# 2.7 The Relative Reactivity of Species as Oxidizing Agents or Reducing Agents Electrode potential values, specifically standard reduction potentials ($E^\circ$), determine the relative reactivity of chemical species as oxidizing or reducing agents. These values quantify the tendency of a species to gain electrons and be reduced. ### Understanding Reduction Potentials **Standard Reduction Potential ($E^\circ$)**: A measure of the tendency of a chemical species to be reduced, meaning to gain electrons. It is measured in volts (V) against a standard hydrogen electrode (SHE) as the reference. **Oxidizing Agent**: A species that causes another species to be oxidized by gaining electrons itself. The oxidizing agent is reduced in the process. **Reducing Agent**: A species that causes another species to be reduced by losing electrons itself. The reducing agent is oxidized in the process. ### Relating Reduction Potential to Reactivity The magnitude and sign of the standard reduction potential ($E^\circ$) directly indicate the strength of a species as an oxidizing or reducing agent: * **Stronger Oxidizing Agents**: Species with higher (more positive) reduction potentials have a greater tendency to gain electrons. These species are strong oxidizing agents. * **Stronger Reducing Agents**: Species with lower (more negative) reduction potentials have a greater tendency to lose electrons. Their reduced form is easily oxidized, making them strong reducing agents. This relationship is summarized in the table below: | Characteristic | High $E^\circ$ Value (More Positive) | Low $E^\circ$ Value (More Negative) | | :------------- | :-------------------------------- | :------------------------------ | | **Tendency to** | Gain electrons (be reduced) | Lose electrons (be oxidized) | | **As Oxidizing Agent** | Strong oxidizing agent | Weak oxidizing agent | | **As Reducing Agent** | Weak reducing agent | Strong reducing agent | --- ## Worked Examples The following examples demonstrate how to compare the relative strengths of oxidizing and reducing agents using standard reduction potentials. ### Example 1: Comparing Copper and Zinc Ions Consider the standard reduction potentials for the following half-reactions: $$ \begin{array}{llll} \mathrm{Cu^{2+}}(aq) + 2\mathrm{e^-} & \longrightarrow & \mathrm{Cu}(s) & E^\circ = +0.34\ \text{V} \\ \mathrm{Zn^{2+}}(aq) + 2\mathrm{e^-} & \longrightarrow & \mathrm{Zn}(s) & E^\circ = -0.76\ \text{V} \end{array} $$ 1. **Identify the species and their reduction potentials:** * $\mathrm{Cu^{2+}}$ has $E^\circ = +0.34\ \text{V}$ * $\mathrm{Zn^{2+}}$ has $E^\circ = -0.76\ \text{V}$ 2. **Compare the reduction potentials:** * $E^\circ(\mathrm{Cu^{2+}}) > E^\circ(\mathrm{Zn^{2+}})$ since $+0.34\ \text{V} > -0.76\ \text{V}$ 3. **Determine relative reactivity:** * $\mathrm{Cu^{2+}}$ ions have a higher reduction potential, so they have a greater tendency to gain electrons and undergo reduction. * Therefore, $\mathrm{Cu^{2+}}$ ions act as a **stronger oxidizing agent** than $\mathrm{Zn^{2+}}$ ions. * Conversely, $\mathrm{Zn}$ metal (the reduced form of $\mathrm{Zn^{2+}}$) is a **stronger reducing agent** than $\mathrm{Cu}$ metal. This means $\mathrm{Zn}$ is more easily oxidized than $\mathrm{Cu}$. ### Example 2: Potassium Permanganate and Iron(II) Sulfate Consider the reaction between potassium permanganate ($\mathrm{KMnO_4}$) and iron(II) sulfate ($\mathrm{FeSO_4}$) in acidic solution. The relevant half-reactions and their standard potentials are: $$ \begin{array}{llll} \mathrm{MnO_4^-}(aq) + 8\mathrm{H^+}(aq) + 5\mathrm{e^-} & \longrightarrow & \mathrm{Mn^{2+}}(aq) + 4\mathrm{H_2O}(l) & E^\circ = +1.52\ \text{V} \\ \mathrm{Fe^{3+}}(aq) + \mathrm{e^-} & \longrightarrow & \mathrm{Fe^{2+}}(aq) & E^\circ = +0.77\ \text{V} \end{array} $$ *Note: The second half-reaction is written as a reduction for comparison. When $\mathrm{Fe^{2+}}$ acts as a reducing agent, it gets oxidized to $\mathrm{Fe^{3+}}$: $\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-}$. The oxidation potential for this is $-E^\circ(\mathrm{Fe^{3+}}/\mathrm{Fe^{2+}}) = -0.77\ \text{V}$. For direct comparison of oxidizing strength, we compare the reduction potentials.* 1. **Identify the species and their reduction potentials:** * $\mathrm{MnO_4^-}$ has $E^\circ = +1.52\ \text{V}$ * $\mathrm{Fe^{3+}}$ has $E^\circ = +0.77\ \text{V}$ 2. **Compare the reduction potentials:** * $E^\circ(\mathrm{MnO_4^-}) > E^\circ(\mathrm{Fe^{3+}}/\mathrm{Fe^{2+}})$ since $+1.52\ \text{V} > +0.77\ \text{V}$ 3. **Determine relative reactivity:** * $\mathrm{MnO_4^-}$ ions have a significantly higher reduction potential, so they have a very strong tendency to gain electrons and be reduced. * Therefore, $\mathrm{MnO_4^-}$ ions act as a **strong oxidizing agent** in this reaction. * $\mathrm{Fe^{2+}}$ ions have a relatively lower reduction potential for the $\mathrm{Fe^{3+}}/\mathrm{Fe^{2+}}$ couple, so they are readily oxidized to $\mathrm{Fe^{3+}}$ (they tend to lose electrons). * Thus, $\mathrm{Fe^{2+}}$ ions act as a **strong reducing agent**. --- ## Possible Questions and Answers **Q:** How do standard reduction potentials help in determining the strength of an oxidizing agent? **A:** A higher (more positive) standard reduction potential indicates a greater tendency for a species to gain electrons and be reduced. Therefore, species with higher $E^\circ$ values are stronger oxidizing agents. --- **Q:** Given the following standard reduction potentials, identify the strongest oxidizing agent and the strongest reducing agent: $$ \begin{array}{ll} \mathrm{A^{2+}} + 2\mathrm{e^-} \longrightarrow \mathrm{A} & E^\circ = +0.80\ \text{V} \\ \mathrm{B^+} + \mathrm{e^-} \longrightarrow \mathrm{B} & E^\circ = -0.15\ \text{V} \\ \mathrm{C^{3+}} + 3\mathrm{e^-} \longrightarrow \mathrm{C} & E^\circ = -1.66\ \text{V} \end{array} $$ **A:** * **Strongest Oxidizing Agent:** $\mathrm{A^{2+}}$ (has the highest $E^\circ = +0.80\ \text{V}$) * **Strongest Reducing Agent:** $\mathrm{C}$ (is the reduced form of $\mathrm{C^{3+}}$, which has the lowest $E^\circ = -1.66\ \text{V}$, indicating $\mathrm{C}$ is most easily oxidized) --- **Q:** What is the relationship between the reduction potential of a species and its ability to act as a reducing agent? **A:** Species that have a low (more negative) reduction potential are less likely to be reduced. Their reduced forms (the element or ion on the product side of the reduction half-reaction) have a strong tendency to lose electrons and be oxidized, thus acting as strong reducing agents. ---

2.7 The Relative Reactivity Of Species As Oxidizing Agents Or Reducing Agents | Electrochemistry | Chemistry 12