Question

Complete and balance the following equations, and identify the oxidizing and reducing agents: (a) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{I}^{-}(a q)-\cdots \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{IO}_{3}^{-}(a q)\) (acidic solution) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{CH}_{3} \mathrm{OH}(a q)-\longrightarrow\) \(\mathrm{Mn}^{2+}(a q)+\mathrm{HCO}_{2} \mathrm{H}(a q)\) (acidic solution) (c) \(\mathrm{I}_{2}(\mathrm{~s})+\mathrm{OCl}^{-}(a q)-\mathrm{IO}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) (acidic solution) (d) \(\mathrm{As}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{NO}_{3}^{-}(a q)-\longrightarrow\) \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{N}_{2} \mathrm{O}_{3}(a q)\) (acidic solution) (e) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Br}^{-}(a q)-\mathrm{MnO}_{2}(s)+\mathrm{BrO}_{3}^{-}(a q)\) (basic solution) (f) \(\mathrm{Pb}(\mathrm{OH})_{4}{ }^{2-}(a q)+\mathrm{ClO}^{-}(a q)-\cdots \mathrm{PbO}_{2}(s)+\mathrm{Cl}^{-}(a q)\) (basic solution)

Short answer

Short Answer: (a) Balanced reaction: \(Cr_2O_7^{2-}(aq) + 6I^-(aq) + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 2IO_3^-(aq) + 7H_2O(l)\), Oxidizing agent: \(Cr_2O_7^{2-}(aq)\), Reducing agent: \(I^-(aq)\). (b) \(MnO_4^-(aq) + 5CH_3OH(aq) + 2H^+(aq) \rightarrow Mn^{2+}(aq) + 5HCO_2H(aq) + 4H_2O(l)\), Oxidizing agent: \(MnO_4^-(aq)\), Reducing agent: \(CH_3OH(aq)\). (c) \(I_2(s) + OCl^-(aq) + 6H_+(aq) \rightarrow 2IO_3^-(aq) + Cl^-(aq) + 3H_2O(l)\), Oxidizing agent: \(OCl^-(aq)\), Reducing agent: \(I_2(s)\). (d) \(2As_2O_3(s) + 2NO_3^-(aq) + 12H^+(aq) \rightarrow 4H_3AsO_4(aq) + N_2O_3(aq) + 3H_2O(l)\), Oxidizing agent: \(NO_3^-(aq)\), Reducing agent: \(As_2O_3(s)\). (e) \(2MnO_4^-(aq) + 5Br^-(aq) + 6H_2O(l) \rightarrow 2MnO_2(s) + 5BrO_3^-(aq) + 12OH^-(aq)\), Oxidizing agent: \(MnO_4^-(aq)\), Reducing agent: \(Br^-(aq)\). (f) \(2Pb(OH)_4^{2-}(aq) + 2ClO^-(aq) \rightarrow 2PbO_2(s) + 4OH^-(aq) + 2Cl^-(aq) + 2H_2O(l)\), Oxidizing agent: \(ClO^-(aq)\), Reducing agent: \(Pb(OH)_4^{2-}(aq)\).

Step-by-step solution

Identify the oxidation states of each element

For the first reaction, these are the oxidation states: \(Cr_2O_7^{2-}: Cr(+6), O(-2) \\ I^-(aq): I(-1) \\ Cr^{3+}(aq): Cr(+3) \\ IO_3^-(aq): I(+5), O(-2)\)

Write down the oxidation and reduction half-reactions

Oxidation half-reaction: \(I^-(aq) \rightarrow IO_3^-(aq) \\ Reduction half-reaction: Cr_2O_7^{2-}(aq) \rightarrow Cr^{3+}(aq)\)

Balance the atoms in each half-reaction (consider acidic solution)

Oxidation half-reaction: \(6I^-(aq) \rightarrow 2IO_3^-(aq) + 12H^+(aq) + 6e^-\) Reduction half-reaction: \(Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)\)

Add the half-reactions together to obtain the balanced redox reaction

Balanced redox reaction: \(Cr_2O_7^{2-}(aq) + 6I^-(aq) + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 2IO_3^-(aq) + 7H_2O(l)\)

Identify the oxidizing and reducing agents

Oxidizing agent: \(Cr_2O_7^{2-}(aq)\) Reducing agent: \(I^-(aq)\) Now, follow the same steps to balance and identify the agents for the remaining reactions (b) to (f).

Key concepts

Oxidation States

Understanding oxidation states is crucial when diving into redox reactions. In simple terms, oxidation states, often called oxidation numbers, reflect the degree of oxidation of an element. They are hypothetical charges that an atom would have if all bonds were completely ionic.When working through a redox reaction, begin by identifying the oxidation state of each atom in the reactants and products. The change in these numbers will point out which species are oxidized and which are reduced. For instance, when an element's oxidation state increases, it's being oxidized, and when it decreases, it's being reduced.Remember that there are some rules to determine the oxidation states. For example, the oxidation state of oxygen is usually -2 (except in peroxides or in combination with fluorine), and for hydrogen, it's usually +1 when bonded to nonmetals and -1 when bonded to metals. The oxidation states of ions are equal to their charge, and the sum of oxidation states in a neutral compound must be zero, while in ions it equals the charge of the ion.

Half-Reactions

Redox reactions can be broken down into two simpler reactions: oxidation half-reactions and reduction half-reactions. This step-by-step approach makes balancing complex redox reactions more manageable.The oxidation half-reaction involves the loss of electrons, hence the term 'oxidation'. Here, the oxidation state of an element increases. Conversely, the reduction half-reaction involves the gain of electrons, with a resultant decrease in the oxidation state. Electrons should be explicitly shown in these half-reactions to clarify their transfer.After writing the half-reactions, the next step is to balance them, usually starting with the atoms of elements other than oxygen and hydrogen, then balancing oxygen atoms by adding water molecules, and hydrogen atoms by adding hydrogen ions. In acidic solutions, you'll add H+ ions, and in basic solutions, OH- ions come into play. Remember, the aim is to equalize the number of electrons lost in oxidation with those gained in reduction by multiplying the half-reactions by appropriate factors before adding them up to get the overall balanced reaction.

Oxidizing and Reducing Agents

Identifying oxidizing and reducing agents is a pivotal part of analyzing redox reactions. The oxidizing agent is the species that gets reduced by accepting electrons, while the reducing agent is the species that gets oxidized by donating electrons.In the reaction from our exercise, dichromate ion, \(Cr_2O_7^{2-}\), is the oxidizing agent because it accepts electrons and is reduced. Conversely, the iodide ion, \(I^-\), is the reducing agent since it donates electrons and is oxidized.These agents are essential because they drive the redox process by causing the oxidation or reduction of other substances without being permanently changed or consumed in the overall reaction. After determining the role of these agents, balancing the reaction involves making sure the total number of electrons lost by the oxidizing agent equals those gained by the reducing agent.

Acidic and Basic Solutions

The pH of the solution—a measure of its acidity or basicity—plays a significant role in balancing redox reactions. In an acidic solution, you balance the charge by adding \(H^+\) ions. For basic solutions, you first balance the equation as if it were in an acidic solution, then add \(OH^-\) ions to both sides to neutralize the excess \(H^+\) ions, finally forming water molecules.In certain redox reactions, like those found in parts (e) and (f) of our exercise, the presence of a basic solution requires the additional step of combining \(H^+\) ions and \(OH^-\) ions to make water, followed by the removal of excess water molecules to achieve balance. The environment, whether acidic or basic, can also influence the type of products formed, as some compounds are stable only under specific pH conditions.