6.5 Redox Reactions

In the previous sections, we explored the formation of complexes and ligand substitution reactions. Now, we turn our attention to another crucial aspect of transition metal chemistry: redox reactions. Transition metals are renowned for their ability to exist in multiple oxidation states, which makes them highly effective at participating in redox reactions. This section focuses on using standard electrode potentials to predict the feasibility and direction of redox reactions involving transition elements and their ions.

Main Idea

The main idea is that transition elements can exist in various oxidation states. In redox reactions, the oxidation states of transition elements are changed when they undergo chemical reaction under suitable conditions. The feasibility of these reactions can be predicted by comparing the standard electrode potentials of the half-reactions involved. A reaction is feasible (spontaneous) if the overall cell potential is positive.

Core Concept

The core concept is that a redox reaction will occur spontaneously if the species with the higher (more positive) standard reduction potential acts as the oxidizing agent (undergoes reduction) and the species with the lower (more negative) standard reduction potential acts as the reducing agent (undergoes oxidation). This is based on the principle that electrons flow from the species with the lower reduction potential to the species with the higher reduction potential.

Detailed Explanation

The textbook discusses redox reactions involving transition elements, using standard electrode potential data to determine their feasibility.

1. Reaction of Acidified Manganate(VII) ions with Ethanedioate ions

The reaction between ethanedioate ions (C_2O_4^{2-}) and acidified manganate(VII) ions (MnO_4^-) is analyzed to find which way it will take place. The standard electrode potentials of the individual reactions are taken into consideration to check the feasibility of the overall reaction. Both reactions are taken in reduction mode according to IUPAC recommendations, and then we compare their standard reduction potentials.

2CO_{2(g)} + 2e^- ⟶ C_2O_4^{2-}_{(aq)} E^° = +0.49 V \] \[ MnO_{4(aq)}^- + 8H^+_{(aq)} + 5e^- ⟶ Mn^{2+}_{(aq)} + 4H_2O_{(l)} E^° = +1.52 V

• The question arises: does manganate(VII) oxidize ethanedioate to carbon dioxide, or does carbon dioxide oxidize manganese(II) ion to manganate(VII)?
• This problem can be resolved by comparing the standard reduction potentials of the two reactions. The reaction with the higher positive value of standard reduction potential is more likely to occur and proceed to the right and vice versa.
• The given reduction potential values indicate that manganate(VII) ion must reduce to manganese(II) ion and should act as the oxidizing agent, while oxalate should oxidize to carbon dioxide and act as the reducing agent.

Writing the Overall Equation: - For writing the overall chemical equation, the reaction with the lower standard electrode potential must be reversed and taken in oxidation mode because it is the oxidation reaction. The sign of its standard electrode potential is reversed while its magnitude remains the same. - The standard reduction potential values of reactions do not change with changing the coefficients. C_2O_4^{2-}_{(aq)} ⟶ 2CO_{2(g)} + 2e^- E^° = -0.49 V

2. Reaction of Acidified Manganate(VII) ions with Iron(II) ions

The textbook provides another example of a redox reaction involving transition metals.

The half-reactions are: MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O E^° = +1.52 V Fe^{2+} \longrightarrow Fe^{3+} + e^- E^° = +0.77 V

• MnO_4^- has a higher reduction potential (+1.52 V) than Fe^{2+} (+0.77 V).
• Therefore, MnO_4^- will be reduced (act as the oxidizing agent), and Fe^{2+} will be oxidized (act as the reducing agent).

3. Reaction of Iodide ions with Copper(II) ions

The textbook also provides an example involving copper.

The half-reactions are: 12I_{2(aq)} + e^- ⟶ I^-_{(aq)} E^° = +0.54 V Cu^{2+}_{(aq)} + e^- ⟶ Cu^+_{(aq)} E^° = +0.15 V

Analysis: - The higher standard reduction potential of the I_\frac{2}{I}^- couple (+0.54 V) indicates that this reaction is more likely to occur. - On the contrary, the lower standard reduction potential of the Cu^{2+}/Cu^+ reaction (+0.15 V) indicates that this reaction is less inclined to proceed in the forward direction, making copper(I) a better reducing agent. - However, this reaction proceeds in the forward direction because of the continuous precipitation of copper(I) iodide in the reaction, although the overall standard potential value is negative as shown: 2I^-_{(aq)} ⟶ I_{2(aq)} + 2e^- E^° = -0.54 V 2Cu^{2+}_{(aq)} + 2e^- ⟶ 2Cu^+_{(aq)} E^° = +0.15 V 2Cu^{2+}_{(aq)} + 2I^-_{(aq)} ⟶ 2Cu^+_{(aq)} + I_{2(aq)} E^° = -0.39 V - The amount of aqueous solution of iodine formed in this reaction can be determined by titrating it with sodium thiosulphate solution.

