Ionic liquids have many applications in engineering and materials science. The dissolution of the metavanadate ion in chloroaluminate ionic liquids has been studied: $$\mathrm{VO}_{3}^{-}+\mathrm{AlCl}_{4}^{-} \longrightarrow \mathrm{VO}_{2}\mathrm{Cl}_{2}^{-}+\mathrm{AlOCl}_{2}^{-}$$ (a) What is the oxidation number of \(\mathrm{V}\) and \(\mathrm{Al}\) in each ion? (b) In reactions of \(\mathrm{V}_{2} \mathrm{O}_{5}\) with \(\mathrm{HCl}\), acid concentration affects the product. At low acid concentration, \(\mathrm{VO}_{2} \mathrm{Cl}_{2}^{-}\) and \(\mathrm{VO}_{3}^{-}\) form: $$\mathrm{V}_{2} \mathrm{O}_{5}+\mathrm{HCl} \longrightarrow \mathrm{VO}_{2}\mathrm{Cl}_{2}^{-}+\mathrm{VO}_{3}^{-}+\mathrm{H}^{+}$$ At high acid concentration, \(\mathrm{VOCl}_{3}\) forms: $$\mathrm{V}_{2} \mathrm{O}_{5}+\mathrm{HCl} \longrightarrow \mathrm{VOCl}_{3}+\mathrm{H}_{2} \mathrm{O}$$ Balance each equation, and state which, if either, is a redox process. (c) What mass of \(\mathrm{VO}_{2} \mathrm{Cl}_{2}^{-}\) or \(\mathrm{VOCl}_{3}\) can form from \(12.5 \mathrm{~g}\) of \(\mathrm{V}_{2} \mathrm{O}_{5}\) and the appropriate concentration of acid?

Step by step solution

01

Determining Oxidation Numbers

Find the oxidation numbers for V and Al in each ion.For \mathrm{VO}_{3}^{-}, assume O has an oxidation number of -2.Let the oxidation number of V be x:\[ x + (-2) \times 3 = -1 \ x - 6 = -1 \ x = 5. \] So, V in \mathrm{VO}_{3}^{-} is +5.For \mathrm{AlCl}_{4}^{-}, assume Cl has an oxidation number of -1.Let the oxidation number of Al be y:\[ y + (-1) \times 4 = -1 \ y - 4 = -1 \ y = 3. \] So, Al in \mathrm{AlCl}_{4}^{-} is +3.For \mathrm{VO}_{2}\text{Cl}_{2}^{-}, assume O is -2 and Cl is -1.Let the oxidation number of V be z:\[ z + (-2) \times 2 + (-1) \times 2 = -1 \ z - 4 - 2 = -1 \ z = 5. \] So, V in \mathrm{VO}_{2}\text{Cl}_{2}^{-} is +5.For \mathrm{AlOCl}_{2}^{-}, assume O is -2 and Cl is -1.Let the oxidation number of Al be w:\[ w + (-2) + (-1) \times 2 = -1 \ w - 2 - 2 = -1 \ w = 3. \] So, Al in \mathrm{AlOCl}_{2}^{-} is +3.

02

Balancing Equation for Low Acid Concentration

Balance the reaction \mathrm{V}_{2} \mathrm{O}_{5} + \mathrm{HCl} \rightarrow \mathrm{VO}_{2}\text{Cl}_{2}^{-} + \mathrm{VO}_{3}^{-} + \mathrm{H}^{+}.Inspection shows:\[ \mathrm{V}_{2}\text{O}_{5} + 10\mathrm{HCl} = 2\mathrm{VO}_{2}\text{Cl}_{2}^{-} + 2\mathrm{VO}_{3}^{-} + 4\mathrm{H}^{+}. \]This is not a redox process because there is no change in oxidation numbers.

03

Balancing Equation for High Acid Concentration

Balance the reaction \mathrm{V}_{2} \text{O}_{5} + \mathrm{HCl} \rightarrow \mathrm{VOCl}_{3} + \mathrm{H}_{2} \mathrm{O}.Inspection shows:\[ \mathrm{V}_{2}\text{O}_{5} + 6\mathrm{HCl} \to 2\mathrm{VOCl}_{3} + 3\mathrm{H}_{2}\text{O}. \]This is a redox process because V changes oxidation state from +5 (in \mathrm{V}_{2}\text{O}_{5}) to +5 (in \mathrm{VOCl}_{3}).

04

Calculating Molar Masses

Compute the molar masses.For \mathrm{V}_{2} \mathrm{O}_{5}:\[ 2(51) + 5(16) = 182 \text{g/mol}. \]For \mathrm{VO}_{2}\text{Cl}_{2}^{-}:\[ 51 + 2(16) + 2(35.5) = 154 \text{g/mol}. \]For \mathrm{VOCl}_{3}:\[ 51 + 3(35.5) = 157.5 \text{g/mol}. \]

