# Balancing Redox Equations This section explores two fundamental methods for balancing oxidation-reduction (redox) reactions: the Oxidation Number Method and the Ion-Electron Method. Both techniques ensure that mass and charge are conserved throughout a chemical transformation. ## Introduction to Balancing Redox Reactions Redox reactions involve the transfer of electrons, where one species is oxidized (loses electrons) and another is reduced (gains electrons). Balancing these reactions is crucial to accurately represent stoichiometry and predict product yields. There are two primary methods for balancing redox equations: * **i. Oxidation Number Method:** Focuses on the change in oxidation states. * **ii. Ion-Electron Method (Half-Reaction Method):** Separates the reaction into oxidation and reduction half-reactions. --- ## Oxidation Number Method This method is based on the principle that in any redox reaction, the total number of electrons lost during oxidation must be equal to the total number of electrons gained during reduction. ### Steps for Balancing Equations by Oxidation Number Method 1. **Assign Oxidation Numbers:** Determine the oxidation number for each atom in the unbalanced equation. 2. **Identify Changes:** Identify elements whose oxidation numbers have changed. * An increase in oxidation number indicates oxidation (loss of electrons). * A decrease in oxidation number indicates reduction (gain of electrons). 3. **Indicate Electron Transfer:** Show the number of electrons gained or lost for each changing atom. 4. **Equalize Electron Transfer:** Multiply the species undergoing oxidation and reduction by appropriate whole numbers to make the total electrons lost equal to the total electrons gained. These numbers become the stoichiometric coefficients for those species. 5. **Balance by Inspection:** Balance the remaining atoms (usually oxygen and hydrogen) by inspection. * First, balance atoms other than O and H. * Then, balance oxygen atoms, typically by adding $\mathrm{H_2O}$ molecules. * Finally, balance hydrogen atoms, typically by adding $\mathrm{H^+}$ ions (in acidic medium) or $\mathrm{OH^-}$ ions (in basic medium, followed by $\mathrm{H_2O}$ adjustments). --- ## Ion-Electron Method (Half-Reaction Method) This method separates the overall redox reaction into two half-reactions: one for oxidation and one for reduction. Each half-reaction is balanced independently for mass and charge, and then they are combined. ### Steps for Balancing Half-Reactions (Acidic Medium) 1. **Separate into Half-Reactions:** Divide the overall unbalanced equation into an oxidation half-reaction and a reduction half-reaction. 2. **Balance Atoms Other Than O and H:** Balance all atoms except oxygen and hydrogen. 3. **Balance Oxygen Atoms:** Add $\mathrm{H_2O}$ molecules to the side deficient in oxygen. 4. **Balance Hydrogen Atoms:** Add $\mathrm{H^+}$ ions to the side deficient in hydrogen. 5. **Balance Charge:** Add electrons ($\mathrm{e^-}$) to the more positive side to balance the net charge. * For oxidation half-reactions, electrons will appear on the product side. * For reduction half-reactions, electrons will appear on the reactant side. ### Steps for Combining Half-Reactions 1. **Equalize Electrons:** Multiply each half-reaction by an appropriate integer so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. 2. **Add Half-Reactions:** Sum the two balanced half-reactions. 