The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}(a q) \longrightarrow\) \(4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(t)\) This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

First, we will assign oxidation numbers to each element before and after the reaction. The rules to assign oxidation numbers are: - The oxidation number of an atom in its elemental form is 0. - The oxidation number of a monatomic ion is equal to its charge. - Oxygen generally has an oxidation number of -2, except in peroxides, where it is -1. - Hydrogen has an oxidation number of +1 when bonded to a nonmetal and -1 when bonded to a metal. - The sum of the oxidation numbers in a compound must add up to the charge on the compound. - The sum of oxidation numbers in a polyatomic ion must add up to the charge on the ion. Applying the rules, we can list oxidation numbers for each element in the reactants and products: Reactants: \(H^+\): +1 \(Cr_2O_7^{2-}\): \(Cr\) is +6, and \(O\) is -2 \(C_2H_3OH\): \(H\) is +1, \(O\) is -2, \(C\) is -2 (for CH3 group) and -1 (for CH2OH group) Products: \(Cr^{3+}\): \(Cr\) is +3 \(CO_2\): \(C\) is +4, \(O\) is -2 \(H_2O\): \(H\) is +1, \(O\) is -2 Now that we have the oxidation numbers, let's identify the species being reduced and oxidized.

In a redox reaction, reduction is the process in which a species gains electrons (thus, its oxidation number decreases), while oxidation is the process in which a species loses electrons (its oxidation number increases). Comparing the initial and final oxidation numbers for each element, we can find that: - Chromium (Cr): Its oxidation number goes from +6 to +3, indicating a decrease (gain of electrons). Therefore, Cr is being reduced. - Carbon (C): Its oxidation number goes from -2 (in CH3 group) and -1 (in CH2OH group) to +4 in CO2, indicating an increase (loss of electrons). Therefore, C is being oxidized. So, the species being reduced in this reaction is the \(\text{Cr}_2\text{O}_7^{2-}\) ion, and the species being oxidized is the \(\text{C}_2\text{H}_3\text{OH}\) molecule.

Now, we will calculate the number of electrons involved in the balanced equation. We can analyze the change in oxidation numbers for the species being reduced and oxidized: - Chromium (Cr) undergoes a change of +6 to +3, thus gaining 3 electrons. - Carbon (C) undergoes a change of -2 (in CH3 group) and -1 (in CH2OH group) to +4 in CO2, thus losing 6 electrons. In the balanced equation, we have two \(\text{Cr}_2\text{O}_7^{2-}\) ions and one \(\text{C}_2\text{H}_3\text{OH}\) molecule. Therefore, the total number of electrons transferred is: \(2 (\text{Cr}_2\text{O}_7^{2-} \text{ ions}) \times 3 (\text{electrons gained per Cr ion}) = 6 (\text{electrons})\) Hence, the balanced equation above involves the transfer of 6 electrons.