Q44 P - Chemistry: Molecular Nature Of Matter And Change

Question

Solutions of KMnO₄ are used in redox titrations because dark purple MnO₄^- oxidizing agent changes to the faint pink Mn²⁺ as it is reduced. The end point occurs when a slight purple remains as more KMnO₄ is added. If a sample that has reached this end point stands for a long time, the purple fades and a suspension of a small amount of brown, muddy MnO₂ appears. Use standard electrode potentials to explain this result.

Step-by-Step Solution

Answer

The  $E_{half-cell}^o$for the formation of MnO₄ to Mn²⁺ is +1.51V. While the $E_{half-cell}^o$ of MnO₂ to Mn²⁺ is 1.23V. Given that the $E_{half-cell}^o$ for manganate is more positive than that of MnO₂, it is reduced as it is the cathode, while the latter is oxidized as the anode, therefore its corresponding reaction will be reversed. Mn²⁺ will be oxidized to form MnO₂.

Step 1: Definition of Coordination Compound

A coordination complex consists of a central atom or ion, which is sometimes metallic and is termed the coordination centre, and a surrounding array of bound molecules or ions, that are successively referred to as ligands or complexing agents.

Step 2: Solve the given statements

In a redox reaction, dark-colored $Mn{O}_4^{-}$ is reduced to Mn²⁺ which is faint pink in solution. After some time, at endpoint, Mn²⁺ is oxidized to MnO₂ this is a brown solid. At endpoint, there is an ample amount of MnO²⁺ that has accumulated.

 

The $E_{half-cell}^o$ for the reaction of MnO₄ to Mn²⁺ is +1.51V. While the $E_{half}^{-anl}$ of MnO₂ to Mn²⁺ is 1.23V. Given that the E_half-enl for manganate is more positive than that of MnO₂, it is reduced as it is the cathode, while the latter is oxidized as the anode, therefore its corresponding reaction will be reversed. Mn²⁺ will be oxidized to form MnO₂, just as noted in the given.