3. **Cancel Duplications:** Cancel any identical species (electrons, $\mathrm{H_2O}$, $\mathrm{H^+}$) that appear on both sides of the combined equation. 4. **Verify:** Check that both mass and charge are balanced. --- ## Worked Examples ### Example 2.3: Balancing by Oxidation Number Method Balance the following equation by the oxidation number method: $$\mathrm{P}+\mathrm{HNO}_{3}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{H}_{3}\mathrm{PO}_{4}+\mathrm{NO}$$ **Solution:** **Step 1: Assign oxidation numbers to all the atoms involved in the equation.** $$\stackrel{0}{\mathrm{P}}+\stackrel{+1+5-2}{\mathrm{HNO}_{3}}+\stackrel{+1-2}{\mathrm{H}_{2}\mathrm{O}} \longrightarrow \stackrel{+1+5-2}{\mathrm{H}_{3}\mathrm{PO}_{4}}+\stackrel{+2-2}{\mathrm{NO}}$$ **Step 2: Identify the elements changing oxidation number.** Phosphorus ($\mathrm{P}$) changes from oxidation state 0 to +5 in $\mathrm{H_3PO_4}$. This is an increase of 5, so it loses 5 electrons (oxidation). Nitrogen ($\mathrm{N}$) in $\mathrm{HNO_3}$ changes from oxidation state +5 to +2 in $\mathrm{NO}$. This is a decrease of 3, so it gains 3 electrons (reduction). **Step 3: Indicate the electron transfer for each changing element.** * $\mathrm{P^0}$ to $\mathrm{P^{+5}}$: loss of $5\mathrm{e^-}$ * $\mathrm{N^{+5}}$ to $\mathrm{N^{+2}}$: gain of $3\mathrm{e^-}$ **Step 4: Equalize the number of electrons lost and gained.** To balance a $3\mathrm{e^-}$ gain against a $5\mathrm{e^-}$ loss, multiply the $3\mathrm{e^-}$ gain by 5 and the $5\mathrm{e^-}$ loss by 3. This equalizes the total electrons to 15. $3\mathrm{P} + 5\mathrm{HNO}_{3} + \mathrm{H}_{2}\mathrm{O} \longrightarrow 3\mathrm{H}_{3}\mathrm{PO}_{4} + 5\mathrm{NO}$ **Step 5: Balance the rest of the equation by inspection.** *Balance O atoms:* Left side has 16 O atoms; Right side has 17 O atoms. Multiply $\mathrm{H_2O}$ by 2. $$3\mathrm{P} + 5\mathrm{HNO}_{3} + 2\mathrm{H}_{2}\mathrm{O} \longrightarrow 3\mathrm{H}_{3}\mathrm{PO}_{4} + 5\mathrm{NO}$$ *Balance H atoms:* Both sides now have 9 H atoms. **Final Balanced Equation:** $$3\mathrm{P} + 5\mathrm{HNO}_{3} + 2\mathrm{H}_{2}\mathrm{O} \longrightarrow 3\mathrm{H}_{3}\mathrm{PO}_{4} + 5\mathrm{NO}$$ --- ### Example 2.4: Constructing a Redox Equation by Ion-Electron Method Construct a redox equation using the following half-equations in acidic medium: Reduction half-reaction: $\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}$ Oxidation half-reaction: $\mathrm{H}_{2}\mathrm{SO}_{3} \longrightarrow \mathrm{HSO}_{4}^{-}$ **Solution:** **Step 1: Balance each half-reaction.** #### a) Balancing the Reduction Half-Reaction 1. **Balance Cr:** $\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+}$ 2. **Balance O:** $\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+}+7\mathrm{H}_{2}\mathrm{O}$ 3. **Balance H:** $\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+14\mathrm{H}^{+} \longrightarrow 2\mathrm{Cr}^{3+}+7\mathrm{H}_{2}\mathrm{O}$ 4. **Balance Charge:** Add $6\mathrm{e^-}$ to the left. $$\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+14\mathrm{H}^{+}+6\mathrm{e}^{-} \longrightarrow 2\mathrm{Cr}^{3+}+7\mathrm{H}_{2}\mathrm{O} \quad \text{(Eq 1)}$$ #### b) Balancing the Oxidation Half-Reaction 1. **Balance S:** Already balanced. 