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?

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.

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.

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 \(\mathrm{Sn}^{2+}\).

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 \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\).

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.