Q21.133CP - Chemistry: Molecular Nature Of Matter And Change

Question

In addition to reacting with gold (see Problem ), aqua regia is used to bring other precious metals into the solution. Balance the skeleton equation for the reaction with Pt:

Step-by-Step Solution

Answer

The balanced equation obtained is –

$3{\mathrm{P}}_{\left(\mathrm{s}\right)}{+4\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}{+18\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}{+16\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}\to 4{\mathrm{NO}}_{\left(\mathrm{g}\right)}{+3\mathrm{PtCl}}_{6\left(\mathrm{aq}\right)}^{2-}+8{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{aq}\right)}$

Step 1: Concept Introduction

A chemical equation is a representation of a chemical process in the form of components. A balanced chemical equation is one in which the number of atoms in all molecules is the same on both sides of the equation.

Step 2: Balancing the Equation

To balance each reaction, divide it into half-reactions and balance the redox through the half-reaction method.

${\mathrm{Pt}\left(\mathrm{s}\right)+\mathrm{NO}}_3^{-}\left(\mathrm{aq}\right)+\mathrm{C}{\mathrm{l}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{PtCl}}_6^{2-}\left(\mathrm{aq}\right)+\mathrm{NO}\left(\mathrm{g}\right)$

Separate the half-reactions.

$$\begin{gathered} \mathrm{OHR}:\mathrm{Pt}+{\mathrm{Cl}}^{-}\to {\mathrm{PtCl}}_6^{2-} \\ {\mathrm{RHR}:\mathrm{NO}}_3^{-}\to \mathrm{NO} \end{gathered}$$

Balance electrons and non-O atoms.

$\begin{array}{c}\mathrm{OHR}:\mathrm{Pt}+6{\mathrm{Cl}}^{-}\to {\mathrm{PtCl}}_6^{2-}+4{\mathrm{e}}^{-}\\ {\mathrm{RHR}:\mathrm{NO}}_3^{-}+3{\mathrm{e}}^{-}\to \mathrm{NO}\end{array}$

Balance reaction charges with ${\mathrm{H}}^{+}$(acidic environment).

$\begin{array}{c}\mathrm{OHR}:\mathrm{Pt}+6{\mathrm{Cl}}^{-}\to {\mathrm{PtCl}}_6^{2-}+4{\mathrm{e}}^{-}\\ {\mathrm{RHR}:\mathrm{NO}}_3^{-}+3{\mathrm{e}}^{-}+4{\mathrm{H}}_2\mathrm{O}\to \mathrm{NO}\end{array}$

Step 3: Balancing the Equation

Balance O atoms by adding ${\mathrm{H}}_2\mathrm{O}$ to the opposite side.

$\begin{array}{c}\mathrm{OHR}:\mathrm{Pt}+6{\mathrm{Cl}}^{-}\to {\mathrm{PtCl}}_6^{2-}+4{\mathrm{e}}^{-}\\ {\mathrm{RHR}:\mathrm{NO}}_3^{-}+3{\mathrm{e}}^{-}+4{\mathrm{H}}^{+}\to \mathrm{NO}+2{\mathrm{H}}_2\mathrm{O}\end{array}$

Multiply reactions by the least common factor and combine half-reactions, and cancel common ions/compounds on both sides accordingly.

$\begin{array}{c}3\times \mathrm{OHR}:3\mathrm{Pt}+18{\mathrm{Cl}}^{-}\to {3\mathrm{PtCl}}_6^{2-}+12{\mathrm{e}}^{-}\\ {4\times \mathrm{RHR}:4\mathrm{N}}_3^{-}+12{\mathrm{e}}^{-}+16{\mathrm{H}}^{+}\to 4\mathrm{NO}+8{\mathrm{H}}_2\mathrm{O}\\ \mathrm{Reaction}:3{\mathrm{P}}_{\left(\mathrm{s}\right)}{+4\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}{+18\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}{+16\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}\to 4{\mathrm{NO}}_{\left(\mathrm{g}\right)}{+3\mathrm{PtCl}}_{6\left(\mathrm{aq}\right)}^{2-}+8{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{aq}\right)}\end{array}$

 

Therefore, the final reaction is –

$3{\mathrm{P}}_{\left(\mathrm{s}\right)}{+4\mathrm{NO}}_{3\left(\mathrm{aq}\right)}^{-}{+18\mathrm{Cl}}_{\left(\mathrm{aq}\right)}^{-}{+16\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}\to 4{\mathrm{NO}}_{\left(\mathrm{g}\right)}{+3\mathrm{PtCl}}_{6\left(\mathrm{aq}\right)}^{2-}+8{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{aq}\right)}$