05

Calculating Mass Produced

Find the mass of products.From 12.5 g of \mathrm{V}_{2} \mathrm{O}_{5}:\[ \text{Mols of } \mathrm{V}_{2} \mathrm{O}_{5} = \frac{12.5 \text{g}}{182 \text{g/mol}} = 0.0687 \text{mol}. \]For \mathrm{VO}_{2}\text{Cl}_{2}^{-}:Use stoichiometry:\[0.0687 \text{mol}_{V_{2}\text{O}_{5}} \rightarrow 2(0.0687) \text{mol}_{\text{product}} = 0.1374 \text{mol}. \]mass = 0.1374 mol \times 154 g/mol = 21.14 gFor \mathrm{VOCl}_{3}:\[\text{Moles}\ = 0.0687 \rightarrow 2(0.0687) = 0.1374 mol. \]mass = 0.1374 mol \times 157.5 g/mol = 21.65 g

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Oxidation Numbers

Oxidation numbers are essential in understanding chemical reactions. They represent the hypothetical charge an atom would have if all bonds were ionic. To determine the oxidation numbers in the provided exercise:

• For \(\text{VO}_{3}^{-}\): Assume each oxygen (O) has an oxidation number of -2. The sum of oxidation numbers in \(\text{VO}_{3}^{-}\) must be -1. If the oxidation number of vanadium (V) is \(x\), then \[x + (3 \times -2) = -1\] giving us \(x = 5\). So, V is +5.
• For \(\text{AlCl}_{4}^{-}\): Each chlorine (Cl) is -1. The sum of oxidation numbers must be -1. If the oxidation number of aluminum (Al) is \(y\), then \[y + (4 \times -1) = -1\] giving us \(y = 3\). So, Al is +3.

Understanding these changes helps to determine whether a redox reaction has occurred.

Balancing Chemical Equations

Balancing equations ensures the conservation of mass and charge in a chemical reaction. Let's balance the given reactions:

• For low acid concentration:
  \(\text{V}_{2}\text{O}_{5} + 10\text{HCl} \rightarrow 2\text{VO}_{2}\text{Cl}_{2}^{-} + 2\text{VO}_{3}^{-} + 4\text{H}^{+}\)
  This shows that for each mole of \(\text{V}_{2}\text{O}_{5}\), you need 10 moles of \(\text{HCl}\).
• For high acid concentration:
  \(\text{V}_{2}\text{O}_{5} + 6\text{HCl} \rightarrow 2\text{VOCl}_{3} + 3\text{H}_{2}\text{O}\)
  This ensures all atoms and charges are balanced accordingly.

These balanced equations reflect the stoichiometric relationships among reactants and products.

Redox Processes

Redox (reduction-oxidation) reactions involve changes in oxidation states of elements. A process is classified as redox if there's a shift in oxidation numbers. In the exercise:

• For low acid concentration: \(\text{V}_{2}\text{O}_{5} + 10\text{HCl} \rightarrow 2\text{VO}_{2}\text{Cl}_{2}^{-} + 2\text{VO}_{3}^{-} + 4\text{H}^{+}\), no change in oxidation state for V or any other element occurs, hence it’s not a redox reaction.
• For high acid concentration: \(\text{V}_{2}\text{O}_{5} + 6\text{HCl} \rightarrow 2\text{VOCl}_{3} + 3\text{H}_{2}\text{O}\), V changes oxidation state from +5 (in \(\text{V}_{2}\text{O}_{5}\)) to a different form. This indicates a redox process as V is oxidized and Cl is reduced.

Recognizing redox reactions is crucial for understanding electron exchange processes.

Molecular Mass Calculation

Calculating molecular masses aids in converting between grams and moles. Here's how to find the molecular masses of compounds in the exercise:

• For \(\text{V}_{2}\text{O}_{5}\): \(\text{Molecular mass} = 2 \times 51 + 5 \times 16 = 182 \text{ g/mol}\).
• For \(\text{VO}_{2}\text{Cl}_{2}^{-}\): \(\text{Molecular mass} = 51 + 2 \times 16 + 2 \times 35.5 = 154 \text{ g/mol}\).
• For \(\text{VOCl}_{3}\): \(\text{Molecular mass} = 51 + 3 \times 35.5 = 157.5 \text{ g/mol}\).

This information is fundamental for pursuing accurate stoichiometric calculations.

Stoichiometry

Stoichiometry allows you to relate quantities of reactants and products in a chemical reaction. Given 12.5 g of \(\text{V}_{2}\text{O}_{5}\), let's determine the mass of products:

• First, calculate moles of \(\text{V}_{2}\text{O}_{5}\): \[ \text{Moles of } \text{V}_{2}\text{O}_{5} = \frac{12.5 \text{ g}}{182 \text{ g/mol}} = 0.0687 \text{mol}\].
• For \(\text{VO}_{2}\text{Cl}_{2}^{-}\):\
  Use stoichiometry: \[0.0687 \text{mol}_{\text{V}_{2}\text{O}_{5}} \rightarrow 2 \times 0.0687 \text{mol}_{\text{product}} = 0.1374 \text{mol}\].
  Then, the mass: \[0.1374 \text{ mol} \times 154 \text{ g/mol} = 21.14 \text{ g}\].
• For \(\text{VOCl}_{3}\):\
  Using similar stoichiometry: \[0.0687 \text{ mol} \rightarrow 2 \times 0.0687 \text{ mol} = 0.1374 \text{ mol}\].
  Thus, the mass: \[0.1374 \text{ mol} \times 157.5 \text{ g/mol} = 21.65 \text{ g}\].

This ensures you measure the correct amount of products formed from a given reactant mass.