Q128P - Chemistry: Molecular Nature Of Matter And Change

Question

Dark green manganate salts contain the  $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$ ion. The ion is stable in basic solution but disproportionates in acid to $\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$  and $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}$  . 

(a) What is the oxidation state of $Mn$  in  $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$,   $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{-}}$ , and  $\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}$?

(b) Write a balanced equation for the reaction of $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$  in acidic solution.1

Step-by-Step Solution

Answer

1. The oxidation state of $Mn$ in $Mn{O_4}^{2-}$  is  $+ 6$  , in $Mn{O_4}^{-}$  is  $+7$ and in $Mn{O}_2$  is  $+4$ .
2. The balanced chemical reaction is given by

       $\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{+}\mathit{M}\mathit{n}{{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}}_{\left(aq\right)}\mathbf{ }\mathbf{+}\mathbf{4}{{\mathit{H}}^{{}^{\mathbf{+}}}}_{\left(aq\right)}\mathbf{ }\mathbf{\to }\mathbf{ }\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}\left(s\right)}\mathbf{ }\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\left(l\right)}$

Step 1: Given information

The composition of dark green manganate salt is $Mn{O_4}^{2-}$  ions. This ion stays stable in the basic solution but shows disproportion reaction in acidic solution of  $Mn{O_4}^{-}$ and $Mn{O}_2$   .

Step 2: Definition of the oxidation process and the loss of oxygen atoms involved in the process

The loss of electrons during a chemical reaction which a molecule, atom or ion undergoes is known as oxidation.

The electrons are taken up by  $\mathit{M}\mathit{n}{{\mathbf{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}$ as it is reduced to $\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}$  in the presence of an acid. The reaction involves losing of two oxygen atoms and this produces water molecules with protons in solution.

 

Step 3: The oxidation state of M n   f o r     M n O 4 2 -

(a)The oxidation state of $Mn$ for $Mn{O_4}^{2-}$  is given by:

The overall charge is  -2  .

The charge of each oxygen atom is also  -2  .

The charge of   is calculated as:

 $$\begin{gathered} =-2-4(-2) \\ =-2+8 \\ =+6 \end{gathered}$$

Step 4: The oxidation state of M n   f o r   M n O 4 -

The oxidation state of $Mn$ for  $Mn{O_4}^{-}$ is given by:

The overall charge is  -1  .

The charge of each oxygen atom is also  -2  .

The charge of   is calculated as:

 $$\begin{gathered} =-1-4(-2) \\ =-1+8 \\ =+7 \end{gathered}$$

 

Step 5: The oxidation state of M n   f o r   M n O 2

The oxidation state of $Mn$ for $Mn{O}_2$  is given by:

The overall charge is  0  .

The charge of each oxygen atom is also  -2  .

The charge of   is calculated as:

 $$\begin{gathered} =0-2(-2) \\ =0+4 \\ =+4 \end{gathered}$$

Step 6: The chemical reaction involved in acidic solution

c) The reaction for this is given by:

 ­$\mathbf{2}{\mathit{e}}^{\mathbf{-}}\mathbf{+}\mathbf{ }\mathit{M}\mathit{n}{{{\mathit{O}}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}}_{\left(aq\right)}\mathbf{ }\mathbf{+}\mathbf{4}{{\mathit{H}}^{\mathbf{+}}}_{\left(aq\right)}\mathbf{ }\mathbf{\to }\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}\left(s\right)}\mathbf{ }\mathbf{+}\mathbf{2}{\mathit{H}}_{\mathbf{2}}{\mathit{O}}_{\left(l\right)}$