Question

In the reaction between acidified \[KMn{O_4}\] and hot oxalic acid, the species that gains electrons are:
A) \[{K^ + }\]
 B)\[MnO_4^ - \]
 C)\[{C_2}{0_4}^ - \] -
 D)\[C{O_2}\]

Answer 1

Hint: To solve this problem, we must first find the oxidation state of Mn in \[KMn{O_4}\] . Then we must find the manganese-based products formed by the reaction between \[KMn{O_4}\] and oxalic acid \[({H_2}{C_2}{0_4})\] . Finally, we will find the oxidation state of Mn in the corresponding product and compare the two values.

Complete step by step answer:
The reaction between \[KMn{O_4}\] and oxalic acid takes place in an acidic medium. To create this acidic environment, sulphuric acid is used. Hence, this reaction can be represented as:
 \[2KMn{O_4} + 5{H_2}{C_2}{O_4}^ + 3{H_2}S{O_4} \to 2MnS{O_4}\,\, + {K_2}S{O_4} + 10C{O_{2}} + 8{H_{2}}O.\]
Before we proceed with the question, let us first calculate the oxidation states of manganese in potassium permanganate.
Let O. S. of Mn in \[KMn{O_4}\] = x. We know that the oxidation states of Potassium = \[K = + 1\] ; and that of oxygen = \[O = - 2\] . Also, the net charge on the compound is zero. Hence, the oxidation state of potassium permanganate can be represented as follows:
 \[
  O.S. = O.S.\left( K \right) + O.S.\left( {Mn} \right) + \left[ {O.S.\left( O \right)} \right] \times \left( 4 \right) \\
  \begin{array}{*{20}{l}}
  {0 = \left( { + 1} \right) + \left( x \right) + 4 \times \left( { - 2} \right)} \\
  {x = 8-1 = + 7}
\end{array} \\
 \]
Hence, the oxidation state of Mn in \[KMn{O_4}\] is +7.
 The manganese-based product formed in the reaction is \[MnS{O_4}\] . Let the oxidation state of Mn in \[MnS{O_4}\] be y. we know that the oxidation state of \[S{O_4}\] the molecule is \[\left( { - 2} \right)\] . Also, the net charge on this compound is zero. Hence, the oxidation state of \[MnS{O_4}\] being represented as:
 \[
  O.S.(MnS{O_4}) = O.S.\left( {Mn} \right) + O.S.(S{O_4}) \\
  \begin{array}{*{20}{l}}
  {0 = y + \left( { - 2} \right)} \\
  {y = + 2}
\end{array} \\
 \]
Now, the change of oxidation state is, \[ + 7 - 2 = 5\] . Therefore, reduction takes place. So, the species that gains electrons is \[MnO_4^ - \] .
The correct answer is, B


Note:\[KMn{O_4}\] or potassium permanganate is a crystalline solid which is usually purplish in color. On the other hand, oxalic acid is an organic compound that is found in the form of a white crystalline solid. In this reaction \[KMn{O_4}\] is an oxidizing reagent and oxalic acid (\[{H_2}{C_2}{0_4}\]) is a reducing agent.