Unit 6: Redox Reactions & Agents
Combining half-equations and identifying the roles of oxidising and reducing agents.
6.5 Deriving Equations for Redox Reactions
Since oxidation and reduction always happen together, we can combine their respective half-equations to create the overall equation for a redox reaction. The key principle is the conservation of electrons: the number of electrons lost in the oxidation half-equation must equal the number of electrons gained in the reduction half-equation.
Procedure:
- Write the two balanced half-equations: One for the oxidation process and one for the reduction process.
- Equalise the electrons: Multiply one or both half-equations by an integer so that the number of electrons ($e^-$) is the same in both.
- Combine the equations: Add the two new half-equations together.
- Simplify: Cancel out any species that appear on both the reactant and product sides of the final equation (usually electrons, but sometimes also $H^+$ or $H_2O$).
Example: Reaction between iron(II) ions ($Fe^{2+}$) and permanganate ions ($MnO_4^-$) in acidic solution.
- Oxidation: $Fe^{2+} \rightarrow Fe^{3+} + e^-$
- Reduction: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
- The reduction involves 5 electrons, while the oxidation involves 1. We must multiply the oxidation half-equation by 5 to equalise the electrons:
$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-$ - Now, add the modified oxidation half-equation to the reduction half-equation:
$5Fe^{2+} + MnO_4^- + 8H^+ + 5e^- \rightarrow 5Fe^{3+} + 5e^- + Mn^{2+} + 4H_2O$ - Finally, cancel the 5 electrons ($5e^-$) from both sides to get the overall balanced redox equation:
$5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$
Solved Examples:
- Combine the half-equations: $Zn \rightarrow Zn^{2+} + 2e^-$ and $Cu^{2+} + 2e^- \rightarrow Cu$.
Solution: The electrons are already balanced (2 on each side). Adding them together and cancelling the electrons gives: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$. - Derive the overall equation for the reaction between iodide ions ($I^-$) and dichromate ions ($Cr_2O_7^{2-}$). Half-equations: $2I^- \rightarrow I_2 + 2e^-$ and $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$.
Solution: Multiply the oxidation half-equation by 3 to get 6 electrons: $6I^- \rightarrow 3I_2 + 6e^-$. Combine and cancel electrons: $6I^- + Cr_2O_7^{2-} + 14H^+ \rightarrow 3I_2 + 2Cr^{3+} + 7H_2O$. - What is the main principle when combining half-equations?
Solution: The number of electrons lost in oxidation must equal the number of electrons gained in reduction. - Combine: $Mg \rightarrow Mg^{2+} + 2e^-$ and $O_2 + 4e^- \rightarrow 2O^{2-}$.
Solution: Multiply the oxidation half-equation by 2: $2Mg \rightarrow 2Mg^{2+} + 4e^-$. Combine and cancel electrons: $2Mg + O_2 \rightarrow 2Mg^{2+} + 2O^{2-}$ (which simplifies to $2MgO$). - Why are electrons cancelled out in the final equation?
Solution: Electrons are transferred in the reaction; they are not free reactants or products. The overall equation shows only the initial reactants and final products. - Combine the disproportionation of copper(I): $Cu^+ + e^- \rightarrow Cu$ and $Cu^+ \rightarrow Cu^{2+} + e^-$.
Solution: Electrons are already balanced. Add the equations: $Cu^+ + Cu^+ \rightarrow Cu + Cu^{2+}$. This gives the final equation: $2Cu^+ \rightarrow Cu + Cu^{2+}$. - Derive the equation for the reaction between $SO_2$ and $Br_2$. Half-equations: $SO_2 + 2H_2O \rightarrow SO_4^{2-} + 4H^+ + 2e^-$ and $Br_2 + 2e^- \rightarrow 2Br^-$.
Solution: Electrons are balanced. Combine and cancel: $SO_2 + 2H_2O + Br_2 \rightarrow SO_4^{2-} + 4H^+ + 2Br^-$. - What species might need to be cancelled from both sides besides electrons?
Solution: Hydrogen ions ($H^+$) and water molecules ($H_2O$). - Combine: $Al \rightarrow Al^{3+} + 3e^-$ and $Fe^{2+} + 2e^- \rightarrow Fe$.
