Oxidation in terms of addition of oxygen or removal of hydrogen: Oxidation: The process where a substance gains oxygen or loses hydrogen*. Example (addition of oxygen)*: $\text{C(s)} + \text{O}_2\text{(g)} \to \text{CO}_2\text{(g)}$ (Carbon is oxidized) Example (removal of hydrogen)*: $\text{CH}_3\text{CH}_2\text{OH(l)} \to \text{CH}_3\text{CHO(l)} + \text{H}_2\text{(g)}$ (Ethanol is oxidized to ethanal) Reduction: The process where a substance loses oxygen or gains hydrogen*. Example (removal of oxygen)*: $\text{CuO(s)} + \text{H}_2\text{(g)} \to \text{Cu(s)} + \text{H}_2\text{O(l)}$ (Copper(II) oxide is reduced) Example (addition of hydrogen)*: $\text{CH}_2\text{=CH}_2\text{(g)} + \text{H}_2\text{(g)} \to \text{CH}_3\text{CH}_3\text{(g)}$ (Ethene is reduced to ethane) Oxidation and reduction in terms of electron transfer: Oxidation: The process of losing electrons*. Reduction: The process of gaining electrons*. Mnemonic*: OIL RIG (Oxidation Is Loss, Reduction Is Gain of electrons). Use of Oxidation Numbers: An oxidation number* (or oxidation state) is a hypothetical charge an atom would have if all bonds were purely ionic. It helps track electron transfer in redox reactions. Rules for assigning oxidation numbers: 1. An atom in its elemental form has an oxidation number of $0$ (e.g., $\text{O}_2$, $\text{Na}$, $\text{Cl}_2$). 2. For a monatomic ion, the oxidation number is equal to its charge (e.g., $\text{Na}^+$ is $+1$, $\text{O}^{2-}$ is $-2$). 3. Oxygen usually has an oxidation number of $-2$ in compounds, except in peroxides (e.g., $\text{H}_2\text{O}_2$) where it is $-1$, and with fluorine (e.g., $\text{OF}_2$) where it is $+2$. 4. Hydrogen usually has an oxidation number of $+1$ in compounds, except in metal hydrides (e.g., $\text{NaH}$) where it is $-1$. 5. Group 1 metals are always $+1$, Group 2 metals are always $+2$. 6. Fluorine is always $-1$. Other halogens are usually $-1$, except when bonded to oxygen or a more electronegative halogen. 7. The sum of oxidation numbers in a neutral compound is $0$. 8. The sum of oxidation numbers in a polyatomic ion equals the charge of the ion. Oxidation and reduction treated as change in oxidation numbers: Oxidation: An increase* in oxidation number. Reduction: A decrease* in oxidation number. Use of oxidation numbers in balancing simple equations: Step 1: Assign oxidation numbers to all atoms in the unbalanced equation. Step 2: Identify atoms that undergo a change in oxidation number. Step 3: Determine the total change in oxidation number for the oxidized and reduced species. Step 4: Balance the changes in oxidation numbers by using coefficients in front of the species. Step 5: Balance the remaining atoms (usually oxygen and hydrogen) by inspection, adding $\text{H}_2\text{O}$ and $\text{H}^+$ (for acidic solutions) or $\text{OH}^-$ (for basic solutions). Example (Acidic solution): Balance $\text{MnO}_4^- + \text{Fe}^{2+} \to \text{Mn}^{2+} + \text{Fe}^{3+}$ 1. Assign oxidation numbers: $\text{Mn}$ in $\text{MnO}_4^-$ is $+7$. $\text{O}$ is $-2$. $\text{Fe}^{2+}$ is $+2$. $\text{Mn}^{2+}$ is $+2$. $\text{Fe}^{3+}$ is $+3$. 