple oxidation states

• Main group elements (N, halogens, O, S...) also have multiple oxidation states and important redox chemistry

• Many inorganic compounds catalyze redox reactions (which is especially useful in industrial and biological applications)

• Energy conversion (solar, batteries, fuel cells) relies on inorganic redox reactions

• Not all elements are equally reactive in redox reactions: there are strong and weak oxidizers and reducers

As stated, not all oxidizers and reducers were created equal. Strong oxidizing agents are typically compounds with elements in high oxidation states or with high electronegativity (electrons are gained from oxidizing compounds). Examples of oxidizers include hydrogen peroxide, permanganate, and osmium tetroxide2. For reducing agents, electropositive elements such as lithium, sodium, iron, and aluminum are common because of their willingness to donate electrons2. Additionally, sodium borohydride and lithium aluminum hydride are often used to reduce carbonyl compounds2.

4.1 Balancing redox reactionsEdit

I- is oxidized to IO3- by MnO4-(→Mn2+). How can this reaction be balanced? The following example demonstrates proper balancing technique, and is known as the ion-electron method.

Step 1: Write out the unbalanced reaction and identify the elements that are undergoing redox.

MnO4- + I-IO3- + Mn2+ (The elements undergoing redox are Mn and I. The oxygen is not undergoing redox since there is no change in its oxidation state.)

Step 2: Separate the original reaction into two half reactions

• MnO4-Mn2+
• I-IO3

Step 2A: Balance the Oxygen

• MnO4-→ Mn2+ + 4H2O
• 3H2O + I- → IO3-

Step 2B: Balance the Hydrogen

• 8H+ + MnO4-→ Mn2+ + 4H2O

The left side has a charge of +7 while the right side has a charge of +2

• 3H2O + I- → IO3- + 6H+

The left side has a charge of -1 while the right side has a charge of +5

Step 2C: Balance the charges with Electrons

• 8H+ + 5e- + MnO4-→ Mn2+ + 4H2O

The left side has a charge of +2 while the right side has a charge of +2. They are balanced.

• 3H2O + I- → IO3- + 6H+ + 6e-

The left side has a charge of -1 while the right side has a charge of -1. They are balanced.
Note: We did not need to explicitly determine the oxidation states of Manganese or Iodine

Step 3: Combine the half reactions so that there are equal numbers of electrons on the left and right sides

6 (8H+ + 5e- + MnO4-→ Mn2+ + 4H2O)
5 (3H2O + I- → IO3- + 6H+ + 6e-)
______________________________

48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+

Cancel the hydrogens, electrons, and water:

48H+ + 30e- + 15H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 24H2O + 30e- + 30H+

The final equation is therefore:
18H+ + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O
Make sure the elements and charges are balanced for BOTH sides.

Step 4: If the reaction occurs in basic conditions, add OH- to each side to cancel H+
18H+ + 18OH- + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 9H2O + 18OH-

The 18H+ + 18OH- will become 18H2O so the final equation would be:

9H2O + 6MnO4- + 5I- → 5IO3- + 6Mn2+ + 18OH-
________________________________________________________________________

An Abbreviated Example:
S2O32- + H2O2S4O62- + H2O
Which elements are undergoing redox? S and O

2S2O32-S4O62- + 2e-
H2O2 + 2H+ + 2e-2H2O
________________________

2S2O32- + H2O2 + 2H+S4O62- + 2H2O
Note: Again, we did not need to know oxidation states of Sulfur or Oxygen

4.2 Electrochemical equilibriaEdit

In electrochemical cells, chemical species migrate to different areas depending on the electrochemical potential. Species move from areas of high potential to areas of low potential3. The equilibrium reduction potential of a half-cell (reduction or oxidation), as well as the total voltage of a full cell, can be determined by the Nernst equation3. The chemical species in a cell have electrochemical potentials given their location, and the potentials are an indication of how easy it would be to add more of the species to that location3. In equilibrium, the potential of a species is constant everywhere. For calculations, standard reduction potentials for most or all species are available in many pieces of chemical literature.

