Question

A solution contains \(\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{3}^{2-}\) at a concentration of \(0.0150 \mathrm{M}\). It was found that in a titration, \(25.00 \mathrm{~mL}\) of this solution reacted completely with \(23.44 \mathrm{~mL}\) of 0.032 \(M \mathrm{FeSO}_{4}\) solution. The reaction gave \(\mathrm{Fe}^{3+}\) as a product in the solution. In this reaction, what is the final oxidation state of the Ce?

Short answer

The final oxidation state of Ce after the reaction is +3.

Step-by-step solution

Write out the balanced chemical equation

Determine the oxidation states of cerium and iron in their respective compounds before and after the reaction. Cerium starts in the +4 oxidation state in \(\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{2}^{2-}\) and iron starts in the +2 oxidation state in \(\mathrm{FeSO}_{4}\). The reaction is a redox reaction where iron is oxidized to \(\mathrm{Fe}^{3+}\) and cerium is reduced. The balanced equation is: \[\mathrm{Ce}^{4+} + \mathrm{Fe}^{2+} \rightarrow \mathrm{Ce}^{3+} + \mathrm{Fe}^{3+}\]

Determine the oxidation state change

From the reaction equation, it is clear that cerium is reduced from +4 to +3 oxidation state.

Calculate the moles of \(\mathrm{FeSO}_{4}\) used

Using the molarity and volume of the \(\mathrm{FeSO}_{4}\) solution, calculate the moles of \(\mathrm{FeSO}_{4}\) that react. Moles of \(\mathrm{FeSO}_{4}\) = Molarity \(\times\) Volume (in liters) = 0.032 M \(\times\) 0.02344 L = 0.00075008 moles.

Apply the stoichiometry of the reaction

For every mole of \(\mathrm{FeSO}_{4}\) that reacts, one mole of \(\mathrm{Ce}^{4+}\) is reduced to \(\mathrm{Ce}^{3+}\). Therefore, the moles of \(\mathrm{Ce}^{4+}\) that reacted is equal to the moles of \(\mathrm{FeSO}_{4}\) used, which is 0.00075008 moles.

Confirm the complete reaction

Check that there was enough \(\mathrm{Ce}^{4+}\) to react completely with the given amount of \(\mathrm{FeSO}_{4}\). Calculate the moles of \(\mathrm{Ce}^{4+}\) present in the original solution: Moles of \(\mathrm{Ce}^{4+}\) = Molarity of \(\mathrm{Ce}^{4+}\) \(\times\) Volume of solution (in liters) = 0.0150 M \(\times\) 0.025 L = 0.000375 moles. Since the required number of moles of \(\mathrm{Ce}^{4+}\) (0.00075008 moles) is double the actual moles present (0.000375 moles), this indicates that all of the \(\mathrm{Ce}^{4+}\) present was reduced to \(\mathrm{Ce}^{3+}\) and confirms the complete reaction.

Key concepts

Oxidation States

Understanding oxidation states, or oxidation numbers, is critical when studying redox (reduction-oxidation) reactions. The oxidation state of an atom in a compound represents the charge it would have if all bonds to atoms of different elements were completely ionic. It's an indicator of the degree of oxidation or reduction an atom undergoes during a chemical reaction.

In the given exercise, we explored the oxidation states of cerium (Ce) and iron (Fe) in a redox reaction. Initially, cerium in \(\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{2}^{2-}\) has an oxidation state of +4, while iron in \(\mathrm{FeSO}_{4}\) has an oxidation state of +2. During the reaction, the iron is oxidized (its oxidation state increases from +2 to +3), while the cerium is reduced (its oxidation state decreases from +4 to +3).

This change is dictated by the transfer of electrons: oxidation involves losing electrons while reduction involves gaining electrons. Keeping track of oxidation states in a redox reaction makes it easier to balance the chemical equation and understand the flow of electrons.

Stoichiometry

Stoichiometry is the section of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In the world of chemical equations, stoichiometry serves as the 'recipe' providing the exact amounts of each substance required to react without any waste. Here, a balanced chemical equation with known reactant concentrations allows us to calculate the amount of chemicals participating in the reaction.

In the provided exercise, we applied stoichiometry to determine the amount of \(\mathrm{FeSO}_{4}\) that reacts. By multiplying the molarity and volume (converted to liters), we found that 0.00075008 moles of \(\mathrm{FeSO}_{4}\) were used in the reaction. Since the reaction stoichiometry is 1:1 between \(\mathrm{FeSO}_{4}\) and \(\mathrm{Ce}^{4+}\), this means the same amount of \(\mathrm{Ce}^{4+}\) reacts. Stoichiometry also confirms that in our reaction mixture, the cerium was entirely consumed, as the amount of cerium present was half the amount required for reaction with the available \(\mathrm{FeSO}_{4}\).

Chemical Titration

Chemical titration is an analytical technique used to determine the concentration of a reactant in a solution. It typically involves the gradual addition of a solution of known concentration, called the titrant, to a known volume of the solution being analyzed, referred to as the analyte. The point at which the reaction between the two solutions is complete is known as the equivalence point.

In our exercise, a titration was performed to react a known volume of \(\mathrm{Ce}\left(\mathrm{SO}_{4}\right)_{2}^{2-}\) solution (the analyte) with a \(\mathrm{FeSO}_{4}\) solution (the titrant). By measuring the volume of titrant needed to reach the equivalence point where the reaction was complete, we could calculate the moles of \(\mathrm{FeSO}_{4}\) and in turn determine the final oxidation state of the Ce. Titration is an elegant and precise method to analyze concentration and understand reaction stoichiometry, often used in various fields including medicine, environmental monitoring, and food science.