Question

Consider the following balanced redox reaction in basic medium. \(3 \mathrm{Sn}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(3 \mathrm{Sn}^{4+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+8 \mathrm{OH}^{-}(a q)\) (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

Question: Analyze the balanced redox reaction involving \(\mathrm{Sn}^{2+}\) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), and: (a) Identify the oxidizing agent. (b) Determine the species with the element that increases its oxidation number. (c) Determine the species containing the element with the highest oxidation number. (d) Identify the species that would not be included in the reaction if it took place in an acidic medium. Answer: (a) The oxidizing agent is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). (b) The species with the element that increases its oxidation number is \(\mathrm{Sn}^{2+}\). (c) The species containing the element with the highest oxidation number is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). (d) The species that would not be included in the reaction if it took place in an acidic medium is \(\mathrm{OH}^-\).

Step-by-step solution

Determine oxidation numbers for elements involved in the reaction

First, we must find the oxidation numbers of all elements in each species. Recall the basic rules for assigning oxidation numbers: 1. The oxidation number of an atom in an elemental state is 0. 2. The oxidation number of a monoatomic ion is equal to its charge. 3. Oxygen usually has an oxidation number of -2 (except in peroxides, where it is -1). 4. Hydrogen generally has an oxidation number of +1. 5. The sum of oxidation numbers of all atoms in a compound is equal to the charge of the compound. Using these rules, we can find the oxidation numbers of the elements in the given reaction: \( \mathrm{Sn}^{2+}\): The oxidation number of Sn is +2. \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\): Oxygen has an oxidation number of -2. By rule 5, the sum of oxidation numbers must be equal to the charge of the compound. In this case: \(2 \times (oxidation\ number\ of\ Cr) + 7 \times (-2) = -2 \) Solving for the oxidation number of Cr, we get: \((oxidation\ number\ of \ Cr) = +6\) \( \mathrm{H}_2\mathrm{O}\): Oxygen has an oxidation number of -2 and hydrogen has an oxidation number of +1. Since water is a neutral compound, the sum of oxidation numbers of all atoms in water is 0, confirming our assignments. \( \mathrm{Sn}^{4+}\): The oxidation number of Sn is +4. \( \mathrm{Cr}_2 \mathrm{O}_3\): Oxygen has an oxidation number of -2. In this compound, chromium's oxidation number can be calculated as: \(2 \times (oxidation\ number\ of\ Cr) + 3 \times (-2) = 0 \) This gives us: \((oxidation\ number\ of\ Cr) = +3\) \( \mathrm{OH}^{-}\): Oxygen has an oxidation number of -2, and hydrogen has an oxidation number of +1. The sum of the oxidation numbers of atoms in the ion must equal its charge, so the assigned values are consistent.

(a) Identify the oxidizing agent

The oxidizing agent is the species that accepts electrons and gets reduced during the reaction. By comparing the initial and final oxidation numbers of each element, we can see that Sn goes from +2 to +4 (which means it loses electrons and is oxidized) and Cr goes from +6 to +3 (which means it gains electrons and is reduced). Therefore, the species containing Cr in its higher oxidation state, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), is the oxidizing agent.

(b) Identify the species with the element that increases its oxidation number

As we already determined in part (a), Sn is the element that increases its oxidation number, going from +2 to +4. Therefore, the species with the element that increases its oxidation number is \(\mathrm{Sn}^{2+}\).

(c) Identify the species containing the element with the highest oxidation number

Comparing the oxidation numbers for Sn and Cr, we see that the highest oxidation number is +6, which belongs to Cr. The species containing Cr in its highest oxidation state is \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\).

(d) Determine the species not included in the reaction if it took place in an acidic medium

The reaction given in the exercise is already balanced in a basic medium, as evidenced by the presence of \(\mathrm{OH}^-\) ions. If the reaction were to take place in an acidic medium, we would have to balance using \(\mathrm{H}^+\) ions instead. In this case, \(\mathrm{OH}^-\) ions would not be involved in the reaction.

Key concepts

Oxidation Numbers

In a redox reaction, oxidation numbers play a crucial role in determining which elements are getting oxidized or reduced. Oxidation numbers are hypothetical charges assigned to atoms, allowing us to track electron transfer. Here are some rules you can use to assign oxidation numbers:

• The oxidation number for an atom in its elemental form is always 0.
• For monoatomic ions, the oxidation number equals the ion's charge.
• Oxygen usually has an oxidation number of -2, except in specific compounds like peroxides, where it is -1.
• Hydrogen generally has an oxidation number of +1, except when it forms compounds with metals (like in hydrides, where it is -1).
• The sum of oxidation numbers in a compound must equal the compound's total charge.

These rules help us identify which species in a reaction gain or lose electrons. By analyzing the initial and final oxidation states in the reaction, we can conclude the following:- Sn changes from +2 in \(\mathrm{Sn}^{2+}\) to +4 in \(\mathrm{Sn}^{4+}\).- Cr changes from +6 in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to +3 in \(\mathrm{Cr}_{2} \mathrm{O}_{3}\).These changes in oxidation numbers help us identify the oxidizing and reducing agents.

Oxidizing Agent

The oxidizing agent in a redox reaction is the substance that is reduced, meaning it gains electrons. This might sound counterintuitive, but remember:

• The oxidizing agent takes electrons from another species.
• As it gains electrons, its oxidation number decreases.

In the given reaction, \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) is the oxidizing agent. You can see this by noting that the oxidation number of chromium decreases from +6 to +3. Why is this important?- It shows us which element in the compound is essential for pulling electrons away from others.- Understanding oxidizing agents helps in predicting reaction outcomes, especially in synthesis and environmental chemistry contexts.

Basic and Acidic Medium Reactions

Redox reactions can occur in different environments, either in a basic or acidic medium. The presence of \(\mathrm{OH}^-\) ions indicates a basic medium, while \(\mathrm{H}^+\) ions indicate an acidic medium. This distinction can change the composition and balance of the reaction.In a basic medium:- Hydroxide ions \(\mathrm{OH}^-\) are present.- Reactions in basic medium often involve adding \(\mathrm{OH}^-\) ions to balance oxygen and hydrogen atoms.If you need to adjust a reaction to take place in an acidic medium:- Replace the \(\mathrm{OH}^-\) ions with \(\mathrm{H}^+\) ions.- You might need to adjust the balance of oxygen and hydrogen by adding water molecules and \(\mathrm{H}^+\) ions.The given reaction is in a basic medium. If you were to perform the same reaction and balance it for an acidic medium, \(\mathrm{OH}^-\) ions would be replaced with \(\mathrm{H}^+\) ions. This alteration is crucial in electrochemistry and biological systems where pH affects reaction pathways.