Redox Reactions and Pourbaix Diagrams

(a) How many chemical species (element, ion, or compound) are shown for this element?

Answer. (a). 4: Eu, Eu²⁺, Eu³⁺, Eu(OH)₃

(b) What oxidation states are represented on the diagram?

Answer. There are three oxidation states represented: 0, +2, +3

 

(a) organic-rich waterlogged soils (pH = 4.5, Eh = -0.10 V) Answer: Eu³⁺

(b) organic-rich salt water (pH = 9.0, Eh = -0.35 V) Answer: Eu(OH)₃

(c) oceans, lakes, streams near the surface of the water (pH = 7.0, Eh = 0.55 V) Answer: Eu³⁺ / Eu(OH)₃

(d) streams and lakes contaminated by acid mine drainage pH = 3.0, Eh = 0.80 V Answer: Eu³⁺

Answer: Any species whose stability field lies entirely below the H₂O/H₂ line will react with water to generate hydrogen gas. Thus both Eu metal and Eu²⁺ willl be oxidized by water and produce H₂ gas. Reactions:

2 Eu²⁺ (aq) + 2 H⁺(aq)---> 2 Eu³⁺ (aq) + H₂ (g) (acid solution)

2 Eu²⁺ (aq) + 4 OH^- (aq) + 2 H₂O ---> 2 Eu(OH)₃ + H₂ (g) (basic solution)

Eu (s) + 2 H⁺ (aq) ---> Eu²⁺ (aq) + H₂ (g) (acid solution)

Eu (s) + 2 H₂O ---> Eu²⁺ (aq) + H₂ (g) + 2 OH- (aq)

Answer: Any species whose stability field lies entirely above the O₂/H₂O line will react with water to generate oxygen gas. Since no species lies entirely above the O₂/H₂O line, none will react with water.
 
 

(a) What is E at pH = 0? Answer: pH 0 corresponds to hydrogen ion activity of 1. Since H⁺ doesn't appear in the half-reaction, and precipitation of Eu(OH)₃ doesn't begin until higher pH values, E = -0.48 V.

(b) At what pH does Eu(OH)₃ begin to precipitate when the metal ion concentration is 1 M? What is E at this pH?

Answer. The equilibrium constant K_sp refers to the chemical reaction

Eu(OH)₃ (s) ---> Eu³⁺ (aq) + 3 OH^- (aq)

K = K_sp = 2.1 x 10^-21 = [Eu³⁺][OH^-]³

(remember that solids don't appear in equilibrium constant expressions because their concentrations don't change).

When [Eu³⁺] = 1 M,

2.1 x 10^-21 = [1][OH^-]³

So when [OH^-] = 1.28 x 10^-7 M (pH = 7.1), Eu(OH)₃ begins to precipitate. The vertical line on the Pourbaix diagram divides the diagram for Eu(III) into two regions: the region below pH 7.1 where Eu³⁺ is the dominant species, and the region above pH 7.1 where Eu(OH)₃ is the dominant species.

Thus, at pH 7.1, just as precipitation begins, the E^o value is still -0.48 V

(c) What is the Eu³⁺ concentration at pH 14? What is E at pH 14?

Answer.

The Eu concentration at pH 14 is found from the K_sp for Eu(OH)₃.

Eu(OH)₃ (s) ---> Eu³⁺ (aq) + 3 OH^- (aq)

K = K_sp = 2.1 x 10^-21 = [Eu³⁺][OH^-]³

At pH 14, [OH^-] = 1 M so Eu³⁺ = 2.1 x 10^-21 M.

Now we have to use the Nernst equation since, as Eu(OH)₃ is precipitating, the potential for the reduction of Eu(III) to Eu(II) changes accordingly.

The reaction quotient Q has the same form as the equilibrium constant for the reaction, but refers to the actual (rather than equilibrium) concentrations available. For the reaction Eu³⁺ + e^- ---> Eu²⁺, Q is [ Eu²⁺] / [ Eu³⁺]. Thus,

Eu(OH)₂ is a soluble compound even at pH 14 so Eu²⁺ is still 1 M. However, Eu³⁺ = 2.1 x 10^-21 M. Substituting these values into the above equation, we find that E at pH 14 is -1.7 V.

