Question

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

Short answer

Short answer: The balanced reaction for (a) is \(4\mathrm{I}^{-}(aq) + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 42\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 12\mathrm{Cr}^{3+}(aq) + 2\mathrm{IO}_{3}^{-}(aq) + 60\mathrm{H}^{+}(aq)\). The oxidizing agent is \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\), and the reducing agent is \(\mathrm{I}^{-}\).

Step-by-step solution

Identify oxidation states and half-reactions

Oxidation states: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\) -> Cr(+6), I(-1) -> Cr(+3), \(\mathrm{IO}_{3}^{-}\) -> I(+5) Half-reactions: 1. Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{Cr}^{3+}\) 2. Oxidation: \(\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-}\)

Balance half-reactions for mass and charge

1. Reduction: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+}\) (balance Cr) \(7\mathrm{H}_{2}\mathrm{O} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+} + 14\mathrm{H}^{+}\) (balance O with H2O and H with H+) 2. Oxidation: \(2\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-}\) + \(6\mathrm{H}_{2}\mathrm{O}\) (balance I) \(2\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-}\) + \(6\mathrm{H}_{2}\mathrm{O}+12\mathrm{H}^{+}+ 6\mathrm{e^{-}}\) (balance O with H2O, H with H+ and charge with e-)

Add half-reactions and identify oxidizing/reducing agents

First, make the electrons equal in both half-reactions: \(6[\) \(7\mathrm{H}_{2}\mathrm{O} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \longrightarrow 2\mathrm{Cr}^{3+} + 14\mathrm{H}^{+}\) \(]\) \(2[\) \(2\mathrm{I}^{-} \longrightarrow \mathrm{IO}_{3}^{-} + 6\mathrm{H}_{2}\mathrm{O}+12\mathrm{H}^{+}+ 6\mathrm{e^{-}}\) \(]\) Now add the half-reactions: \(42\mathrm{H}_{2}\mathrm{O} + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 4\mathrm{I}^{-} \longrightarrow 12\mathrm{Cr}^{3+} + 84\mathrm{H}^{+} + 2\mathrm{IO}_{3}^{-} + 12\mathrm{H}_{2}\mathrm{O}+24\mathrm{H}^{+}+ 12\mathrm{e^{-}}\) Simplify the reaction: \(4\mathrm{I}^{-} + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 42\mathrm{H}_{2}\mathrm{O}\longrightarrow 12\mathrm{Cr}^{3+}+\mathrm{IO}_{3}^{-}+ 60\mathrm{H}^{+}\) So the balanced reaction is: \(4\mathrm{I}^{-}(aq) + 6\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 42\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 12\mathrm{Cr}^{3+}(aq) + 2\mathrm{IO}_{3}^{-}(aq) + 60\mathrm{H}^{+}(aq)\) Oxidizing agent: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\), Reducing agent: \(\mathrm{I}^{-}\)

Key concepts

Oxidation States

Understanding oxidation states is a fundamental aspect of analyzing redox reactions. An oxidation state, often referred to as oxidation number, is an indicator of the degree of oxidation (loss of electrons) or reduction (gain of electrons) an atom undergoes during a chemical reaction.

It's key to note that, in a neutral molecule, the sum of the oxidation states equals zero, while in ions, it equals the ionic charge. For instance, in the reaction \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\), chromium (Cr) has an oxidation state of +6. When changes in oxidation states occur, such as Cr going from +6 to +3, we can deduce that a redox process is taking place.

These changes are what drive the reaction forward, and by identifying the shifts in oxidation states, we can determine which species are oxidized and which are reduced, a necessary step before balancing the overall reaction.

Half-Reactions

Dividing a redox reaction into half-reactions helps simplify the process of balancing the overall equation. Each half-reaction represents either the oxidation or reduction process happening in a redox reaction.

An oxidation half-reaction shows the loss of electrons, hence an increase in the oxidation state. In contrast, a reduction half-reaction shows the gain of electrons, indicated by a decrease in the oxidation state. For instance, in the reduction half-reaction \(\mathrm{Cr}_2\mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}\), electrons are gained, signifying a reduction. Balancing these half-reactions for mass and charge is crucial before they can be recombined to provide the overall balanced chemical equation.

Oxidizing Agents

An oxidizing agent, or oxidant, is the substance that gets reduced by gaining electrons in a redox reaction and, in the process, causes the oxidation of another substance. It essentially 'takes' electrons from other species. Therefore, it must be a good electron acceptor.

In the given exercise, the dichromate ion \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) is the oxidizing agent because it gains electrons and its oxidation state decreases from +6 to +3 as it is converted into \(\mathrm{Cr}^{3+}\). The ability to identify oxidizing agents is crucial as it determines which species is causing the other to lose electrons, and thereby getting oxidized itself.

Reducing Agents

Opposite to oxidizing agents, reducing agents, or reductants, are substances that lose electrons in a redox reaction and, as a result, cause the reduction of another substance. Reducing agents undergo oxidation themselves by supplying electrons to the species that is being reduced.

In the provided example, \(\mathrm{I}^{-}\) is the reducing agent since it donates electrons, increasing its oxidation state from -1 to +5 to form \(\mathrm{IO}_3^{-}\). Recognizing the role of reducing agents is key to interpreting redox reactions, as it highlights the substance that is giving away electrons and facilitating the reduction of the oxidizing agent.