Problem 87 - Chemistry and Chemical Reactivity

Question

A 0.213 -g sample of uranyl(VI) nitrate, \(\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2},\) is dissolved in \(20.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and shaken with Zn. The zinc reduces the uranyl ion, \(\mathrm{UO}_{2}^{2+},\) to a uranium ion, \(\mathrm{U}^{n+}\). To determine the value of \(n,\) this solution is titrated with \(\mathrm{KMnO}_{4} .\) Permanganate is reduced to \(\mathrm{Mn}^{2+}\) and \(\mathrm{U}^{n+}\) is oxidized back to \(\mathrm{UO}_{2}^{2+}\) (a) In the titration, \(12.47 \mathrm{mL}\) of \(0.0173 \mathrm{M} \mathrm{KMnO}_{4}\) was required to reach the equivalence point. Use this information to determine the charge on the ion \(\mathrm{U}^{n+}\). (b) With the identity of \(\mathrm{U}^{n+}\) now established, write a balanced net ionic equation for the reduction of \(\mathrm{UO}_{2}^{2+}\) by zinc (assume acidic conditions). (c) Write a balanced net ionic equation for the oxidation of \(\mathrm{U}^{n+}\) to \(\mathrm{UO}_{2}^{2+}\) by \(\mathrm{MnO}_{4}^{-}\) in acid.

Step-by-Step Solution

Answer

(a) n = 4, (b) \( \text{Zn} + 2\text{H}^+ + \text{UO}_2^{2+} \rightarrow \text{Zn}^{2+} + \text{U}^{4+} + \text{H}_2 \), (c) \( 5\text{U}^{4+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5 \text{UO}_2^{2+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \).

Step 1: Calculate moles of KMnO4 used

The concentration of KMnO4 is given as 0.0173 M and the volume used is 12.47 mL (which needs to be converted to liters). First, convert the volume of KMnO4 from mL to L:\[ 12.47 ext{ mL} = 12.47 imes 10^{-3} ext{ L} \]Now calculate the moles of KMnO4:\[ ext{Moles of KMnO}_4 = 0.0173 ext{ M} imes 12.47 imes 10^{-3} ext{ L} = 2.1553 imes 10^{-4} ext{ mol} \]

Step 2: Determine moles of MnO4- reacting

In the redox reaction between MnO4- and U^{n+}, each mole of KMnO4 provides one mole of MnO4-. Therefore, the moles of MnO4- participating in the reaction is equal to the moles of KMnO4 used, which is \[ 2.1553 imes 10^{-4} ext{ mol MnO}_4^- \]

Step 3: Relate moles of U^n+ oxidized

In acidic media, the reduction of MnO4- to Mn^{2+} involves the transfer of 5 electrons:\[ ext{MnO}_4^- + 8 ext{H}^+ + 5 ext{e}^- \rightarrow ext{Mn}^{2+} + 4 ext{H}_2 ext{O} \]The stoichiometry of the reaction suggests that \[ 1 ext{ mol of MnO}_4^- \equiv 5 ext{ mol of electrons} \]Thus, the total moles of electrons provided from MnO4- is:\[ 2.1553 imes 10^{-4} imes 5 = 1.07765 imes 10^{-3} ext{ mol electrons} \]These electrons are used to oxidize U^{n+} back to UO_2^{2+}, allowing us to deduce the charge of U^{n+}.

Step 4: Determine the charge on U^n+ (n)

The amount of uranium in the original prepared solution from the 0.213 g of uranyl nitrate can be calculated from its molar mass (394.04 g/mol):\[ ext{Moles of UO}_2( ext{NO}_3)_2 = \frac{0.213 ext{ g}}{394.04 ext{ g/mol}} = 5.405 imes 10^{-4} ext{ mol} \]The U^{n+} ions receive the same total amount of electrons, so:\[ 5.405 imes 10^{-4} ext{ mol U}^{n+} \times x ext{ electrons/mol} = 1.07765 imes 10^{-3} ext{ mol electrons} \]Solving for x gives us:\[ x = \frac{1.07765 imes 10^{-3}}{5.405 imes 10^{-4}} = 2 \]Here, "x" refers to the oxidation state difference upon reduction, meaning U^{n+} becomes U^{4+}, indicating the charge n=4.