Intermediate Level

Transition metals participate in redox reactions by changing their oxidation states. The feasibility of these reactions is determined by comparing the standard reduction potentials of the half-reactions. The species with the higher (more positive) reduction potential acts as the oxidizing agent (is reduced), while the species with the lower reduction potential acts as the reducing agent (is oxidized). The overall cell potential is calculated using E^°_cell = E^°_cathode - E^°_anode. A positive value indicates a spontaneous reaction. For example, in the reaction between MnO_4^- and Fe^{2+}, MnO_4^- (+1.52 V) is reduced and Fe^{2+} (+0.77 V) is oxidized. The overall potential is 1.52 - 0.77 = 0.75 V, which is positive, so the reaction is spontaneous.

Important Points

• - Redox Reactions: Involve changes in oxidation states of transition metals.
• - Feasibility: Determined by standard reduction potentials.
• - Higher E^°: More likely to be reduced (act as oxidizing agent).
• - Lower E^°: More likely to be oxidized (act as reducing agent).
• - Positive E^°_cell: Spontaneous reaction.
• - Reaction of MnO_4^- with C_2O_4^{2-}: MnO_4^- is reduced, oxalate is oxidized.
• - Reaction of MnO_4^- with Fe^{2+}: MnO_4^- is reduced, Fe^{2+} is oxidized.
• - Reaction of Cu^{2+} with I^-: Proceeds due to precipitation of CuI, even though E^° is negative.

Remember

• - Redox reactions involve electron transfer.
• - Standard reduction potentials predict feasibility.
• - Higher E^° = stronger oxidizing agent.
• - Lower E^° = stronger reducing agent.
• - Positive E^°_cell = spontaneous.
• - MnO_4^- is a strong oxidizing agent.
• - Fe^{2+} is a reducing agent.
• - Precipitation can drive a non-spontaneous reaction.

Common Mistakes

• - Forgetting to reverse the sign of the reduction potential for the oxidation half-reaction. When a half-reaction is reversed to represent oxidation, the sign of its E^° must be changed.
• - Adding the potentials instead of subtracting them. Use E^°_cell = E^°_cathode - E^°_anode.
• - Assuming that a reaction with a negative E^° will not occur. It may still occur if the products are removed (e.g., by precipitation), shifting the equilibrium.
• - Confusing the oxidizing and reducing agents. The species with the higher reduction potential is the oxidizing agent (gets reduced).
• - Not balancing the electrons. The number of electrons lost must equal the number of electrons gained.

Quick Revision

Transition metals participate in redox reactions because they can exist in multiple oxidation states. The feasibility of these reactions is predicted by comparing standard reduction potentials. The species with the higher (more positive) reduction potential is reduced (acts as the oxidizing agent), while the species with the lower reduction potential is oxidized (acts as the reducing agent). A positive overall cell potential indicates a spontaneous reaction. The textbook provides examples such as the reaction of MnO_4^- with C_2O_4^{2-} and with Fe^{2+}, where MnO_4^- is reduced. In the reaction of Cu^{2+} with I^-, the reaction proceeds despite a negative standard potential because the precipitation of CuI removes the product and drives the equilibrium forward.

60-Second Revision

Transition metal redox reactions are predicted using standard reduction potentials. The species with the higher E^° is reduced (oxidizing agent), and the species with the lower E^° is oxidized (reducing agent). A positive E^°_cell means the reaction is spontaneous. Examples include MnO_4^- oxidizing Fe^{2+}. A reaction with a negative E^° can still occur if a product precipitates, as in the Cu^{2+} + I^- reaction.

Frequently Asked Questions (FAQs)

Q1. How can you deduce the feasibility of a redox reaction using electrode potentials? Answer: The feasibility of a redox reaction can be deduced by calculating the standard cell potential (E^°_cell). This is done by identifying the half-reaction with the higher (more positive) reduction potential (which will be the cathode, where reduction occurs) and the half-reaction with the lower (more negative) reduction potential (which will be the anode, where oxidation occurs). The cell potential is calculated using E^°_cell = E^°_cathode - E^°_anode. If E^°_cell is positive, the reaction is feasible (spontaneous) in the forward direction. If it is negative, the reaction is not feasible as written.

