Chapter 23, Problem 52QP

Interpretation Introduction

Interpretation:

The balanced equations for the reaction of $\text{Al}$ in a basic solution are to be represented, given that the aluminum metal is a strong reducing agent in a basic solution.

Concept Introduction:

Aluminum is an abundant metal.

It is an electropositive element.

Its principal ore is bauxite, and it has low density.

It has high tensile strength.

It exhibits a $+3$ oxidation state.

It is malleable, can be rolled into thin foils.

It is ductile, can be drawn into thin wires.

It is excellent conductor of electricity.

An element or compound having the tendency to lose an electron to another element or compound in a redox chemical reaction is known as the reducing agent.

When an aqueous solution’svalue of $\text{pH}$ is more than $7$, it will be called a basic solution.

Answer to Problem 52QP

Solution: The half reaction for the oxidation of $\text{Al}$ to ${\text{AlO}}_2{}^{-}$ that takes place in the basic solution is represented as:

$\text{Al}\left(s\right)+4{\text{OH}}^{-}\left(aq\right)\to {\text{AlO}}_2{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3e^{-}$.

(a) $\text{8Al}\left(s\right)+5{\text{OH}}^{-}\left(aq\right)+3{\text{NO}}_3{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 8{\text{AlO}}_2{}^{-}\left(aq\right)+3{\text{NH}}_3\left(aq\right)$.

(b) $\text{Al}\left(s\right)+{\text{OH}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{AlO}}_2{}^{-}\left(aq\right)+\frac{3}{2}{\text{H}}_{\text{2}}\left(g\right)$.

(c) $\text{4Al}\left(s\right)+3{\text{SnO}}_3{}^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\to {\text{AlO}}_2{}^{-}\left(aq\right)+3\text{Sn}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)$.

Explanation of Solution

a)Aluminum in a basic solution with ${\text{NaNO}}_{\text{3}}$ to give ammonia.

The half reaction for the oxidation of $\text{Al}$ to ${\text{AlO}}_2{}^{-}$ that takes place in the basic solution is represented as:

$\text{Al}\left(s\right)+4{\text{OH}}^{-}\left(aq\right)\to {\text{AlO}}_2{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3e^{-}$.

The half reaction that takes place for the nitrate–ammonia is represented below.

${\text{NO}}_{\text{3}}{}^{-}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)+8e^{-}\to {\text{NH}}_{\text{3}}\left(aq\right)+9{\text{OH}}^{-}\left(aq\right)$.

Both equations combined together to give the complete equation, which is represented below:

$\text{8Al}\left(s\right)+5{\text{OH}}^{-}\left(aq\right)+3{\text{NO}}_3{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 8{\text{AlO}}_2{}^{-}\left(aq\right)+3{\text{NH}}_3\left(aq\right)$.

b)Aluminum in a basic solution with water to give hydrogen.

The half reaction for the oxidation of $\text{Al}$ to ${\text{AlO}}_2{}^{-}$ that takes place in the basic solution is represented.

The reaction is as

$\text{Al}\left(s\right)+4{\text{OH}}^{-}\left(aq\right)\to {\text{AlO}}_2{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3e^{-}$.

The half reaction that takes place for the water–hydrogen is represented below.

${\text{H}}_{\text{2}}\text{O}\left(l\right)+e^{-}\to {\text{OH}}^{-}\left(aq\right)+\frac{1}{2}{\text{H}}_{\text{2}}\left(g\right)$.

Both equations combined together give the complete equation, which is represented below:

$\text{Al}\left(s\right)+{\text{OH}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{AlO}}_2{}^{-}\left(aq\right)+\frac{3}{2}{\text{H}}_{\text{2}}\left(g\right)$.

c)Aluminum in a basic solution with ${\text{Na}}_{\text{2}}{\text{SnO}}_{\text{3}}$ to give metallic tin.

The half reaction for the oxidation of $\text{Al}$ to ${\text{AlO}}_2{}^{-}$ that takes place in the basic solution is represented.

The reaction is as:

$\text{Al}\left(s\right)+4{\text{OH}}^{-}\left(aq\right)\to {\text{AlO}}_2{}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3e^{-}$.

The half reaction that takes place for the ${\text{SnO}}_3{}^{2-}$-$\text{Sn}$ is represented below:

${\text{SnO}}_3{}^{2-}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)+4e^{-}\to \text{Sn}\left(s\right)+6{\text{OH}}^{-}\left(aq\right)$.

Both equations combined together to give the complete equation, which is represented below:

$\text{4Al}\left(s\right)+3{\text{SnO}}_3{}^{2-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\to {\text{AlO}}_2{}^{-}\left(aq\right)+3\text{Sn}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)$.