For systems that are in equilibrium, ΔG°= -nFE°cell where o indicates that the electric potential is a standard state
More generally , $Delta G^o = -nFE$ , where $E = E^o-\frac{RT}{nF} * ln Q$ then $E = E^o-\frac{0.0592}{n} * log Q$

For example, if we have the equation 2H+ + 2e- = H2 then E°1/2 = 0.000 Volts (by definition)

Question: What is E1/2 at pH 5 and PH2 = 1 atm?
$pH = - log [H^+]=5$ where $H^+ = 10^-5 M$

$E = E^o-\frac{0.0592}{n} * log Q$ where n=2 and Q is then equal to $\frac{P (H2)}{(H^+)^2}$

So $E = E^o -\frac{0.0592}{2} * log$ $\frac{P(H2)}{(H^+)^2} = E^o -\frac{0.0592}{2} * log (10)^(10) = E^o -\frac{0.0592}{2} (10)$

$= 0.000 - 0.296 = -0.296V$

Question: What is the difference between the H2/ H+ and O2/H2O couples at pH5?
2H+ + 2e- → H2 that has an E°1/2 = 0.000 Volts for (2H+, 2e-)
O2 + 4H+ + 4e- → 2H2O that has an E°1/2 = +1.229 Volts for (4H+, 4e-)

The total E°1/2
0.000 V
+1.229 V

cell = +1.229 Volts

This value does not change with pH (since both couples shift -59.2 mV/pH in the H2O Pourbaix diagram)

4.3 Latimer, Frost, and Pourbaix diagramsEdit

There are three kinds of redox stability diagrams known as Latimer, Frost, and Pourbaix diagrams. Each of these diagrams contain similar information, but one representation may be more useful for a given situation.

Latimer and Frost diagrams help predict stability relative to higher and lower oxidation states. Pourbaix diagrams help understand pH-dependent equilibria and corrosion (which will be talked about more later).

Latimer diagrams:
Electric potential values are given for successive redox reactions (from highest to lowest oxidation state). The difference in oxidation states for each potential value is not necessarily one as in the provided diagram. For example, potentials in a Latimer diagram can be given for a +1 to +3 oxidation state change, not just for a +1 to +2 or +3 to +4. Identifying the oxidation state for each species is important if it's not already provided.

Example:

Mn in Acid
The Latimer Diagram for Mn in acid illustrates the standard reduction potentials using the different oxidation states.

Example:

Mn2++2e-→Mn, E1/2°=-1.18V
MnO2+4H++e-→Mn3++2H2O, :E1/2°=+0.95V

We can also easily calculate values for multi-electron reactions by adding ΔG°(=-nFE°)
Example:

MnO4-→Mn2+

E° = 1(0.564)+1(0.274)+1(4.27)+1(0.95)+1(1.51)5 = +1.51

MnO4-→MnO2

E° = 1(0.564)+1(0.274)+1(4.27)3 = +1.70

Divide by the number of electrons involved in the oxidation number change (5 and 3 for the above equations).

An Unstable species on a Latimer diagram will have a lower electric potential to the left than to the right.

Example:

2MnO43-→MnO2+MnO42-; the MnO43- species is unstable
E°=+4.27-0.274=+3.997 (spontaneous)

Which Mn species are unstable with regards to disproportionation?

MnO43-; 5+→6+,4+
Mn3+; 3+→4+,2+

So stable species are: MnO4-, MnO42-, MnO2, Mn2+, and Mn0.

But MnO42- is also unstable, why?

MnO42-→MnO2 ;

2MnO42- → 2MnO4- + MnO2 ; E° = 2.272 - 0.564 = +1.708

Moral: All possible disproportionation reactions must be considered in order to determine stability (this is often more convenient with a Frost diagram).

Note: Thermodynamically unstable ions can be quite stable kinetically. Also, most N-containing molecules (NO2,NO,N2H4) are unstable relative to the elements (O2, N2, H2)

Frost diagrams:
Stable and unstable oxidation states can be easily identified by plotting ΔG°F (=nE°) vs. oxidation number.

• Contains same information as in a Latimer diagram, but graphically shows stability and oxidizing power.
• Lowest species on diagram are most stable (Mn2+, MnO2)
• Highest species on diagram are strongest oxidizers (MnO4-)

Pourbaix Diagrams:
Plot electrochemical equilibria as a function of pH.

Example: Iron Pourbaix Diagram

Pourbaix-plot E vs. pH:
• pure redox reactions are horizontal lines, these reactions are not pH dependent
• pure acid-base reactions are vertical lines, these reactions are pH dependent
• reactions that are acid-base and redox have a slope of -0.0592 (# H+# e-)
Key Equilibria:

1. Fe2+ + 2e- → Fe(s) (pure redox- no pH dependence)
2. Fe3+ + e- → Fe2+ (pure redox)
3. Fe3+ + 3OH- → Fe(OH)3(s) (pure acid-base, no redox)
4. Fe2+ + 2OH- → Fe(OH)2(s) (pure acid-base, no redox)
5. Fe(OH)3 + e- + 3H+ → Fe2+ + 3H2O ; E1/2 = E1/2° + 0.05921 log[Fe2+]/[H+] → slope = 3 × 0.0592V / per pH unit
6. Fe(OH)3 + H+ + e- → Fe(OH)2 + H2O

Key Points:

• Fe(s) is unstable at all pH values in H2O
• Iron can be catholically protected by applying a potential below its stability line
for example, at pH of 7, a potential less than or equal to -0.44 need to be applied
• Fe corrodes in pH/potential range where Fe2+ is stable (in acid, below purple line)
• Fe is passivated by a solid film of Fe(OH)3, Fe(OH)2, or Fe3O4 in base

4.4 Redox reactions with coupled equilibriaEdit

Redox Reactions with Coupled Equilibria

• Coupled equilibria influence E°.

e.g.

