Problem 65 - General Chemistry: Principles and Modern Applications

Question

Both \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})\) can be used to titrate \(\mathrm{Fe}^{2+}(\mathrm{aq})\) to \(\mathrm{Fe}^{3+}(\mathrm{aq}) .\) Suppose you have available as titrants two solutions: \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})\) and \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}(\mathrm{aq})\). (a) For which solution would the greater volume of titrant be required for the titration of a particular sample of \(\mathrm{Fe}^{2+}(\text { aq }) ?\) Explain. (b) How many \(\mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}(\) aq) would be required for a titration if the same titration requires \(24.50 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) ?\)

Step-by-Step Solution

Answer

For a given sample of \(\mathrm{Fe}^{2+}\) ions, more volume of \(\mathrm{MnO}_{4}^{-}\) is required than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), and the volume of \(0.1000 \mathrm{M} \mathrm{MnO}_{4}^{-}\) required would be 20.42 mL.

Step 1: Analyze the Stoichiometric Relationship

The first step is to write down the reduction half equations for \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} and \(\mathrm{MnO}_{4}^{-}\), and the oxidation half equation for \(\mathrm{Fe}^{2+}\). Then combine the appropriate reduction and oxidation half equations to obtain the full redox equations. For \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\), the full redox equation is \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 6\mathrm{Fe}^{2+} + 14H^{+} \rightarrow 2Cr^{3+} + 6\mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}\), and for \(\mathrm{MnO}_{4}^{-}\), the equation is \( \mathrm{MnO}_{4}^{-} + 5\mathrm{Fe}^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5\mathrm{Fe}^{3+} + 4 \mathrm{H}_{2} \mathrm{O}\).

Step 2: Determine the Volume of Titrant

We will now explore the stoichiometric ratio between the titrant and the sample, which is \(\mathrm{Fe}^{2+}\) ions. For part (a), comparing the stoichiometric ratios, one can see that the same amount of \(\mathrm{Fe}^{2+}\) ions requires more moles of \(\mathrm{MnO}_{4}^{-}\) than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) since 5 moles of \(\mathrm{Fe}^{2+}\) ions react with 1 mole \(\mathrm{MnO}_{4}^{-}\), while 6 moles of \(\mathrm{Fe}^{2+}\) ions react with 1 mole of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\). Therefore, the solution of \(\mathrm{MnO}_{4}^{-}\) would require greater volume for the titration.

Step 3: Convert Volume of one Titrant to Another

For part (b), since the molarity of the two solutions is the same (0.1000 M), the required volume of \(\mathrm{MnO}_{4}^{-}\) will be proportional to the stoichiometric ratio of the two species. The stoichiometric ratio of \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{MnO}_{4}^{-}\) is 6:5 (from step 1). Therefore, the required volume of \(\mathrm{MnO}_{4}^{-}\) = \(24.50 mL* \frac{5}{6} = 20.42 mL\).

Key Concepts

Stoichiometric Ratio

In a redox titration, understanding the stoichiometric ratio is crucial. It tells us how many moles of a titrant react with the moles of a substance in a chemical reaction. For the case of using

• der \( ext{Cr}_{2} ext{O}_{7}^{2-}\)
• \( ext{MnO}_{4}^{-}\)

to titrate \( ext{Fe}^{2+}\) to \( ext{Fe}^{3+}\), the stoichiometric ratio is derived from balancing the redox equations. Let's observe how stoichiometry plays a role here.
For the reaction with \( ext{Cr}_{2} ext{O}_{7}^{2-}\), six moles of \( ext{Fe}^{2+}\) react with one mole of \( ext{Cr}_{2} ext{O}_{7}^{2-}\).
Meanwhile, with \( ext{MnO}_{4}^{-}\), five moles of \( ext{Fe}^{2+}\) react with one mole of \( ext{MnO}_{4}^{-}\).
This ratio is crucial for determining how much titrant is needed to react with the sample entirely. In this particular titration, since the stoichiometric ratio for \( ext{MnO}_{4}^{-}\) involves fewer moles of \( ext{Fe}^{2+}\) per mole of titrant compared to \( ext{Cr}_{2} ext{O}_{7}^{2-}\), more volume of \( ext{MnO}_{4}^{-}\) solution is required.

Oxidation-Reduction Reactions

Redox titrations are a particular category of oxidation-reduction reactions where an oxidizing agent reacts with a reducing agent. These reactions involve the transfer of electrons between chemical species. Here's a simple way to understand oxidation-reduction reactions in the context of this exercise.In the reaction:- \( ext{Fe}^{2+}\) is oxidized to \( ext{Fe}^{3+}\). Oxidation is the loss of electrons.- Meanwhile, \( ext{Cr}_{2} ext{O}_{7}^{2-}\) or \( ext{MnO}_{4}^{-}\) is being reduced. Reduction is the gain of electrons.Breaking down these terms:

• **Oxidizing Agent:** The substance that gains electrons and is reduced, like \( ext{Cr}_{2} ext{O}_{7}^{2-}\)
• **Reducing Agent:** The substance that loses electrons and is oxidized, like \( ext{Fe}^{2+}\)

Understanding these fundamental roles in redox reactions will help predict the direction of the reaction and the changes in oxidation states.

Volumetric Analysis

Volumetric analysis is a quantitative analytical method used in chemistry, primarily for determining concentrations of unknown solutions. Redox titrations, such as the one in the exercise, are a special type of volumetric analysis. This technique involves delivering one solution from a burette and allowing it to react with a known volume of another to reach the equivalence point.Steps in a Redox Volumetric Analysis include:

• **Standardization:** Ensuring the titrant has a known concentration.
• **Titration:** Slowly adding the titrant to the analyte until the reaction reaches an endpoint.
• **Indicator Selection:** Often a color change indicates an endpoint; however, in some redox titrations, the change in color can be from the titrant or the analyte itself.

In this redox titration example:- \( ext{Cr}_{2} ext{O}_{7}^{2-}\) or \( ext{MnO}_{4}^{-}\) is the titrant added to the solution containing \( ext{Fe}^{2+}\).- The reaction ends when all \( ext{Fe}^{2+}\) has been converted to \( ext{Fe}^{3+}\).The volumetric analysis is complete when the exact volume of a titrant needed to complete the reaction is determined, enabling calculation of the analyte concentration. This makes volumetric analysis a powerful tool for chemical quantification.