Q2. In the reaction between acidified manganate(VII) ions and ethanedioate ions, which species is the oxidizing agent? Answer: The oxidizing agent is the acidified manganate(VII) ion (MnO_4^-). Its standard reduction potential is +1.52 V, which is higher than that of the ethanedioate ion (-0.49 V for the oxidation half-reaction). Therefore, MnO_4^- is reduced to Mn^{2+} and causes the ethanedioate ion to be oxidized to carbon dioxide.

Q3. Why does the reaction between Cu^{2+} and I^- proceed even though the standard cell potential is negative? Answer: The reaction proceeds because the product, copper(I) iodide (CuI), is insoluble and precipitates out of the solution. This continuous removal of the product shifts the equilibrium to the right, according to Le Chatelier's principle, allowing the reaction to proceed despite the negative standard cell potential.

• Q4. What is the standard reduction potential of the MnO_4^-/Mn^{2+} half-reaction?
• Answer: The standard reduction potential of the MnO_4^-/Mn^{2+} half-reaction in acidic solution is +1.52 V.

Q5. In the reaction between MnO_4^- and Fe^{2+}, which species is oxidized? Answer: The Fe^{2+} ion is oxidized to Fe^{3+}. This is because MnO_4^- has a higher reduction potential (+1.52 V) than Fe^{3+}/Fe^{2+} (+0.77 V), so MnO_4^- is reduced and Fe^{2+} is oxidized.

Q6. What is the significance of the standard reduction potential value of a half-reaction? Answer: The standard reduction potential value indicates the tendency of a species to gain electrons and be reduced. A higher (more positive) value means a greater tendency to be reduced, making the species a stronger oxidizing agent. A lower (more negative) value means a lower tendency to be reduced, making the species a stronger reducing agent.

Q7. How do you write the overall cell reaction from two half-reactions? Answer: The overall cell reaction is obtained by adding the reduction half-reaction (at the cathode) and the oxidation half-reaction (the reverse of the reduction half-reaction at the anode). The electrons on both sides must cancel out. If necessary, the half-reactions are multiplied by integers to equalize the number of electrons before adding them.

• Q8. What is the standard reduction potential of the Cu^{2+}/Cu^+ half-reaction?
• Answer: The standard reduction potential of the Cu^{2+}/Cu^+ half-reaction is +0.15 V.

• Q9. What is the standard reduction potential of the I_2/I^- half-reaction?
• Answer: The standard reduction potential of the I_2/I^- half-reaction is +0.54 V.

Multiple Choice Questions (MCQs)

• Q1. In a redox reaction, which species is the oxidizing agent?
• A. The species with the lower reduction potential.
• B. The species with the higher reduction potential.
• C. The species that loses electrons.
• D. The species that is oxidized.
• Correct Answer: B
• Explanation: The species with the higher reduction potential has a greater tendency to gain electrons and is therefore the oxidizing agent (it gets reduced).

• Q2. What is the standard reduction potential of the MnO_4^-/Mn^{2+} half-reaction?
• A. +0.77 V
• B. +1.52 V
• C. +0.15 V
• D. +0.54 V
• Correct Answer: B
• Explanation: The standard reduction potential of MnO_4^- to Mn^{2+} is +1.52 V.

• Q3. In the reaction between MnO_4^- and Fe^{2+}, which species is oxidized?
• A. MnO_4^-
• B. Fe^{2+}
• C. Mn^{2+}
• D. Fe^{3+}
• Correct Answer: B
• Explanation: Fe^{2+} is oxidized to Fe^{3+} because MnO_4^- has a higher reduction potential and is reduced.

• Q4. Which of the following reactions is feasible based on standard electrode potentials?
• A. Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn
• B. Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu
• C. Ag + Cu^{2+} \rightarrow Ag^+ + Cu
• D. Au + Zn^{2+} \rightarrow Au^{3+} + Zn
• Correct Answer: B
• Explanation: Zinc has a lower reduction potential than copper, so zinc can reduce copper ions. This reaction has a positive cell potential and is spontaneous.

