Question

The standard reduction potential for the reduction of \(\mathrm{RuO}_{4}^{-}(a q)\) to \(\mathrm{RuO}_{4}^{2-}(a q)\) is \(+0.59 \mathrm{~V}\). By using Appendix E, which of the following substances can oxidize \(\mathrm{RuO}_{4}^{2-}(a q)\) to \(\mathrm{RuO}_{4}^{-}(a q)\) under standard conditions: \(\mathrm{Br}_{2}(l), \mathrm{BrO}_{3}^{-}(a q)\). \(\mathrm{Mn}^{2+}(a q), \mathrm{O}_{2}(g), \mathrm{Sn}^{2+}(a q) ?\)

Short answer

The substances that can oxidize \(\mathrm{RuO}_{4}^{2-}(aq)\) to \(\mathrm{RuO}_{4}^{-}(aq)\) under standard conditions are \(\mathrm{Br}_{2}(l)\), \(\mathrm{BrO}_{3}^{-}(aq)\), and \(\mathrm{O}_{2}(g)\), as their standard reduction potentials are greater than \(+0.59~V\).

Step-by-step solution

Identify the standard reduction potential of the given reaction

The standard reduction potential for the reduction of \(\mathrm{RuO}_{4}^{-}(aq)\) to \(\mathrm{RuO}_{4}^{2-}(aq)\) is given as \(+0.59~V\). This means that any substance with a standard reduction potential greater than \(+0.59~V\) will be able to oxidize \(\mathrm{RuO}_{4}^{2-}(aq)\) to \(\mathrm{RuO}_{4}^{-}(aq)\).

Find the standard reduction potentials for the given substances

We will use Appendix E to find the standard reduction potentials for the given substances. 1. \(\mathrm{Br}_{2}(l) + 2e^{-} \rightarrow 2\mathrm{Br}^{-}(aq)\): \(E^0 = +1.09~V\) 2. \(\mathrm{BrO}_{3}^{-}(aq) + 6\mathrm{H}^{+}(aq) + 6e^{-} \rightarrow \mathrm{Br}^{-}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\): \(E^0 = +1.53~V\) 3. \(\mathrm{Mn}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Mn}(s)\): \(E^0 = -1.18~V\) 4. \(\mathrm{O}_{2}(g) + 4\mathrm{H}^{+}(aq) + 4e^{-} \rightarrow 2\mathrm{H}_{2}\mathrm{O}(l)\): \(E^0 = +1.23~V\) 5. \(\mathrm{Sn}^{2+}(aq) + 2e^{-} \rightarrow \mathrm{Sn}(s)\): \(E^0 = -0.14~V\)

Compare the standard reduction potentials

Now that we have the standard reduction potentials for all given substances, we can compare them to the standard reduction potential of the \(\mathrm{RuO}_{4}^{-}(aq)\) to \(\mathrm{RuO}_{4}^{2-}(aq)\) reaction (\(+0.59~V\)). If the substance's standard reduction potential is greater than \(+0.59~V\), it can oxidize \(\mathrm{RuO}_{4}^{2-}(aq)\) to \(\mathrm{RuO}_{4}^{-}(aq)\). 1. \(\mathrm{Br}_{2}(l)\): \(E^0 = +1.09~V\) (greater than \(+0.59~V\)) 2. \(\mathrm{BrO}_{3}^{-}(aq)\): \(E^0 = +1.53~V\) (greater than \(+0.59~V\)) 3. \(\mathrm{Mn}^{2+}(aq)\): \(E^0 = -1.18~V\) (less than \(+0.59~V\)) 4. \(\mathrm{O}_{2}(g)\): \(E^0 = +1.23~V\) (greater than \(+0.59~V\)) 5. \(\mathrm{Sn}^{2+}(aq)\): \(E^0 = -0.14~V\) (less than \(+0.59~V\))

Identify the substances that can oxidize \(\mathrm{RuO}_{4}^{2-}(aq)\) to \(\mathrm{RuO}_{4}^{-}(aq)\)

Based on the comparison of standard reduction potentials, the following substances can oxidize \(\mathrm{RuO}_{4}^{2-}(aq)\) to \(\mathrm{RuO}_{4}^{-}(aq)\) under standard conditions: - \(\mathrm{Br}_{2}(l)\) - \(\mathrm{BrO}_{3}^{-}(aq)\) - \(\mathrm{O}_{2}(g)\)

Key concepts

Oxidation-reduction reactions

Oxidation-reduction reactions, also known as redox reactions, are processes in which electrons are transferred from one substance to another. These reactions are central to electrochemistry and are characterized by two main processes: oxidation and reduction.
Oxidation is the process where a substance loses electrons, while reduction is the gain of electrons. The substance that donates electrons is known as the reducing agent, and the one that accepts electrons is the oxidizing agent. These reactions always occur together; when one species is oxidized, another is simultaneously reduced.
For instance, in the exercise provided, we are examining the ability of various substances to oxidize \( \mathrm{RuO}_4^{2-}(aq) \) to \( \mathrm{RuO}_4^{-}(aq) \). This means \( \mathrm{RuO}_4^{2-}(aq) \) will undergo oxidation, suggesting it loses electrons and thus is the reducing agent in the reaction.

Electrochemistry

Electrochemistry is a fascinating branch of chemistry that explores the relationship between electrical energy and chemical reactions. It plays a critical role in various applications, such as batteries and electroplating, but can also be quite intricate in its conceptual foundations.
Electrochemistry focuses on redox reactions where oxidation and reduction occur, leading to an exchange of electrons between substances. This exchange causes an electric current, which can be harnessed for various uses. In adjustable processes called half-reactions, one substance undergoes oxidation, losing electrons, while another undergoes reduction, gaining electrons.
In our exercise, understanding electrochemistry involves recognizing that the capacity of different substances to act as oxidizing agents depends on their respective standard electrode potentials. These potentials determine how readily a substance will either lose or gain electrons during a redox reaction.

Standard electrode potentials

Standard electrode potentials, symbolized as \( E^0 \), are a measure of the tendency of a chemical species to be reduced, measured under standard conditions. This potential depends on the substance in question and its environment, usually involving an aqueous solution at 25°C and a concentration of 1 mol/L.
These potentials are crucial in predicting the direction of electrochemical reactions. A higher \( E^0 \) suggests a greater propensity to gain electrons and undergo reduction, making the substance a strong oxidizing agent.
In the context of the exercise, a substance with a higher standard reduction potential than \(+0.59 \mathrm{~V}\) can effectively oxidize \( \mathrm{RuO}_4^{2-}(aq) \). By comparing the given \( E^0 \) values, we identified \( \mathrm{Br}_2(l) \), \( \mathrm{BrO}_3^-(aq) \), and \( \mathrm{O}_2(g) \) as capable oxidizers due to their higher potentials.