Question

Balance each of the following oxidation-reduction reactions, which take place in acidic solution. a. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{O}_{2}(g)\) b. \(\operatorname{Br} \mathrm{O}_{3}^{-}(a q)+\mathrm{Cu}^{+}(a q) \rightarrow \mathrm{Br}^{-}(a q)+\mathrm{Cu}^{2+}(a q)\) c. \(\mathrm{HNO}_{2}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{NO}(g)+\mathrm{I}_{2}(a q)\)

Short answer

The balanced equations for the given reactions are: a. \(MnO_4^{-}(aq)+H_2O_2(aq)+2H^+ \rightarrow Mn^{2+}(aq)+O_2(g)+2H_2O(l)+5e^-\) b. \(3BrO_3^-(aq)+6Cu^{+}(aq)+12H^+ \rightarrow 3Br^-(aq)+6Cu^{2+}(aq)+6H_2O(l)\) c. \(2HNO_2(aq)+I^-(aq) \rightarrow 2NO(g)+I_2(aq)+2H^+\)

Step-by-step solution

Identify the oxidation states of each element

Assign the oxidation states to each of the elements in the given reaction, following standard rules. \(MnO_4^-\): Mn = +7, O = -2 \(H_2O_2\): H = +1, O = -1 \(Mn^{2+}\): Mn = +2 \(O_2\): O = 0

Determine which are the reducing and oxidizing agents

Mn changes its oxidation state from +7 to +2, which is a reduction. O changes its oxidation state from -1 to 0, which is an oxidation. Therefore, \(MnO_4^-\) is the oxidizing agent and \(H_2O_2\) is the reducing agent.

Balance atoms other than hydrogen and oxygen

We have 1 Mn atom on each side of the equation, so Mn atoms are already balanced.

Balance oxygen atoms by adding water molecules

We have 4 O atoms from \(MnO_4^-\) and 2 O atoms from \(H_2O_2\) on the left side. On the right side, there are no additional oxygen atoms to add, but we need 4 O atoms, so we add 2 water molecules (because 2 x 2 O = 4 O) to the right side: \(MnO_4^{-}(aq)+H_2O_2(aq) \rightarrow Mn^{2+}(aq)+O_2(g)+2H_2O(l)\)

Balance hydrogen atoms by adding protons (H+)

Now, we balance the hydrogen atoms. There are 2 hydrogen atoms on the left side from \(H_2O_2\) and 4 hydrogen atoms on the right side from 2\(H_2O\). To balance, we add 2 H+ ions to the left side: \(MnO_4^{-}(aq)+H_2O_2(aq)+2H^+ \rightarrow Mn^{2+}(aq)+O_2(g)+2H_2O(l)\)

Balance the charges by adding electrons

Finally, we balance the charges. On the left side, we have a total charge of -1 while on the right side, we have a total charge of +2. To balance the charges, we add 5 electrons to the right side: \(MnO_4^{-}(aq)+H_2O_2(aq)+2H^+ \rightarrow Mn^{2+}(aq)+O_2(g)+2H_2O(l)+5e^-\) ## Reaction b: ## Follow the same steps as reaction a, balancing atoms, oxidation states, and charges. The balanced equation for reaction b will be: \(3BrO_3^-(aq)+6Cu^{+}(aq)+12H^+ \rightarrow 3Br^-(aq)+6Cu^{2+}(aq)+6H_2O(l)\) ## Reaction c: ## Similarly, for reaction c, balance atoms, oxidation states, and charges. The balanced equation for reaction c will be: \(2HNO_2(aq)+I^-(aq) \rightarrow 2NO(g)+I_2(aq)+2H^+\)

Key concepts

Oxidation States

Understanding oxidation states is fundamental to mastering the art of balancing redox reactions. An oxidation state is essentially the 'charge' of an element when it forms specific compounds or ions. For instance, in the compound \(MnO_4^-\), manganese (Mn) has an oxidation state of +7 while the oxygen has a typical -2 state. These assigned values follow certain rules, such as the sum of oxidation states in a neutral molecule being zero, or equal to the charge of the ion in the case of ionic species.

During a redox reaction, the changes in these oxidation states reveal which elements are gaining or losing electrons. A decrease in oxidation state signifies a gain of electrons, termed as reduction, while an increase indicates loss of electrons, known as oxidation. In our educational resources, we emphasize recognizing these changes, as they directly lead to identifying oxidizing and reducing agents in a chemical reaction.

Oxidizing and Reducing Agents

Identifying oxidizing and reducing agents is a pivotal step in understanding redox reactions. An oxidizing agent takes electrons from another substance, therefore it becomes reduced during the reaction. Conversely, a reducing agent donates electrons, oxidizing in the process. For instance, balancing the reaction \(MnO_4^- + H_2O_2 \rightarrow Mn^{2+} + O_2\), we observe manganese's state changing from +7 in \(/mathrm{MnO}_{4}^{-}(aq)\) to +2 in \(/mathrm{Mn}^{2+}(aq)\), indicating \(MnO_4^-\) is reduced and therefore, is the oxidizing agent.

On the other side, oxygen in \(H_2O_2\) goes from an oxidation state of -1 to 0 in \(O_2\), indicating oxidation. Consequently, \(H_2O_2\) serves as the reducing agent. Highlighting these agent roles help students deduce the flow of electrons and provides a deeper grasp of the redox process.

Stoichiometry in Acidic Solution

Stoichiometry in acidic solutions adds an extra layer to balancing redox equations. It involves the careful accounting for hydrogen ions (\(H^+\)) and water molecules to ensure both mass and charge balance. After balancing the atoms of other elements, you'll often need to add \(H^+\) ions to equalize the number of hydrogen atoms on both sides of the reaction. This step is vividly seen in the balancing of our example reaction:

\(/mathrm{MnO}_{4}^{-}(aq)+/mathrm{H}_{2}/mathrm{O}_{2}(aq)+2H^+ \rightarrow /mathrm{Mn}^{2+}(aq)+/mathrm{O}_{2}(g)+2/mathrm{H}_{2}/mathrm{O}(l)\)

Finally, ensuring the charges on both sides are balanced often requires adding electrons. These electrons are critical in representing the actual electron transfer happening in the reaction. Our educational platform showcases each step with detailed explanations, assisting students to confidently solve similar problems on their own and deepen their comprehension of stoichiometry in the context of redox reactions.