Q5. Why does the reaction between Cu^{2+} and I^- proceed? A. Because the standard cell potential is positive. B. Because the product CuI precipitates, shifting the equilibrium. C. Because Cu^{2+} is a stronger reducing agent than I^-. D. Because the reaction is exothermic. Correct Answer: B Explanation: The reaction proceeds because the precipitation of CuI removes the product, shifting the equilibrium to the right according to Le Chatelier's principle, even though the standard cell potential is negative.

• Q6. What is the standard reduction potential of the Fe^{3+}/Fe^{2+} half-reaction?
• A. +0.34 V
• B. +0.77 V
• C. +1.52 V
• D. +0.54 V
• Correct Answer: B
• Explanation: The standard reduction potential of Fe^{3+} to Fe^{2+} is +0.77 V.

• Q7. If E^°_cell is negative, the reaction is:
• A. Spontaneous
• B. Non-spontaneous
• C. At equilibrium
• D. Always explosive
• Correct Answer: B
• Explanation: A negative E^°_cell indicates that the reaction is non-spontaneous in the forward direction.

• Q8. In the reaction between MnO_4^- and C_2O_4^{2-}, which species is the reducing agent?
• A. MnO_4^-
• B. C_2O_4^{2-}
• C. Mn^{2+}
• D. CO_2
• Correct Answer: B
• Explanation: C_2O_4^{2-} is oxidized to CO_2, so it is the reducing agent.

• Q9. What is the standard reduction potential of the I_2/I^- half-reaction?
• A. +0.15 V
• B. +0.54 V
• C. +0.77 V
• D. +1.52 V
• Correct Answer: B
• Explanation: The standard reduction potential of I_2 to I^- is +0.54 V.

Q10. Which of the following is a correct statement about the reaction between Cu^{2+} and I^-? A. The standard cell potential is positive. B. The reaction is non-spontaneous. C. The reaction proceeds because CuI precipitates. D. The reaction is not a redox reaction. Correct Answer: C Explanation: The reaction proceeds because CuI precipitates, which shifts the equilibrium and makes the reaction feasible despite a negative standard cell potential.

Multiple Choice Questions (MCQs)

• Q1. In a redox reaction, which species is the oxidizing agent?
• A. The species with the lower reduction potential.
• B. The species with the higher reduction potential.
• C. The species that loses electrons.
• D. The species that is oxidized.
• Correct Answer: B
• Explanation: The species with the higher reduction potential has a greater tendency to gain electrons and is therefore the oxidizing agent.

• Q2. What is the standard reduction potential of the MnO_4^-/Mn^{2+} half-reaction?
• A. +0.77 V
• B. +1.52 V
• C. +0.15 V
• D. +0.54 V
• Correct Answer: B
• Explanation: The standard reduction potential of MnO_4^- to Mn^{2+} is +1.52 V.

• Q3. In the reaction between MnO_4^- and Fe^{2+}, which species is oxidized?
• A. MnO_4^-
• B. Fe^{2+}
• C. Mn^{2+}
• D. Fe^{3+}
• Correct Answer: B
• Explanation: Fe^{2+} is oxidized to Fe^{3+}.

• Q4. Which of the following reactions is feasible based on standard electrode potentials?
• A. Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn
• B. Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu
• C. Ag + Cu^{2+} \rightarrow Ag^+ + Cu
• D. Au + Zn^{2+} \rightarrow Au^{3+} + Zn
• Correct Answer: B
• Explanation: Zinc has a lower reduction potential than copper, so zinc can reduce copper ions.

• Q5. Why does the reaction between Cu^{2+} and I^- proceed?
• A. Because the standard cell potential is positive.
• B. Because the product CuI precipitates, shifting the equilibrium.
• C. Because Cu^{2+} is a stronger reducing agent than I^-.
• D. Because the reaction is exothermic.
• Correct Answer: B
• Explanation: The precipitation of CuI removes the product and shifts the equilibrium.

• Q6. What is the standard reduction potential of the Fe^{3+}/Fe^{2+} half-reaction?
• A. +0.34 V
• B. +0.77 V
• C. +1.52 V
• D. +0.54 V
• Correct Answer: B
• Explanation: The standard reduction potential of Fe^{3+} to Fe^{2+} is +0.77 V.

• Q7. If E^°_cell is negative, the reaction is:
• A. Spontaneous
• B. Non-spontaneous
• C. At equilibrium
• D. Always explosive
• Correct Answer: B
• Explanation: A negative E^°_cell indicates a non-spontaneous reaction.