(i) (acidic solution) Rh (s) ---> RhO₂ (s)

Answer. The rules for balancing redox half reactions in acid solution are:

(i) First balance the number of non-H and non-O atoms in the half reaction given (in black) below

(ii) Next balance the O's by adding water molecules as needed. (in red below)

(iii) Then balance the H's with H⁺ (in blue below)

(iv) Finally, balance the charge on each side of the equation by adding electrons (in green below)

Thus for the reaction above,

Rh (s) + 2 H₂O ---> RhO₂ (s) + 4 H⁺ + 4 e^-

(ii) (basic solution) Rh (s) ---> RhO₂ (s)

Answer. The simplest way to balance half-reactions in basic solution is to balance in acid, then convert all H⁺ ions to water by adding hydroxide ions.

Above we balanced the half-reaction for an acid solution:

Rh (s) + 2 H₂O ---> RhO₂ (s) + 4 H⁺ + 4 e^-

However, H⁺ is not dominant in basic solution, so let's add 4 hydroxide ions to each side of the equation:

Rh (s) + 2 H₂O + 4 OH^- ---> RhO₂ (s) + 4 H⁺ + 4 OH^- + 4 e^-

Combine the hydrogen and hydroxide ions on the right-hand side to give water:

Rh (s) + 2 H₂O + 4 OH^- ---> RhO₂ (s) + 4 H₂O + 4 e^-

Cancel water molecules on each side to get the final balanced half reaction in basic solution:

Rh (s) + 4 OH^- ---> RhO₂ (s) + 2 H₂O + 4 e^-

(iii) (basic solution) RhO₄^2- (aq) ---> Rh₂O₃ (s). Use same color code as established in part (i) to see how the rules work below:

balanced in acid: 2 RhO₄^2- + 10 H⁺ + 6 e^----> Rh₂O₃ + 5 H₂O

to balance in base: add 10 OH^- to each side of the equation, allow H⁺ to react with OH^-, then cancel water molecules:

2 RhO₄^2- + 10 H⁺ + 10 OH^- + 6 e^- ---> Rh₂O₃ + 5 H₂O + 10 OH^-

2 RhO₄^2- + 10 H₂O + 6 e^- ---> Rh₂O₃ + 5 H₂O + 10 OH^-

Answer: 2 RhO₄^2- + 5 H₂O + 6 e^- ---> Rh₂O₃ + 10 OH^-

(b) Give the letters (i, ii, or iii) of the half-reactions in part (a) that are oxidation half-reactions.

Answer: oxidation is defined as the loss of electrons; thus oxidation half-reactions have electrons on the right-hand side of the half-reaction. Reactions (i) and (ii) are oxidation half-reactions. Reaction (iii) is a reduction half-reaction since the reactant gains electrons (is reduced); thus electrons appear on the left-hand side.

(c) Give the formulas of the reactants in part (a) that are oxidizing agents.

Answer: oxidizing agents are the species that accept electrons. Thus RhO₄^2- in part (iii) is an oxidizing agent. [In reactions (i) and (ii), Rh is a reducing agent because it provides electrons]

(a) (acid solution): RhO₄^2- + Eu²⁺ ---> Rh₂O₃ + Eu³⁺

Answer. (i) Write the two half reactions. One will be an oxidation half-reaction and the other will be a reduction half-reaction. (ii) Then multipy each by a number that will result in cancellation of electrons. (iii) Add the two halves. Cancel like terms on each side of the equation.

(2 RhO₄^2- + 10 H⁺ + 6 e^- ---> Rh₂O₃ + 5 H₂O) x 1

(Eu²⁺ ---> Eu³⁺ + e^- ) x 6

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6 Eu²⁺ + 2 RhO₄^2- + 10 H⁺ ---> Rh₂O₃ + 5 H₂O+ 6 Eu³⁺

(b) (basic solution): MnO₄^- + Fe(OH)₂ ---> MnO₂ + Fe(OH)₃

(MnO₄^- + 2 H₂O + 3e^- ---> MnO₂ + 4 OH^- ) x 1

(Fe(OH)₂ + OH^- ---> Fe(OH)₃ + e^- ) x 3

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MnO₄^- + 2 H₂O + 3 Fe(OH)₂ ---> MnO₂ + 3 Fe(OH)₃ + OH^-

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Last revised: March 15 2001 by M. F. Richardson
© Brock University, 2001