Question

One of the classic methods for the determination of the manganese content in steel involves converting all manganese to the deeply colored permanganate ion and then measuring the absorption of light. The steel is first dissolved in nitric acid, producing the manganese (II) ion and nitrogen dioxide gas. This solution is then reacted with an acidic solution containing periodate ion;

the products are the permanganate and iodate ions. Write balanced chemical equations for both of these steps.

Short answer

The balanced equations for the reactions are:

1. $\mathrm{Mn}\left(\mathrm{s}\right)+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{NO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$
   
2. $3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+5{\mathrm{IO}}_4^{-}\left(\mathrm{aq}\right)\to 2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{IO}}_3^{-}\left(\mathrm{aq}\right)$
   

Step-by-step solution

Definition of balanced chemical equation

An equation for a chemical reaction in which the number of atoms for each element and the total charge is the same for both the reactants and the products is defined as the balanced chemical equation.

Balanced chemical equations

The first step is to write the equation for the reaction.

$\mathrm{Mn}\left(\mathrm{s}\right)+{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{NO}}_2\left(\mathrm{g}\right)$

Balancing the first part of the reaction:

The first half-reaction is,

$\mathrm{Mn}\left(\mathrm{s}\right)+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{NO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

For the second part of the reaction:

width="366">Mn2+aq+IO4-aq+H+aq→MnO4-aq+IO3-aq

The half-reaction is,

width="196">Mn2+aq→MnO4-aqIO4-aq+H+aq→IO3-aq

The balanced equations for the reactions are

$$\begin{gathered} \mathrm{Mn}\left(\mathrm{s}\right)+2{\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to {\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{NO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ 2{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+5{\mathrm{IO}}_4^{-}\left(\mathrm{aq}\right)+3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to 2{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{IO}}_3^{-}\left(\mathrm{aq}\right) \end{gathered}$$