Step 5: Write the net ionic equation for reduction by zinc

In the presence of zinc and under acidic conditions, uranyl ion \( \text{UO}_2^{2+} \) is reduced to \( \text{U}^{4+} \). The balanced net ionic equation is:\[ \text{Zn} + 2\text{H}^+ + \text{UO}_2^{2+} \rightarrow \text{Zn}^{2+} + \text{U}^{4+} + \text{H}_2\]

Step 6: Write the net ionic equation for oxidation by MnO4-

The previously reduced uranium ion \( \text{U}^{4+} \) is oxidized back to \( \text{UO}_2^{2+} \) by permanganate ion in acidic conditions:\[ 5\text{U}^{4+} + \text{MnO}_4^- + 8\text{H}^+ \rightarrow 5 \text{UO}_2^{2+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

Key Concepts

Uranyl Ion

The uranyl ion, represented as \( \text{UO}_2^{2+} \), is a common form of uranium in aqueous solutions, especially under oxidizing conditions. This ion is linear in structure, comprising a uranium atom double-bonded to two oxygen atoms. Understanding the uranyl ion is essential when studying redox processes involving uranium compounds.
One of the nuances of dealing with uranyl ions is their behavior in acidic solutions. When dissolved in acid, the uranyl ion can undergo reduction or oxidation depending on the reagents present. In the exercise given, the uranyl ion is reduced before being oxidized again in a titration experiment.

• The initial reduction occurs via a reaction with zinc, reducing the uranyl ion to a uranium ion \( \text{U}^{n+} \).
• Subsequently, in the titration step, the uranium ion \( \text{U}^{n+} \) is re-oxidized back to \( \text{UO}_2^{2+} \).

This transformation highlights the versatility and reactivity of uranyl ions in redox chemistry, especially in analytical chemistry processes, where they are frequently studied using titration techniques.

Oxidation States

Oxidation states, also referred to as oxidation numbers, help describe the transfer of electrons during chemical reactions. They indicate the degree of oxidation or reduction an atom undergoes in a particular compound or reaction.
In the context of the given exercise, understanding the oxidation states of uranium is crucial to solving the problem of identifying \( n \) in the ion \( \text{U}^{n+} \). Initially, uranium in \( \text{UO}_2^{2+} \) is in the +6 oxidation state. When it is reduced to \( \text{U}^{n+} \), the challenge is to determine what this reduced state \( n \) is. The method involves:

• Knowing that through the titration with \( \text{KMnO}_4 \), electrons are transferred which relate directly to the change in oxidation state.
• The electrons transferred in the reduction of \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \) correlate with manganese changing from an oxidation state of +7 to +2.
• From calculations, \( x \), the number of electrons transferred per uranium ion, is found to be 2, indicating a reduction of uranium from a +6 state to a +4 state, thus \( \text{U}^{4+} \).

Mastering oxidation states is fundamental, as this concept forms the backbone of understanding how matter interacts and transforms through redox reactions.

Titration Chemistry

Titration is an essential analytical technique used to determine the unknown concentration of a given solute in solution. It involves adding a titrant, a solution with a known concentration, to a solution with the unknown concentration until the reaction reaches an equivalence point.
In the exercise described, titration using \( \text{KMnO}_4 \) serves as a method to determine the amount of uranium reduced by zinc. Here's how it operates:

• The solution containing \( \text{U}^{n+} \) is titrated with a known concentration of \( \text{KMnO}_4 \) to find the equivalence point, where all \( \text{U}^{n+} \) is re-oxidized to \( \text{UO}_2^{2+} \).
• The volume and concentration of \( \text{KMnO}_4 \) used provides vital information which relates to the volume of \( \text{MnO}_4^{-} \) reacted, indirectly determining the charge on uranium \( n \).
• The stoichiometry of the reaction and the amount of five electrons transferred per molecule of \( \text{MnO}_4^{-} \), when known, aids in determining \( n = 4 \).

Titration chemistry is a wonderful tool for chemists to not only find concentrations but also to analyze complex redox reactions, offering a stoichiometric bridge between reactants and products.