Question

When ammonium vanadate is heated with oxalic acid solution, a substance Z is formed. A sample of Z is treated with a solution in hot acidic solution. The resulting liquid was reduced with , the excess boiled off and the liquid again titrated with same . The ratio of volumes used in the two titrations was . What is the oxidation state of vanadium in the substance Z? Given that oxidize all oxidation states of vanadium to vanadium and reduces to .

Answer 1

Many redox reactions involve a transfer of electrons from one substance to another. These processes have ions as reactants or products. ${\text{KMn}}{{\text{O}}_4}$ is a very strong oxidizing agent and oxalic acid acts as a strong reducing agent which helps in the redox reaction.

Complete step by step solution:
The total reaction is as given below:
Ammonium vanadate, ${\text{N}}{{\text{H}}_4}{\text{V}}{{\text{O}}_3}\xrightarrow{{{{\text{C}}_2}{{\text{H}}_2}{{\text{O}}_4}}}{\text{Z}}\xrightarrow{{{\text{KMn}}{{\text{O}}_4}}}{\text{V}}\left( { + 5} \right)\xrightarrow{{{\text{S}}{{\text{O}}_2}}}{\text{V}}\left( { + 4} \right)\xrightarrow{{{\text{KMn}}{{\text{O}}_4}}}{\text{V}}\left( { + 5} \right)$
${{\text{C}}_2}{{\text{H}}_2}{{\text{O}}_4}$ is the chemical formula for oxalic acid. When ${\text{N}}{{\text{H}}_4}{\text{V}}{{\text{O}}_3}$ is reacted with ${{\text{C}}_2}{{\text{H}}_2}{{\text{O}}_4}$, ${\text{V}}\left( { + 5} \right)$ changes to ${\text{V}}\left( {{\text{x}} + } \right)$. So $5 - {\text{x}}$ electrons are involved in the reaction.
${{\text{V}}^{5 + }} + \left( {5 - {\text{x}}} \right){{\text{e}}^ - } \to {{\text{V}}^{{\text{x}} + }}$
Now let’s consider the first titration of ${\text{Z}}$ with ${\text{KMn}}{{\text{O}}_4}$. When ${\text{Z}}$ is reacted with ${\text{KMn}}{{\text{O}}_4}$, vanadium of ${\text{x}}$ oxidation state is changed to vanadium of oxidation state $5$.
i.e. ${{\text{V}}^{{\text{x}} + }} \to {{\text{V}}^{5 + }} + \left( {5 - {\text{x}}} \right){{\text{e}}^ - }$
Now this ${\text{V}}\left( { + 5} \right)$ is reduced to ${\text{V}}\left( {4 + } \right)$ by action of ${\text{S}}{{\text{O}}_2}$.
i.e. $\begin{gathered} {{\text{V}}^{5 + }} + {{\text{e}}^ - } \to {{\text{V}}^{4 + }} \\\ {{\text{S}}^{4 + }} \to {{\text{S}}^{6 + }} + 2{{\text{e}}^ - } \\\ \end{gathered}$
This ${\text{V}}\left( {4 + } \right)$ is titrated with ${\text{KMn}}{{\text{O}}_4}$ to give ${\text{V}}\left( { + 5} \right)$.
i.e. ${{\text{V}}^{4 + }} \to {{\text{V}}^{5 + }} + {{\text{e}}^ - }$
${\text{M}}{{\text{n}}^{7 + }} + 5{{\text{e}}^ - } \to {\text{M}}{{\text{n}}^{2 + }}$
Assume that the sample contains a millimole of ${\text{V}}{{\text{O}}_4}^{3 - }$ and ${{\text{V}}_1}$ be the volume of ${\text{KMn}}{{\text{O}}_4}$.
So the mass equivalent of ${\text{V}}\left( { + 5} \right)$ $=$ mass equivalent of ${\text{V}}\left( {{\text{x}} + } \right)$ $=$ mass equivalent of ${\text{KMn}}{{\text{O}}_4}$.
Let’s assume that ${\text{a}}$ be the millimoles of vanadium.
Thus mass equivalent can be expressed by:
${\text{a}} \times \left( {5 - {\text{x}}} \right) = 1 \times 5 \times {{\text{V}}_1} \to \left( 1 \right)$
Now consider the second titration.
Mass equivalence of ${\text{V}}\left( {4 + } \right)$ $=$ mass equivalence of ${\text{KMn}}{{\text{O}}_4}$.
${\text{a}} \times 1 = 1 \times 5 \times {{\text{V}}_2} \to \left( 2 \right)$
Combining $\left( 1 \right),\left( 2 \right)$, we get
$\dfrac{{{\text{a}} \times \left( {5 - {\text{x}}} \right)}}{{{\text{a}} \times 1}} = \dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} \Leftrightarrow \dfrac{{5 - x}}{1} = \dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} = \dfrac{5}{1}$
This denotes that ${\text{x = 0}}$

So when ${\text{N}}{{\text{H}}_4}{\text{V}}{{\text{O}}_3}$ is reacted with ${{\text{C}}_2}{{\text{H}}_2}{{\text{O}}_4}$, ${\text{V}}\left( { + 5} \right)$ is changed to ${\text{V}}\left( 0 \right)$.

Note:
We know that number of moles can be calculated by dividing given mass, ${\text{w}}$ by molar mass, ${\text{W}}$, i.e. ${\text{n}} = \dfrac{{\text{w}}}{{\text{W}}}$. Also gram equivalents can be calculated by dividing given mass, ${\text{w}}$ by equivalent mass, ${\text{M}}$, i.e. ${\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{\text{M}}}$. And equivalent mass, ${\text{M}}$ can be calculated by dividing molar mass, ${\text{W}}$ by valency or n factor, ${\text{x}}$, i.e. ${\text{M}} = \dfrac{{\text{W}}}{{\text{x}}}$. Combining all these values, we get the equation to find the gram equivalents with respect to the number of moles and n factor, i.e. ${\text{G}}.{\text{eq}} = {\text{n}} \times {\text{x}}$.