Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is [2007] (a) 3 (b) 4 (c) 5 (d) 6

Step-by-Step Solution

Answer

The correct answer is (d) 6.

Step 1: Understand the Reaction

The titration involves potassium dichromate \( K_2Cr_2O_7 \) reacting with Mohr's salt, which is a compound of \( FeSO_4 \, (NH_4)_2SO_4 \, 6H_2O \). Potassium dichromate in acidic solution acts as an oxidizing agent. Mohr's salt acts as a reducing agent. The typical reaction in an acidic environment is where \( Cr_2O_7^{2-} \) gets reduced to \( Cr^{3+} \) and \( Fe^{2+} \) gets oxidized to \( Fe^{3+} \).

Step 2: Balance the Oxidation Reaction

The reduction half-reaction in acidic medium for dichromate is:\[Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\]This indicates that 6 electrons are needed to fully reduce one mole of dichromate ions.

Step 3: Balance the Reduction Reaction

The oxidation half-reaction for \( Fe^{2+} \) to \( Fe^{3+} \) is:\[Fe^{2+} \rightarrow Fe^{3+} + e^-\]This reaction releases one electron per mole of iron converted from ferrous to ferric.

Step 4: Determine Electron Transfer per Mole

Since dichromate requires 6 electrons as calculated in Step 2, and each \( Fe^{2+} \) provides 1 electron when it is oxidized as stated in Step 3, it takes 6 moles of \( Fe^{2+} \) to provide 6 electrons needed to fully reduce one mole of \( Cr_2O_7^{2-} \).

Step 5: Conclusion and Answer Selection

From our calculations, 6 moles of Mohr's salt (each containing 1 \( Fe^{2+} \) ion) are needed per mole of dichromate. Therefore, the correct answer is option (d) 6.

Key Concepts

Potassium Dichromate

Potassium dichromate is a powerful oxidizing agent, often used in chemical laboratories for various redox reactions. Its chemical formula is \( K_2Cr_2O_7 \). When dissolved in solution, it forms the dichromate ion \( Cr_2O_7^{2-} \), which is responsible for its oxidative properties. In the presence of an acid, this ion can readily accept electrons, making it suitable for redox titrations.
The distinct orange color of potassium dichromate shifts to green as it reduces, which aids in visually observing the progress of the reaction. Its ability to change color depending on its oxidation state is particularly useful in titrations, as it gives a visual cue when the reaction is complete. However, care should be taken while handling it, as it is toxic and can cause harm if inhaled or ingested.
In the context of titrations involving potassium dichromate, its role as an electron acceptor is fundamental. It ensures the accurate determination of the amounts of reducing agents present in the solution through its reaction dynamics.

Mohr's Salt

Mohr's salt, known chemically as ferrous ammonium sulfate \( FeSO_4 \, (NH_4)_2SO_4 \, 6H_2O \), is a double salt containing ferrous ion \( Fe^{2+} \). It acts as a stable source of ferrous ions for redox reactions and is less prone to oxidation by air.Several characteristics make it particularly favorable for redox titrations:

• Stability: Mohr's salt is stable in air compared to other ferrous salts, making it reliable for experiments.
• Solubility: It dissolves easily in water, facilitating a homogeneous reaction.
• Role in Redox: In titrations, Mohr’s salt acts as a reducing agent. Each ferrous ion can donate an electron, converting to the ferric state \( Fe^{3+} \).

Mohr's salt's reliability in providing a consistent flow of electrons makes it ideal for titrations against strong oxidizing agents like potassium dichromate. By maintaining consistency in its reactions, it helps assure accurate titration results.

Oxidation-Reduction Reactions

Oxidation-reduction (redox) reactions involve the transfer of electrons between two substances. In a typical redox reaction, one substance undergoes oxidation (loss of electrons) while the other undergoes reduction (gain of electrons). Understanding the dynamics of such reactions is essential in titrations, where precise stoichiometry is crucial:

• Oxidizing agents: These substances gain electrons. In this exercise, potassium dichromate is the oxidizing agent that accepts electrons, thereby being reduced.
• Reducing agents: These substances lose electrons. Mohr's salt serves as the reducing agent by donating electrons, resulting in oxidation.

Redox titrations involve carefully balancing the reduction and oxidation half-reactions. The ultimate goal is to reach an equivalence point, where the amount of oxidizing agent equals the amount of reducing agent in terms of electron transfer. Knowing how to calculate the number of electrons exchanged and the corresponding moles involved is fundamental to mastering oxidation-reduction dynamics in chemical reactions.