Chapter 21, Problem 21.204QP

Interpretation Introduction

Interpretation:

The reaction between copper and hydrochloric acid, iron and hydrochloric acid have to be written.  Also the difference between the reactions has to be given using potential of electrodes.

Concept introduction:

Standard electrode potential:  At concentration of ${\text{1 mol dm}}^{\text{-3}}$ and pressure of one atmosphere, the potential of individual electrode of a reversible electrode is called standard electrode potential.  It is denoted as $\text{E°}$.

$\begin{array}{l}{\text{ΔG°}}_{\text{cell}}{\text{ = -nFE°}}_{\text{cell}}\\ \\ \text{n}\text{is}\text{number}\text{of}\text{moles}\\ \\ \text{F is Faraday constant}\\ \\ {\text{E°}}_{\text{cell}}\text{is}\text{standard}\text{cell}\text{potential}\text{.}\end{array}$

When, ${\text{E°}}_{\text{cell}}\text{ > 0}$, the reaction is spontaneous

When, ${\text{E°}}_{\text{cell}}\text{ < 0}$, the reaction is non-spontaneous.

Explanation of Solution

To write: The reaction between copper and hydrochloric acid, iron and hydrochloric acid.

The standard potentials and half reactions for iron and copper are as follows

$\begin{array}{l}\begin{array}{l}{\text{Fe}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\stackrel{}{\to }{\text{Fe}}_{\left(\text{s}\right)}\text{E°= -0}\text{.41 V}\\ \end{array}\hfill \\ \begin{array}{l}{\text{ Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\stackrel{}{\to }{\text{Cu}}_{{}_{\left(\text{s}\right)}}\text{E°= 0}\text{.34 V}\\ \end{array}\hfill \\ \begin{array}{l}{\text{ 2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\stackrel{}{\to }{\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\text{E°= 0}\text{.00 V}\\ \end{array}\hfill \\ {\text{NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{ + 4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + 3e}}^{\text{-}}\stackrel{}{\to }{\text{NO}}_{\left(\text{g}\right)}{\text{ + 2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{E°= 0}\text{.96 V}\hfill \end{array}$

The reaction of iron with hydrochloric acid is given as

$\begin{array}{llll}{\text{Fe}}_{\left(\text{s}\right)}\hfill & \stackrel{}{\to }\hfill & {\text{Fe}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\hfill & \text{E°= -}\left(\text{-0}\text{.41 V}\right)\hfill \\ {\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\hfill & \stackrel{}{\to }\hfill & {\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\hfill & \text{E°= 0}\text{.00 V}\hfill \\ {\text{Fe}}_{\left(\text{s}\right)}{\text{ + 2H}}^{\text{+}}{}_{\left(\text{aq}\right)}\hfill & \stackrel{}{\to }\hfill & {\text{Fe}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + H}}_{\text{2}}{}_{\left(\text{g}\right)}\hfill & {\text{E}}_{\text{rxn}}\text{= 0}\text{.41 V}\hfill \end{array}$

The obtained electrode potential is positive which means spontaneous reaction.  Here iron chloride is formed.

Hydrogen ion are not involved in reaction between copper and nitric acid.  The reaction of copper with nitric acid is given as

$\begin{array}{llll}{\text{Cu}}_{\left(\text{s}\right)}{}^{}\hfill & \stackrel{}{\to }\hfill & {\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\hfill & \text{E°= -}\left(\text{0}\text{.34 V}\right)\hfill \\ {\text{2H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\hfill & \stackrel{}{\to }\hfill & {\text{H}}_{\text{2}}{}_{\left(\text{g}\right)}\hfill & \text{E°= 0}\text{.00 V}\hfill \\ {\text{Cu}}_{\left(\text{s}\right)}{\text{ + 2H}}^{\text{+}}{}_{\left(\text{aq}\right)}\hfill & \stackrel{}{\to }\hfill & {\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + H}}_{\text{2}}{}_{\left(\text{g}\right)}\hfill & {\text{E}}_{\text{rxn}}\text{= 0}\text{.34 V}\hfill \end{array}$

The obtained electrode potential is negative which means non-spontaneous reaction.  Here copper does not react with nitric acid and form hydrogen gas.

In the above reaction, nitrate ion reacts with copper as follows

$\begin{array}{llll}\text{3}\text{×}{\text{ Cu}}_{\left(\text{s}\right)}{}^{}\hfill & \stackrel{}{\to }\hfill & {\text{Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + 2e}}^{\text{-}}\hfill & \text{E°= -}\left(\text{0}\text{.34 V}\right)\hfill \\ \text{2}\text{×}{\text{ NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{\text{ + 4H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + 3e}}^{}\hfill & \stackrel{}{\to }\hfill & {\text{NO}}_{\left(\text{g}\right)}{\text{ + 2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\hfill & \text{E°= 0}\text{.96 V}\hfill \\ {\text{3Cu}}_{\left(\text{s}\right)}{\text{ + 8H}}^{\text{+}}{}_{\left(\text{aq}\right)}{\text{ + 2NO}}_{\text{3}}{{}^{\text{-}}}_{\left(\text{aq}\right)}{}^{}\hfill & \stackrel{}{\to }\hfill & {\text{3Cu}}^{\text{2+}}{}_{\left(\text{aq}\right)}{\text{ + 2NO}}_{\left(\text{g}\right)}{\text{ + 4H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\hfill & {\text{E}}_{\text{rxn}}\text{= 0}\text{.62 V}\hfill \end{array}$

The obtained electrode potential is positive which means spontaneous reaction.  Here nitrogen monoxide is formed while copper dissolves in nitric acid.

Conclusion

The reaction between copper and hydrochloric acid, iron and hydrochloric acid was written.  From the electrode potentials, the difference between the reactions was given.