Fe3+/2+ complexation by CN- → Fe(CN)63- (Fe3+), Fe(CN)64- (Fe2+)
Fe3+(aq) + e- = Fe2+(aq) ; E° = +0.77V
Fe(CN)63- + e- = Fe(CN)64- ; E° = +0.42V
• Iron(II) is easier to oxidize when complexed to CN-
Fe3+ + Fe(SN)64- ↔ Fe2+ + Fe(SN)63-
E° = 0.77 - 0.42 = +0.35V = RTnFlnKeq = 0.0592nlogK
→keq = 100.350.0592 ≈ 106 ; ratio of complexation equilibrium constants for Fe3+ + Fe2+

Solubility Equlibria

• Often coupled to electrochemical reactions; sometimes best way to measure ksp values.

e.g:

Ag+ + e- → Ag ; E° = +0.799V
Ag+ + Cl- → AgCl ; Keq = Ksp-1 = 1010
What is E12° for AgCl(s) + e- →Ag(s) + Cl- ?
Under standard conditions, [Cl-] = 1M , [Ag+] = Ksp = 10-10M
→Ag is more easily oxidized in the presence of Cl-

Acid-Base Equilibria

• Many electrochemical reactions involve H+ or OH-, so E1/2 values are pH-dependent.

e.g.

Recall 3MnO42- → 2MnO4<sip>- + MnO2 is spontaneous at pH=0 ([H+] = 1M) from Latimer diagram or Frost plot, but 3MnO42- + 4H+ ↔ 2MnO4- + MnO2 + 2H2O (balanced disproportionation reaction)
• Left side is stabilized in base

4.5 Discussion questionsEdit

• Explain the simplified Pourbaix diagram for copper, shown below. Discuss the reactions that are implied by the lines, explain why they have the slopes they do, and use the diagram to determine the conditions under which Cu is passivated against corrosion.

• Use the information in the Pourbaix diagram to construct a Frost diagram for copper at pH 8.

4.6 ProblemsEdit

1. Balance the following redox reaction in acid solution and predict whether the reaction would become more or less spontaneous at higher pH.

MnO4-(aq) + N2O(g) = Mn2+(aq) + NO3-(aq)

2. Silver metal is not easily oxidized, and does not react with oxygen-saturated water. However, when excess NaCN is added to a suspension of silver particles, some silver dissolves. If oxygen is removed (e.g., by bubbling nitrogen through the solution), the dissolution reaction stops. Write a balanced equation for the dissolution reaction (hint: it is a redox reaction).

3. The Latimer potential diagram for iodine (in acid solutions) is given below.

H5IO6 +1.60 → IO3- +1.13 → IO- +1.44 → I2 +.535 → I-

(a) What is the potential for the IO3-/I- redox couple?

(b) Construct a Frost diagram and identify any species that are unstable with respect to disproportionation.

4. Referring to the Pourbaix diagram for Mn below:

(a) Write out the reaction corresponding to the line separating Mn2+ and Mn2O3. What is the slope of the line (give units)?

(b) Is Mn metal stable in water at any pH? If so, in what range of pH?

(c) What spontaneous reaction would you expect for an aqueous solution of MnO4- at pH 6?

(d) Describe an electrochemical procedure (specifying pH and potential) for making Mn3O4(s) from aqueous Mn2+.

(e) Label the regions of the diagram that correspond to corrosion and passivation of Mn metal.

4.7 ReferencesEdit

1. Redox Reactions in Daily Life: Common Examples. Answers. http://chem.answers.com/reactions/redox-reactions-in-daily-life-common-examples (accessed March 17, 2014).
2. Redox. Wikipedia. https://en.wikipedia.org/wiki/oxidisation (accessed March 17, 2014).
3. Equilibrium Constant and Cell Potential. Boundless. https://www.boundless.com/chemistry/electrochemistry/cell-potentials/equilibrium-constant-and-cell-potential/ (accessed March 17, 2014).