# 2.1 Oxidation - Reduction Concepts This section defines oxidation and reduction in terms of electron transfer and changes in oxidation number. These reactions, called redox reactions, are fundamental to many chemical processes. ## Defining Oxidation and Reduction Historically, oxidation referred to reactions involving oxygen, and reduction to reactions involving hydrogen. A more comprehensive modern definition is based on the transfer of electrons. - **Oxidation** is the process where a substance **loses** electrons. - **Reduction** is the process where a substance **gains** electrons. These two processes always occur simultaneously in a chemical reaction. A reaction involving both oxidation and reduction is called a **redox reaction**. A helpful mnemonic to remember this is **OIL RIG**: - **O**xidation **I**s **L**oss (of electrons) - **R**eduction **I**s **G**ain (of electrons) ## Oxidation Number The **oxidation number** (or oxidation state) is a concept used to track electron transfer in redox reactions. > The **oxidation number** is the apparent charge, positive or negative, that an atom would have in a compound if all bonds were treated as purely ionic. ## Oxidation-Reduction in Terms of Change in Oxidation Number Redox processes can be defined based on how the oxidation number of an element changes during a reaction. - **Oxidation** is an **increase** in oxidation number. - **Reduction** is a **decrease** in oxidation number. **Example: Formation of Sodium Chloride** Consider the reaction between sodium metal and chlorine gas: $$2\mathrm{Na}_{(s)} + \mathrm{Cl}_{2(g)} \longrightarrow 2\mathrm{NaCl}_{(s)}$$ To analyze this as a redox reaction, assign oxidation numbers to each element: $$2\stackrel{0}{\mathrm{Na}}_{(s)} + \stackrel{0}{\mathrm{Cl}}_{2(g)} \longrightarrow 2\stackrel{+1}{\mathrm{Na}}\stackrel{-1}{\mathrm{Cl}}_{(s)}$$ - **Sodium (Na):** The oxidation number of Na increases from **0** to **+1**. This is **oxidation**. Sodium loses an electron. $$2\mathrm{Na}_{(s)} \longrightarrow 2\mathrm{Na}^{+}_{(g)} + 2e^{-} \quad \text{(Oxidation Half-Reaction)}$$ - **Chlorine (Cl):** The oxidation number of each Cl atom decreases from **0** to **-1**. This is **reduction**. Each chlorine atom gains an electron. $$\mathrm{Cl}_{2(g)} + 2e^{-} \longrightarrow 2\mathrm{Cl}^{-}_{(g)} \quad \text{(Reduction Half-Reaction)}$$ ## Disproportionation Reaction A **disproportionation reaction** (or self-oxidation-reduction reaction) is a specific type of redox reaction where the **same element** is simultaneously oxidized and reduced. **Example: Decomposition of Hydrogen Peroxide** In hydrogen peroxide ($\mathrm{H_2O_2}$), oxygen has an oxidation state of -1. When it decomposes, it forms water and oxygen gas. $$2\stackrel{+1}{\mathrm{H}}_2\stackrel{-1}{\mathrm{O}}_{2(l)} \longrightarrow 2\stackrel{+1}{\mathrm{H}}_2\stackrel{-2}{\mathrm{O}}_{(l)} + \stackrel{0}{\mathrm{O}}_{2(g)}$$ - **Reduction:** The oxidation number of some oxygen atoms decreases from **-1** (in $\mathrm{H_2O_2}$) to **-2** (in $\mathrm{H_2O}$). - **Oxidation:** The oxidation number of other oxygen atoms increases from **-1** (in $\mathrm{H_2O_2}$) to **0** (in $\mathrm{O_2}$). Since oxygen is both oxidized and reduced, this is a disproportionation reaction. ## Worked Examples ### Example 2.1: Zinc and Copper Sulfate Reaction When a zinc strip is placed in a blue copper(II) sulfate solution, the zinc becomes coated with a dark brown layer of copper, and the blue color of the solution fades. The overall ionic equation is: $$\mathrm{Zn}_{(s)} + \mathrm{Cu}^{2+}_{(aq)} \longrightarrow \mathrm{Zn}^{2+}_{(aq)} + \mathrm{Cu}_{(s)}$$ This reaction consists of two **half-reactions**: 1. **Oxidation:** Zinc atoms lose two electrons to become zinc ions. $$\mathrm{Zn}_{(s)} \longrightarrow \mathrm{Zn}^{2+}_{(aq)} + 2e^{-}$$ 2. **Reduction:** Copper(II) ions gain two electrons to become solid copper. $$\mathrm{Cu}^{2+}_{(aq)} + 2e^{-} \longrightarrow \mathrm{Cu}_{(s)}$$ ### Example 2.2: Combustion of Methane Identify the elements undergoing oxidation and reduction in the combustion of natural gas ($\mathrm{CH_4}$). $$\mathrm{CH}_{4(g)} + 2\mathrm{O}_{2(g)} \longrightarrow \mathrm{CO}_{2(g)} + 2\mathrm{H_2O}_{(g)}$$ **Solution:** 1. **Assign oxidation numbers** to all atoms in the reaction. - In $\mathrm{CH_4}$: H is +1, so C is **-4**. - In $\mathrm{O_2}$: O is in its elemental form, so its oxidation number is **0**. - In $\mathrm{CO_2}$: O is -2, so C is **+4**. - In $\mathrm{H_2O}$: H is +1, so O is **-2**. The reaction with oxidation numbers is: $$\stackrel{-4}{\mathrm{C}}\stackrel{+1}{\mathrm{H}}_{4(g)} + 2\stackrel{0}{\mathrm{O}}_{2(g)} \longrightarrow \stackrel{+4}{\mathrm{C}}\stackrel{-2}{\mathrm{O}}_{2(g)} + 2\stackrel{+1}{\mathrm{H}}_2\stackrel{-2}{\mathrm{O}}_{(g)}$$ 2. **Identify the changes:** - **Carbon (C):** The oxidation number increases from **-4** in $\mathrm{CH_4}$ to **+4** in $\mathrm{CO_2}$. This is **oxidation**. - **Oxygen (O):** The oxidation number decreases from **0** in $\mathrm{O_2}$ to **-2** in both $\mathrm{CO_2}$ and $\mathrm{H_2O}$. This is **reduction**. ## Concept Assessment Exercise 2.1 Use the oxidation number change method to identify the atoms undergoing oxidation or reduction. **Q1:** $\mathrm{Cl}_{2(g)} + 2\mathrm{NaOH}_{(aq)} \longrightarrow \mathrm{NaCl}_{(aq)} + \mathrm{NaClO}_{(aq)} + \mathrm{H_2O}_{(l)}$ **A:** - Assign oxidation numbers: $$\stackrel{0}{\mathrm{Cl}}_{2(g)} + 2\stackrel{+1}{\mathrm{Na}}\stackrel{-2}{\mathrm{O}}\stackrel{+1}{\mathrm{H}}_{(aq)} \longrightarrow \stackrel{+1}{\mathrm{Na}}\stackrel{-1}{\mathrm{Cl}}_{(aq)} + \stackrel{+1}{\mathrm{Na}}\stackrel{+1}{\mathrm{Cl}}\stackrel{-2}{\mathrm{O}}_{(aq)} + \stackrel{+1}{\mathrm{H}}_2\stackrel{-2}{\mathrm{O}}_{(l)}$$ - **Chlorine (Cl)** is both reduced (from 0 to -1 in NaCl) and oxidized (from 0 to +1 in NaClO). This is a **disproportionation reaction**. **Q2:** $4\mathrm{Fe}_{(s)} + 3\mathrm{O}_{2(g)} \longrightarrow 2\mathrm{Fe_2O}_{3(s)}$ **A:** - Assign oxidation numbers: $$4\stackrel{0}{\mathrm{Fe}}_{(s)} + 3\stackrel{0}{\mathrm{O}}_{2(g)} \longrightarrow 2\stackrel{+3}{\mathrm{Fe}}_2\stackrel{-2}{\mathrm{O}}_{3(s)}$$ - **Iron (Fe)** is oxidized (its oxidation number increases from 0 to +3). - **Oxygen (O)** is reduced (its oxidation number decreases from 0 to -2). **Q3:** $\mathrm{C_6H_{12}O_{6(s)}} + 6\mathrm{O}_{2(g)} \longrightarrow 6\mathrm{CO}_{2(g)} + 6\mathrm{H_2O}_{(l)}$ **A:** - Assign oxidation numbers: $$\stackrel{0}{\mathrm{C}}_6\stackrel{+1}{\mathrm{H}}_{12}\stackrel{-2}{\mathrm{O}}_{6(s)} + 6\stackrel{0}{\mathrm{O}}_{2(g)} \longrightarrow 6\stackrel{+4}{\mathrm{C}}\stackrel{-2}{\mathrm{O}}_{2(g)} + 6\stackrel{+1}{\mathrm{H}}_2\stackrel{-2}{\mathrm{O}}_{(l)}$$ - **Carbon (C)** is oxidized (its average oxidation number increases from 0 to +4). - **Oxygen (O)** from $\mathrm{O_2}$ is reduced (its oxidation number decreases from 0 to -2 in both $\mathrm{CO_2}$ and $\mathrm{H_2O}$). ## Summary - **Redox reactions** involve the transfer of electrons between chemical species. - **Oxidation** is the loss of electrons, resulting in an **increase** in oxidation number. - **Reduction** is the gain of electrons, resulting in a **decrease** in oxidation number. - The **oxidation number** is a tool for tracking electron flow in a reaction. - In a **disproportionation reaction**, the same element is both oxidized and reduced. | Process | Electron Transfer | Change in Oxidation Number | | :---------- | :---------------- | :-------------------------- | | **Oxidation** | Loss of $e^{-}$ | Increase | | **Reduction** | Gain of $e^{-}$ | Decrease | **Significance:** Redox reactions are central to many natural and industrial processes, including cellular respiration, photosynthesis, combustion, batteries, and metallurgy. The concepts of oxidation number are further applied in the and in determining the .

2.1 Oxidation Reduction Concepts | Electrochemistry | Chemistry 12