# Introduction to Inorganic Chemistry/Redox Stability and Redox Reactions

## Chapter 4: Redox Stability and Redox Reactions

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

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

• Transition metals 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 (in industry and biology)

• 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

## 4.1 Balancing redox reactions

I- is oxidized to IO3- by MnO4-(→Mn2+) How do you balance this equation? The most reliable procedure is 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 are Mn and I)

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 Mn7+ or Mn2+)

Step 3: Combine the half reactions

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

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

Cancel out and balance the hydrogens, electrons, and water:

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

The final equation:
18H+ + 6MnO4- + 5I-5IO3- + 6Mn2+ + 9H2O
Make sure the ELEMENTS and CHARGES are balanced for BOTH sides.

Step 4: 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 become

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

Another 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 S or O)

## 4.2 Electrochemical equilibria

For systems that are equilibrium, ΔG°= -nFE°cell where o means that it 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 diagrams

Three kinds of redox stability diagrams:

• Latimer, Frost, and Pourbaix
• All contain similar information, but different representations are useful in different situations.

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: E° values for successive redox reactions (from highest to lowest oxidation state)

Example:

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

e.g.

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

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

e.g.

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

Unstable species have a lower number to the left than to the right.

e.g.

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:

MnO42-→MnO2 ;

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

Moral: we need to consider all possible disproportionation reactions (this is easier with a Frost diagram, which will soon be discussed)

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

Frost diagrams: easily identify stable and unstable oxidation states, 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.
e.g. Fe Pourbaix Diagram

Pourbaix-plot E vs. pH:
• pure redox reactions are horizontal lines
• pure acid-base reactions are vertical lines
• mixed reactions have slope -0.0592 (# H+# e-)
Key Equilibria:
1. Fe2+ + 2e- → Fe(s) (no pH dependence)
2. Fe3+ + e- → Fe2+ (ditto-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
• It can be catholically protected by applying a potential below its stability line
e.g. at pH of 7, it needs to hold a potential ≤ -0.44V
• 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 equilibria

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 questions

• 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 Problems

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.