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 \(0.1000 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) ?\)

Short 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-by-step solution

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 \( \mathrm{MnO}_{4}^{-} + 5\mathrm{Fe}^{2+} + 8H^{+} \rightarrow Mn^{2+} + 5\mathrm{Fe}^{3+} + 4 \mathrm{H}_{2} \mathrm{O}\).

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.

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.