2. Changes: $\text{Mn}$: $+7 \to +2$ (decrease of $5$, reduction) $\text{Fe}$: $+2 \to +3$ (increase of $1$, oxidation) 3. Balance changes: To balance the $5$ decrease in $\text{Mn}$ with the $1$ increase in $\text{Fe}$, we need $5$ $\text{Fe}^{2+}$ ions. $$\text{MnO}_4^- + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+}$$ 4. Balance oxygen (with $\text{H}_2\text{O}$) and hydrogen (with $\text{H}^+$): There are $4$ oxygen atoms on the left in $\text{MnO}_4^-$. Add $4\text{H}_2\text{O}$ to the right. $$\text{MnO}_4^- + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$$ Now there are $8$ hydrogen atoms on the right in $4\text{H}_2\text{O}$. Add $8\text{H}^+$ to the left. $$\boxed{8\text{H}^+ + \text{MnO}_4^- + 5\text{Fe}^{2+} \to \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}}$$ (Check charges: Left: $8(+1) + (-1) + 5(+2) = +8 - 1 + 10 = +17$. Right: $(+2) + 5(+3) + 0 = +2 + 15 = +17$. Charges are balanced.) IUPAC nomenclature of inorganic compounds using oxidation number: For compounds containing metals that can have multiple oxidation states (e.g., transition metals), the oxidation number is indicated by a Roman numeral in parentheses immediately following the metal's name. Example*: $\text{FeCl}_2$: Iron(II) chloride (Iron has an oxidation state of $+2$) $\text{FeCl}_3$: Iron(III) chloride (Iron has an oxidation state of $+3$) $\text{CuO}$: Copper(II) oxide (Copper has an oxidation state of $+2$) $\text{Cu}_2\text{O}$: Copper(I) oxide (Copper has an oxidation state of $+1$) Tests for oxidizing and reducing agents: An oxidizing agent* (or oxidant) causes another substance to be oxidized while itself being reduced. A reducing agent* (or reductant) causes another substance to be reduced while itself being oxidized. Tests for Oxidizing Agents: Oxidizing agents typically oxidize* indicator substances, causing a visible color change. Potassium iodide ($\text{KI}$): Oxidizing agents will oxidize $\text{I}^-$ ions to $\text{I}_2$ (iodine), which is brown in solution or turns starch indicator blue-black. Example*: $\text{MnO}_4^-$ (purple) + $\text{I}^-$ (colorless) $\to$ $\text{Mn}^{2+}$ (colorless) + $\text{I}_2$ (brown) Acidified potassium manganate(VII) ($\text{KMnO}_4$): Strong oxidizing agents will remain purple or change color less readily than when reacting with a reducing agent. (This is more commonly used to test for reducing agents, as it is itself a strong oxidizing agent). Acidified potassium dichromate(VI) ($\text{K}_2\text{Cr}_2\text{O}_7$): Strong oxidizing agents will remain orange or change color less readily. Tests for Reducing Agents: Reducing agents typically reduce* indicator substances, causing a visible color change. Acidified potassium manganate(VII) ($\text{KMnO}_4$): Reducing agents will reduce the purple $\text{MnO}_4^-$ ion to the colorless $\text{Mn}^{2+}$ ion. The purple solution will decolorize. Example*: $\text{Fe}^{2+}$ (pale green) + $\text{MnO}_4^-$ (purple) $\to$ $\text{Fe}^{3+}$ (yellow-brown) + $\text{Mn}^{2+}$ (colorless) Acidified potassium dichromate(VI) ($\text{K}_2\text{Cr}_2\text{O}_7$): Reducing agents will reduce the orange $\text{Cr}_2\text{O}_7^{2-}$ ion to the green $\text{Cr}^{3+}$ ion. The orange solution will turn green. Example*: $\text{SO}_2$ (gas) + $\text{Cr}_2\text{O}_7^{2-}$ (orange) $\to$ $\text{SO}_4^{2-}$ + $\text{Cr}^{3+}$ (green) Fehling's solution or Benedict's solution: Used to test for reducing sugars (which are reducing agents). A blue solution turns brick-red upon heating in the presence of a reducing agent.