Question

In the presence of nitric acid, \(UO\) \(^{2+}\) undergoes a redox process. It is converted to \(\mathrm{UO}_{2}^{2+}\) and nitric oxide (NO) gas is produced according to the following unbalanced equation: $$\begin{aligned}\mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{UO}^{2 *}(a q) & \longrightarrow \\\\\mathrm{NO}(g)+& \mathrm{UO}_{2}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned}$$ If \(2.55 \times 10^{2} \mathrm{mL} \mathrm{NO}(g)\) is isolated at \(29^{\circ} \mathrm{C}\) and \(1.5 \mathrm{atm}\), what amount (moles) of \(UO\) \(^{2+}\) was used in the reaction? (Hint: Balance the reaction by the oxidation states method.)

Short answer

The amount of \(UO^{2+}\) used in the reaction is 0.0189 moles.

Step-by-step solution

Balance the redox equation using the oxidation states method

We will balance the given redox equation step by step: 1. Determine the oxidation states of each element: \(H^+\) : +1 \(NO_3^-\) : N = +5, O = -2 \(UO^{2+}\) : U = +6, O = -2 \(NO\) : N = +2, O = -2 \(UO_2^{2+}\) : U = +4, O = -2 \(H_2O\): H = +1, O = -2 2. Identify the elements undergoing oxidation and reduction: - U is reduced (oxidation state change: +6 to +4) - N in \(NO_3^-\) is oxidized (oxidation state change: +5 to +2) 3. Balance the half-reactions: - Reduction half-reaction: \(UO^{2+} \rightarrow UO_2^{2+}\) - To balance O, add 1 H_2O: \(UO^{2+} \rightarrow UO_2^{2+} + H_2O\) - Oxidation half-reaction: \(NO_3^- \rightarrow NO\) - To balance O, add 2 H^+: \(NO_3^- + 2H^+ \rightarrow NO + H_2O\) 4. Balance the charge by adding electrons and balance the number of electrons transferred in both half-reactions: - Reduction half-reaction: \(UO^{2+} + 2e^{-} \rightarrow UO_2^{2+} + H_2O\) - Oxidation half-reaction: \(NO_3^- + 2H^+ \rightarrow NO + H_2O + 3e^{-}\) - To balance electrons, multiply the reduction half-reaction by 3 and the oxidation half-reaction by 2: - \(3(UO^{2+} + 2e^{-} \rightarrow UO_2^{2+} + H_2O)\) - \(2(NO_3^- + 2H^+ \rightarrow NO + H_2O + 3e^{-})\) 5. Combine the balanced half-reactions and simplify: - \(6H^+ + 2NO_3^- + 3UO^{2+} \rightarrow 3UO_2^{2+} + 4H_2O + 2NO\) The balanced redox equation is: \(6H^+ + 2NO_3^- + 3UO^{2+} \rightarrow 3UO_2^{2+} + 4H_2O + 2NO\)

Calculate moles of NO gas produced

Given the volume, temperature, and pressure of NO gas, use the Ideal Gas Law, \(PV = nRT\), where P is pressure, V is volume, n is moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin. Convert temperature to Kelvin and volume to Liters: - \(T = 29°C + 273.15 = 302.15K\) - \(V = 2.55 × 10^2 mL = 0.255 L\) Solve for n: \(n = \frac{PV}{RT} = \frac{(1.5 atm)(0.255 L)}{(0.0821 L·atm/mol·K)(302.15 K)} = 0.0126 mol\) There are 0.0126 moles of NO gas produced.

Determine moles of UO^{2+} used in the reaction

Using the balanced redox equation and stoichiometry, we can determine the moles of \(UO^{2+}\) used: - From the balanced equation: \(3UO^{2+} \rightarrow 2NO\) - Calculate the moles of \(UO^{2+}\) used based on the moles of NO produced: \(\frac{3 \ moles\ UO^{2+}}{2 \ moles\ NO} × 0.0126 \ moles\ NO = 0.0189 \ moles\ UO^{2+}\) There were 0.0189 moles of \(UO^{2+}\) used in the reaction.

Key concepts

Oxidation States

Understanding oxidation states is key to identifying which atoms are being oxidized or reduced in a reaction. An oxidation state, also known as an oxidation number, represents the hypothetical charge an atom would have if the compound was composed only of ions. By convention, specific rules are followed to assign oxidation states: the oxidation state of an element in its natural form is zero, for ions it equals the charge of the ion, and in compounds, hydrogen usually is +1 and oxygen is -2, while elements in the same group typically have similar oxidation numbers.

In the given exercise, oxidation states were determined for each compound:

• For the reactant \(NO_3^-\), nitrogen has an oxidation state of +5 whereas oxygen is -2.
• For the product \(NO\), nitrogen is in a lower oxidation state of +2 due to reduction.
• The uranium transitions from an oxidation state of +6 in \(UO^{2+}\) to +4 in \(UO_2^{2+}\).

These changes help to identify that:

• The nitrogen is reduced as its oxidation states change from +5 to +2.
• Uranium is also reduced, shifting from +6 to +4.

Half-Reactions

Half-reactions are a way to break down a redox (reduction-oxidation) process into its two components: oxidation and reduction. Each half-reaction separately shows the loss or gain of electrons. Balancing these half-reactions and combining them correctly ensures that the overall redox equation is balanced.

For the problem discussed, separate half-reactions are:

• Reduction half-reaction: \(UO^{2+} \rightarrow UO_2^{2+} + H_2O\)
• Oxidation half-reaction: \(NO_3^- + 2H^+ \rightarrow NO + H_2O + 3e^-\)

To balance these reactions, align the number of electrons exchanged. If more electrons appear on one side, multiply the half-reactions by appropriate factors to match the electron transfer. Only then can they be combined into one overall balanced equation. This method is crucial in solving complex redox reactions accurately.

Ideal Gas Law

The Ideal Gas Law provides a mathematical relationship between the pressure, volume, temperature, and number of moles of a gas. It is expressed as \(PV = nRT\). In this equation, \(P\) represents pressure, \(V\) is volume, \(n\) stands for moles, \(R\) is the ideal gas constant (0.0821 L·atm/mol·K), and \(T\) is the temperature in Kelvin.

In the provided example, the Ideal Gas Law is used to calculate the amount of \(NO\) gas produced:

• First, temperature is converted from Celsius to Kelvin by adding 273.15.
• Volume is converted from milliliters to liters.
• These values are then substituted into the equation to solve for \(n\) (moles of gas).
• As a result, 0.0126 moles of NO are determined to have been produced in the reaction.

This process demonstrates how knowledge of the gas's state conditions and the gas constant can help solve for unknowns in gas reactions.

Stoichiometry

Stoichiometry studies the quantitative relationships between reactants and products in a chemical reaction. It relies on balanced chemical equations to determine the proportions in which different substances react and form products. In stoichiometry, coefficients from a balanced reaction are interpreted as mole ratios.

For the given reaction, stoichiometry helps us understand the relationship between \(UO^{2+}\) and \(NO\): from the balanced equation \(3UO^{2+} \rightarrow 2NO\). This translates into a ratio of 3 moles of \(UO^{2+}\) for every 2 moles of \(NO\) produced.

Thus, when we know the moles of \(NO\) produced, we can calculate the moles of \(UO^{2+}\) consumed:

• Using the given data, 0.0126 moles of NO result from 0.0189 moles of \(UO^{2+}\).

This conversion underlines why stoichiometry is essential for relating measurable quantities in chemical processes.