# Introduction to Redox Reactions Redox reactions are fundamental chemical processes involving the transfer of electrons between atoms, leading to changes in their oxidation states. Transition elements are particularly adept at participating in these reactions due to their ability to exist in multiple oxidation states. * **Oxidation State**: A number assigned to an element in a compound that represents the number of electrons lost or gained by an atom of that element. * **Reduction**: * **Definition**: A process where an atom, ion, or molecule **gains one or more electrons**. * **Effect on Oxidation State**: The oxidation state of the species **decreases**. * **Oxidation**: * **Definition**: A process where an atom, ion, or molecule **loses one or more electrons**. * **Effect on Oxidation State**: The oxidation state of the species **increases**. * **Oxidizing Agent (Oxidant)**: * **Role**: The species that **accepts electrons** (causes oxidation in another species). * **Change**: It undergoes **reduction** itself. * **Reducing Agent (Reductant)**: * **Role**: The species that **donates electrons** (causes reduction in another species). * **Change**: It undergoes **oxidation** itself. ### Feasibility of Redox Reactions using Standard Electrode Potentials ($E^\circ$) The feasibility of a redox reaction under standard conditions can be predicted by comparing the standard electrode potentials ($E^\circ$) of the half-reactions involved. According to IUPAC recommendations, electrode potentials are typically presented as reduction potentials. * **Rule**: The reaction with a **more positive (or less negative) standard reduction potential ($E^\circ$)** is more likely to occur as a **reduction** (i.e., the species will act as an oxidizing agent). Conversely, the reaction with a **less positive (or more negative) standard reduction potential ($E^\circ$)** is more likely to occur as an **oxidation**. * **Overall Cell Potential ($E^\circ_{cell}$)**: For a spontaneous (feasible) redox reaction, the overall standard cell potential, calculated as $E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation}$, must be **positive ($E^\circ_{cell} > 0$)**. ### i. Reaction of Acidified Manganate(VII) ions with Ethanedioate ions Let's analyze the feasibility of the reaction between acidified manganate(VII) ions ($\mathrm{MnO_4^-}$) and ethanedioate ions ($\mathrm{C_2O_4^{2-}}$). **Given Half-Reactions (in reduction mode):** 1. $$ 2 \mathrm{CO}_{2(\mathrm{~g})} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}{ }_{(\mathrm{aq})} \quad E^{\circ} = +0.49 \mathrm{~V} $$ 2. $$ \mathrm{MnO}_{4}{ }^{-}{ }_{(\mathrm{aq})} + 8 \mathrm{H}^{+}{ }_{(\mathrm{aq})} + 5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})} + 4 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \quad E^{\circ} = +1.52 \mathrm{~V} $$ **Determining Oxidizing and Reducing Agents:** * Comparing the $E^\circ$ values, $+1.52 \mathrm{~V}$ for $\mathrm{MnO_4^-}$ is much higher than $+0.49 \mathrm{~V}$ for $\mathrm{CO_2}$. * This indicates that $\mathrm{MnO_4^-}$ acts as a **stronger oxidizing agent**. * $\mathrm{C_2O_4^{2-}}$ will be **oxidized** to $\mathrm{CO_2}$, acting as the **reducing agent**. **Balancing and Combining:** To balance electrons, multiply the oxidation half-equation by 5 and the reduction half-equation by 2. $$ \begin{array}{rlr} 2 \mathrm{MnO}_{4}{ }^{-}{ }_{(\mathrm{aq})}+16 \mathrm{H}^{+}{ }_{(\mathrm{aq})}+10 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+8 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} & E^{\circ}=+1.52 \mathrm{~V} \\ 5 \mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}{ }_{(\mathrm{aq})} & \longrightarrow 10 \mathrm{CO}_{2(\mathrm{~g})}+10 \mathrm{e}^{-} & E^{\circ}=-0.49 \mathrm{~V} \\ \hline 2 \mathrm{MnO}_{4}{ }^{-}{ }_{(\mathrm{aq})}+5 \mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}{ }_{(\mathrm{aq})}+16 \mathrm{H}^{+}{ }_{(\mathrm{aq})} \longrightarrow 2 \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+10 \mathrm{CO}_{2(\mathrm{~g})}+8 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} & E^{\circ}_{cell}=+1.03 \mathrm{~V} \end{array} $$ **Feasibility**: $E^\circ_{cell} = +1.03 \mathrm{~V}$ (Positive), so the reaction is **feasible**. ### ii. Reaction of Manganate(VII) ion with Iron(II) ion Consider the reaction between manganate(VII) ions ($\mathrm{MnO_4^-}$) and iron(II) ions ($\mathrm{Fe^{2+}}$). **Given Half-Reactions:** 1. $$ \mathrm{MnO}_{4}{ }^{-}{ }_{(\mathrm{aq})} + 8 \mathrm{H}^{+}{ }_{(\mathrm{aq})} + 5 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})} + 4 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} \quad E^{\circ} = +1.52 \mathrm{~V} $$ 2. $$ \mathrm{Fe}^{3+}{ }_{(\mathrm{aq})} + \mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}{ }_{(\mathrm{aq})} \quad E^{\circ} = +0.77 \mathrm{~V} $$ **Overall Equation:** $$ \begin{array}{rlr} \mathrm{MnO}_{4}{ }^{-}{ }_{(\mathrm{aq})}+8 \mathrm{H}^{+}{ }_{(\mathrm{aq})}+5 \mathrm{e}^{-} & \longrightarrow \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} & E^{\circ}=+1.52 \mathrm{~V} \\ 5 \mathrm{Fe}^{2+}{ }_{(\mathrm{aq})} & \longrightarrow 5 \mathrm{Fe}^{3+}{ }_{(\mathrm{aq})}+5 \mathrm{e}^{-} & E^{\circ}=-0.77 \mathrm{~V} \\ \hline \mathrm{MnO}_{4}{ }^{-}{ }_{(\mathrm{aq})}+5 \mathrm{Fe}^{2+}{ }_{(\mathrm{aq})}+8 \mathrm{H}^{+}{ }_{(\mathrm{aq})} \longrightarrow \mathrm{Mn}^{2+}{ }_{(\mathrm{aq})}+5 \mathrm{Fe}^{3+}{ }_{(\mathrm{aq})}+4 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{l})} & E^{\circ}_{cell}=+0.75 \mathrm{~V} \end{array} $$ **Feasibility**: $E^\circ_{cell} = +0.75 \mathrm{~V}$ (Positive), so the reaction is **feasible**. ### iii. Reaction Between Copper(II) ion and Iodide ion When aqueous copper(II) sulfate is mixed with excess potassium iodide: $$ 2 \mathrm{Cu}^{2+}{ }_{(\mathrm{aq})}+4 \mathrm{I}^{-}{ }_{(\mathrm{aq})} \longrightarrow \underset{\text{white ppt}}{2 \mathrm{CuI}_{(\mathrm{s})}} + \underset{\substack{\mathrm{I}_{2(\mathrm{aq})} \\ \text{red-brown solution}}}{\mathrm{I}_{2(\mathrm{aq})}} $$ **Feasibility (Special Case)**: The calculated standard cell potential is **negative ($E^\circ_{cell} = -0.39 \mathrm{~V}$)**. However, this reaction proceeds because $\mathrm{Cu}^{+}$ ions immediately precipitate as **copper(I) iodide ($\mathrm{CuI}_{(\mathrm{s})}$)**. According to Le Chatelier's Principle, the removal of $\mathrm{Cu}^{+}$ ions shifts the equilibrium to the right, driving the reaction forward despite the negative standard potential. ### Practical Activity 6.1: Redox Titration (Manganate(VII) and Iron(II)) This reaction is used to determine the concentration of $\mathrm{Fe}^{2+}$ ions. * **Titrant**: Potassium manganate(VII) ($\mathrm{KMnO_4}$) in the burette. * **Analyte**: Acidified $\mathrm{Fe}^{2+}$ solution in the conical flask. * **Endpoint**: The first permanent pale pink color. $\mathrm{KMnO_4}$ acts as its own indicator. ## Worked Examples ### Worked Example 6.1 An excess of potassium iodide solution was added to $30.00 \mathrm{~cm}^{3}$ of copper(II) sulfate solution. The solution was titrated against $0.05 \mathrm{~mol\,dm}^{-3}$ sodium thiosulfate solution. The volume of thiosulfate used was $10 \mathrm{~cm}^{3}$. Calculate the concentration of the copper(II) sulfate solution. **Solution:** 1. **Moles of $\mathrm{S_2O_3^{2-}}$**: $0.05 \times (10/1000) = 0.0005 \mathrm{~mol}$ 2. **Stoichiometry**: $2 \mathrm{Cu}^{2+} \equiv 1 \mathrm{I}_2 \equiv 2 \mathrm{S_2O_3^{2-}}$ Therefore, moles of $\mathrm{Cu}^{2+} = \text{moles of } \mathrm{S_2O_3^{2-}} = 0.0005 \mathrm{~mol}$ 3. **Concentration of $\mathrm{CuSO_4}$**: $M = \frac{0.0005}{0.030} = \mathbf{0.0167 \mathrm{~mol\,dm}^{-3}}$ ## Possible Questions/Answers * **Q:** How does the oxidation state of Chromium change when $Cr^{3+}$ is oxidized to $CrO_4^{2-}$? **A:** In $Cr^{3+}$, the oxidation state is +3. In $CrO_4^{2-}$, let $x$ be the oxidation state of Cr: $x + 4(-2) = -2 \Rightarrow x - 8 = -2 \Rightarrow x = +6$. The oxidation state increases from +3 to +6.