2. **Balance O:** $\mathrm{H}_{2}\mathrm{SO}_{3}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{HSO}_{4}^{-}$ 3. **Balance H:** $\mathrm{H}_{2}\mathrm{SO}_{3}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{HSO}_{4}^{-}+3\mathrm{H}^{+}$ 4. **Balance Charge:** Add $2\mathrm{e^-}$ to the right. $$\mathrm{H}_{2}\mathrm{SO}_{3}+\mathrm{H}_{2}\mathrm{O} \longrightarrow \mathrm{HSO}_{4}^{-}+3\mathrm{H}^{+}+2\mathrm{e}^{-} \quad \text{(Eq 2)}$$ **Step 2: Equalize electrons and add.** Multiply Eq 2 by 3 and add to Eq 1. $$\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 5\mathrm{H}^{+} + 3\mathrm{H}_{2}\mathrm{SO}_{3} \longrightarrow 2\mathrm{Cr}^{3+} + 4\mathrm{H}_{2}\mathrm{O} + 3\mathrm{HSO}_{4}^{-}$$ --- ## Concept Assessment Exercise 2.2 Using the oxidation number method, balance the following equations: **Question 1:** Balance the equation: $$\mathrm{Cu}+\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}(\mathrm{NO}_{3})_{2}+\mathrm{NO}+\mathrm{H}_{2}\mathrm{O}$$ **Answer:** $3\mathrm{Cu}+8\mathrm{HNO}_{3} \rightarrow 3\mathrm{Cu}(\mathrm{NO}_{3})_{2}+2\mathrm{NO}+4\mathrm{H}_{2}\mathrm{O}$ **Question 2:** Balance the equation: $$\mathrm{Cu}+\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}(\mathrm{NO}_{3})_{2}+\mathrm{NO}_{2}+\mathrm{H}_{2}\mathrm{O}$$ **Answer:** $\mathrm{Cu}+4\mathrm{HNO}_{3} \rightarrow \mathrm{Cu}(\mathrm{NO}_{3})_{2}+2\mathrm{NO}_{2}+2\mathrm{H}_{2}\mathrm{O}$ --- ## Concept Assessment Exercise 2.3 Use the half-reaction method to balance the following redox reactions (assuming acidic medium): **(i) Question:** Balance: $\mathrm{Co}+\mathrm{Cr}^{3+} \longrightarrow \mathrm{Co}^{2+}+\mathrm{Cr}^{2+}$ **Answer:** $\mathrm{Co} + 2\mathrm{Cr}^{3+} \longrightarrow \mathrm{Co}^{2+} + 2\mathrm{Cr}^{2+}$ **(ii) Question:** Balance: $\mathrm{Au}^{3+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}^{+}+\mathrm{Au}$ **Answer:** $2\mathrm{Au}^{3+} + 3\mathrm{H}_{2} \longrightarrow 2\mathrm{Au} + 6\mathrm{H}^{+}$ **(iii) Question:** Balance: $\mathrm{Cl}_{2}+\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2}+\mathrm{Cl}^{-}$ **Answer:** $\mathrm{Cl}_{2} + 2\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2} + 2\mathrm{Cl}^{-}$ --- ## Concept Assessment Exercise 2.4 Construct a redox equation from the following half-equations (assuming acidic medium): **Question 1:** Combine $\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$ and $\mathrm{H}_{2}\mathrm{O}_{2} \rightarrow \mathrm{O}_{2}$ **Answer:** $2\mathrm{MnO}_{4}^{-} + 5\mathrm{H}_{2}\mathrm{O}_{2} + 6\mathrm{H}^{+} \longrightarrow 2\mathrm{Mn}^{2+} + 5\mathrm{O}_{2} + 8\mathrm{H}_{2}\mathrm{O}$ **Question 2:** Combine $\mathrm{NO}_{3}^{-} \rightarrow \mathrm{NO}$ and $\mathrm{Br}^{-} \rightarrow \mathrm{Br}_{2}$ **Answer:** $2\mathrm{NO}_{3}^{-} + 6\mathrm{Br^{-} + 8H^{+} \longrightarrow 2NO + 3Br_{2} + 4H_{2}O}$ **Question 3:** Combine $\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}$ and $\mathrm{C}_{2}\mathrm{O}_{4}^{2-} \rightarrow \mathrm{CO}_{2}$ **Answer:** $2\mathrm{MnO}_{4}^{-} + 5\mathrm{C}_{2}\mathrm{O}_{4}^{2-} + 16\mathrm{H}^{+} \longrightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}$ --- ## Balancing in Basic Medium In basic medium, the steps are similar to the acidic medium, but with an additional step to neutralize $\mathrm{H^+}$ ions. 1. Balance the equation as if it were in an acidic medium (using $\mathrm{H^+}$ and $\mathrm{H_2O}$). 2. Add $\mathrm{OH^-}$ ions to both sides of the equation equal to the number of $\mathrm{H^+}$ ions present. 3. Combine $\mathrm{H^+}$ and $\mathrm{OH^-}$ on one side to form $\mathrm{H_2O}$. 4. Cancel any redundant $\mathrm{H_2O}$ molecules from both sides.