Question

Consider the following balanced redox reaction in basic medium. $$ \begin{aligned} 3 \mathrm{Sn}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O} & \longrightarrow \\ 3 \mathrm{Sn}^{4+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+8 \mathrm{OH}^{-}(a q) \end{aligned} $$ (a) What is the oxidizing agent? (b) What species has the element that increases its oxidation number? (c) What species contains the element with the highest oxidation number? (d) If the reaction were to take place in acidic medium, what species would not be included in the reaction?

Short answer

Answer: (a) The oxidizing agent is Cr₂O₇²⁻; (b) The species with an element increasing its oxidation number is Sn²⁺; (c) The species with the highest oxidation number is Cr₂O₇²⁻, and (d) The species that would not be included in the acidic reaction is the OH⁻ ion.

Step-by-step solution

Identify the change in oxidation numbers

Determine the oxidation numbers and changes for each species in the given reaction: For tin (Sn): \(\text{Sn}^{2+} \rightarrow \text{Sn}^{4+}\): Oxidation number increases from +2 to +4. For chromium (Cr): \(\text{Cr}_{2}\text{O}_{7}^{2-} \rightarrow \text{Cr}_{2}\text{O}_{3}\): Oxidation number decreases from +6 to +3.

Determine the oxidizing and reducing agents

Oxidizing agent: The species which is reduced and causes the oxidation of another species. Here, \(\text{Cr}_{2}\text{O}_{7}^{2-}\) is reduced (from +6 to +3 Cr) and Sn is oxidized. So, the oxidizing agent is \(\text{Cr}_{2}\text{O}_{7}^{2-}\). Reducing agent: The species which is oxidized and causes the reduction of another species. Here, \(\text{Sn}^{2+}\) is oxidized (from +2 to +4 Sn) and Cr is reduced. So, the reducing agent is \(\text{Sn}^{2+}\). Now, let's answer the given questions: (a) The oxidizing agent is \(\text{Cr}_{2}\text{O}_{7}^{2-}\). (b) The species that has the element that increases its oxidation number is \(\text{Sn}^{2+}\). (c) The species containing the element with the highest oxidation number is \(\text{Cr}_{2}\text{O}_{7}^{2-}\) (Cr has an oxidation number of +6).

Compare basic and acidic reactions

The given reaction takes place in a basic medium and includes OH⁻ ions. Compare it with the hypothetical reaction in an acidic medium: Basic reaction: 3\(\text{Sn}^{2+}\) + \(\text{Cr}_{2}\text{O}_{7}^{2-}\) + 4\(\text{H}_{2}\text{O}\) → 3\(\text{Sn}^{4+}\) + \(\text{Cr}_{2}\text{O}_{3}\) + 8\(\text{OH}^-\). For an acidic reaction, we would replace OH⁻ ions with H⁺ ions: Hypothetical acidic reaction: 3\(\text{Sn}^{2+}\) + \(\text{Cr}_{2}\text{O}_{7}^{2-}\) + \(\text{H}^+\) → 3\(\text{Sn}^{4+}\) + \(\text{Cr}_{2}\text{O}_{3}\)+ n\(\text{H}^+\) (n would be the number of H⁺ ions required to balance the equation). (d) The species that would not be included in the reaction taking place in acidic medium is \(\text{OH}^-\) ion.

Key concepts

Oxidizing and Reducing Agents

Understanding the roles of oxidizing and reducing agents is crucial in the study of redox (reduction-oxidation) reactions. In the context of a redox reaction, an oxidizing agent, also known as an oxidant, is a substance that accepts electrons from another substance, thus becoming reduced itself. Conversely, a reducing agent, or reductant, donates electrons to another substance, leading to its own oxidation.

For example, in the balanced redox reaction provided, the compound containing chromium, \(\text{Cr}_{2}\text{O}_{7}^{2-}\), is the oxidizing agent because it gains electrons, reducing its oxidation number from +6 to +3. This reduction corresponds to the gain of electrons by the chromium ions within the compound. On the flip side, the tin cations, \(\text{Sn}^{2+}\), act as the reducing agent as they lose electrons and their oxidation number increases from +2 to +4, indicating oxidation.

Identifying these agents is essential for understanding how the electrons are transferred between reactants and the overall process governing the chemical reaction.

Change in Oxidation Numbers

The change in oxidation numbers is a fundamental aspect of redox reactions as it indicates the transfer of electrons between substances involved. The oxidation number, often referred to as oxidation state, helps in determining how many electrons are lost or gained by an element during a reaction.

In the given reaction, \(\text{Sn}^{2+}\) is oxidized as its oxidation number increases from +2 to +4. This increase suggests that tin loses two electrons per ion. Meanwhile, the compound \(\text{Cr}_{2}\text{O}_{7}^{2-}\) sees a decrease in the oxidation number of chromium from +6 to +3, indicating that each chromium atom gains three electrons.

Applying this understanding, we can address the questions posed by observing that \(\text{Sn}^{2+}\) is the species where tin increases its oxidation number, whereas chromium in \(\text{Cr}_{2}\text{O}_{7}^{2-}\) possesses the highest initial oxidation number before the reaction proceeds.

Redox Reactions in Acidic vs Basic Medium

Redox reactions can take place in either acidic or basic mediums, and this environment can impact the reactants and products involved. In a basic medium, hydroxide ions (\(\text{OH}^-\)) are prevalent, and they play a role in the balancing of redox equations by absorbing or releasing protons (\(H^+\)) or electrons.

In our exercise, the reaction occurs in a basic medium, as evidenced by the presence of \(\text{OH}^-\) ions among the products. In contrast, if this reaction were to occur in an acidic medium, \(\text{H}^+\) ions would be present instead of \(\text{OH}^-\) ions. In acidic conditions, we would use \(\text{H}^+\) to balance the charge and mass in the equation, which could change the stoichiometry of the reactants and products.

Answering the question related to what species would not be included if the reaction took place in acidic conditions, \(\text{OH}^-\) ions would be absent, replaced by \(\text{H}^+\) ions which would pair with the anions or interact with other reactants in the acidic environment to maintain charge balance.