Solution: Find the lowest common multiple of 2 and 3, which is 6. Multiply the first equation by 2 and the second by 3: $2Al \rightarrow 2Al^{3+} + 6e^-$ and $3Fe^{2+} + 6e^- \rightarrow 3Fe$. Combine and cancel: $2Al + 3Fe^{2+} \rightarrow 2Al^{3+} + 3Fe$. - What is the purpose of deriving the overall redox equation?
Solution: To get a single, balanced chemical equation that represents the complete reaction, showing the correct stoichiometric ratios of all reactants and products.
6.6 Oxidising Agents and Reducing Agents
In any redox reaction, one substance causes another to be oxidised, while another substance causes the first to be reduced. These are known as oxidising and reducing agents.
Definitions:
- An oxidising agent (or oxidant) is a species that accepts electrons and gets reduced in the process. It causes the oxidation of another species by taking its electrons.
- A reducing agent (or reductant) is a species that donates electrons and gets oxidised in the process. It causes the reduction of another species by giving it electrons.
Think of it this way: The agent is what it does to the other substance. An oxidising agent oxidises something else.
Example: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$
- $Zn$ is oxidised (loses electrons). Therefore, $Zn$ is the reducing agent.
- $Cu^{2+}$ is reduced (gains electrons). Therefore, $Cu^{2+}$ is the oxidising agent.
Common oxidising agents are typically species with elements in high oxidation states (e.g., $MnO_4^-$, $Cr_2O_7^{2-}$) or highly electronegative elements (e.g., $O_2$, $Cl_2$). Common reducing agents are typically metals (e.g., $Na$, $Zn$) or species with elements in low oxidation states (e.g., $I^-$, $Fe^{2+}$).
Solved Examples:
- In the reaction $2Mg + O_2 \rightarrow 2MgO$, identify the oxidising agent.
Solution: Oxygen ($O_2$) is reduced (gains electrons), so it is the oxidising agent. - What is the role of magnesium in the reaction $2Mg + O_2 \rightarrow 2MgO$?
Solution: Magnesium ($Mg$) is oxidised (loses electrons), so it is the reducing agent. - Define a reducing agent in terms of oxidation number.
Solution: A reducing agent is a species that contains an element that increases in oxidation number during a reaction. - Identify the reducing agent in the reaction: $Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$.
Solution: The iodide ion ($I^-$) is oxidised (from -1 to 0), so it is the reducing agent. - Is potassium permanganate ($KMnO_4$) typically an oxidising or reducing agent? Why?
Solution: An oxidising agent. The manganese is in its highest oxidation state (+7) and is therefore readily reduced. - What happens to an oxidising agent during a redox reaction?
Solution: It gets reduced (it gains electrons). - Identify the oxidant and reductant in: $5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$.
Solution: $Fe^{2+}$ is oxidised, so it is the reductant (reducing agent). $MnO_4^-$ is reduced, so it is the oxidant (oxidising agent). - Are active metals like sodium and zinc generally good oxidising or reducing agents?
Solution: Good reducing agents, because they are easily oxidised (they readily lose electrons). - Define an oxidising agent in terms of electron transfer.
Solution: An oxidising agent is an electron acceptor. - In the reaction $H_2 + Cl_2 \rightarrow 2HCl$, what is the oxidising agent?
Solution: Chlorine ($Cl_2$) is reduced (oxidation number goes from 0 to -1), so it is the oxidising agent.
Knowledge Check (20 Questions)
Answer: Reduced.
Answer: The number of electrons lost must equal the number of electrons gained.
Answer: Sodium ($Na$).
Answer: Donor.
Answer: Electrons ($e^-$).
Answer: It increases.
Answer: Oxidising agent.
Answer: Multiply the first half-equation by 2.
Answer: Acceptor.
Answer: The hydrogen ion ($H^+$).
Answer: It reduces the other substance (causes it to gain electrons).
Answer: Simplify the equation by cancelling any species that appear on both sides.
Answer: Oxidising agents.
Answer: It gets oxidised (it loses electrons).
Answer: $4Al + 3O_2 \rightarrow 4Al^{3+} + 6O^{2-}$ (or $2Al_2O_3$).
Answer: An oxidant.
Answer: The iodide ion ($I^-$).
Answer: It decreases.
Answer: A reductant.
Answer: The iron(III) ion ($Fe^{3+}$).