• Q8. In the reaction between MnO_4^- and C_2O_4^{2-}, which species is the reducing agent?
• A. MnO_4^-
• B. C_2O_4^{2-}
• C. Mn^{2+}
• D. CO_2
• Correct Answer: B
• Explanation: C_2O_4^{2-} is oxidized to CO_2, so it is the reducing agent.

• Q9. What is the standard reduction potential of the I_2/I^- half-reaction?
• A. +0.15 V
• B. +0.54 V
• C. +0.77 V
• D. +1.52 V
• Correct Answer: B
• Explanation: The standard reduction potential of I_2 to I^- is +0.54 V.

• Q10. Which of the following is a correct statement about the reaction between Cu^{2+} and I^-?
• A. The standard cell potential is positive.
• B. The reaction is non-spontaneous.
• C. The reaction proceeds because CuI precipitates.
• D. The reaction is not a redox reaction.
• Correct Answer: C
• Explanation: The reaction proceeds because CuI precipitates, shifting the equilibrium.

Multiple Choice Questions (MCQs)

• Q1. Which of the following is a strong oxidizing agent?
• A. MnO_4^-
• B. Fe^{2+}
• C. C_2O_4^{2-}
• D. I^-
• Correct Answer: A
• Explanation: MnO_4^- has a high reduction potential (+1.52 V) and is a strong oxidizing agent.

• Q2. In the reaction MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O, the standard reduction potential is:
• A. +0.77 V
• B. +1.52 V
• C. +0.15 V
• D. +0.54 V
• Correct Answer: B
• Explanation: The standard reduction potential of this half-reaction is +1.52 V.

• Q3. In the reaction Fe^{3+} + e^- \rightarrow Fe^{2+}, the standard reduction potential is:
• A. +0.77 V
• B. +1.52 V
• C. +0.15 V
• D. +0.54 V
• Correct Answer: A
• Explanation: The standard reduction potential of Fe^{3+} to Fe^{2+} is +0.77 V.

• Q4. The reaction between Cu^{2+} and I^- has a negative standard cell potential. Why does it still occur?
• A. Because the temperature is high.
• B. Because CuI is insoluble and precipitates.
• C. Because the reaction is exothermic.
• D. Because I^- is a strong oxidizing agent.
• Correct Answer: B
• Explanation: The precipitation of CuI removes the product and drives the reaction forward.

• Q5. Which of the following is a reducing agent?
• A. MnO_4^-
• B. Fe^{2+}
• C. Cu^{2+}
• D. I_2
• Correct Answer: B
• Explanation: Fe^{2+} is a reducing agent because it can be oxidized to Fe^{3+}.

• Q6. What is the standard reduction potential of the Cu^{2+}/Cu^+ half-reaction?
• A. +0.15 V
• B. +0.34 V
• C. +0.77 V
• D. +1.52 V
• Correct Answer: A
• Explanation: The standard reduction potential of Cu^{2+} to Cu^+ is +0.15 V.

• Q7. In the reaction between MnO_4^- and C_2O_4^{2-}, which species is oxidized?
• A. MnO_4^-
• B. C_2O_4^{2-}
• C. Mn^{2+}
• D. CO_2
• Correct Answer: B
• Explanation: C_2O_4^{2-} is oxidized to CO_2.

• Q8. Which of the following is a correct statement about a spontaneous redox reaction?
• A. It has a negative cell potential.
• B. It has a positive cell potential.
• C. It does not involve electron transfer.
• D. It is always slow.
• Correct Answer: B
• Explanation: A spontaneous redox reaction has a positive cell potential.

• Q9. In the reaction between MnO_4^- and Fe^{2+}, the oxidation state of Mn changes from:
• A. +7 to +2
• B. +2 to +7
• C. +5 to +2
• D. +2 to +5
• Correct Answer: A
• Explanation: In MnO_4^-, Mn is in the +7 oxidation state. It is reduced to Mn^{2+}, which has a +2 oxidation state.

• Q10. The reaction between MnO_4^- and Fe^{2+} is used in:
• A. The Haber process.
• B. Redox titrations.
• C. The Contact process.
• D. Electroplating.
• Correct Answer: B
• Explanation: This reaction is commonly used in redox titrations to determine the concentration of iron(II) or permanganate.