Chapter 4, Problem 4.127AP

(a)

Interpretation Introduction

Interpretation: The balanced equation for the tarnishing of $\text{Ag}$ with ${\text{H}}_{\text{2}}\text{S}$ is to be written. The oxidation numbers of the reactants and products are to be assigned. The number of electrons transferred per mole of silver is to be calculated. The balance equation of ${\text{Ag}}_{\text{2}}\text{S}$ with $\text{Al}$ metal, sodium bicarbonate and water is to be written.

Concept introduction: Oxidation number is defined as the charge present on ions. When an atom forms a chemical bond with another atom, the number of electrons that an atom gain or loses is the oxidation number. It is also known as oxidation state.

The chemical equation is said to be balanced when the number of atoms on the reactant side is equal to the number of atom on the product side.

To determine: The balanced equation for the tarnishing of $\text{Ag}$ with ${\text{H}}_{\text{2}}\text{S}$, the oxidation numbers of the reactants and products are to be assigned and the number of electrons transferred per mole of silver.

Answer to Problem 4.127AP

Solution

The balanced equation for the tarnishing of $\text{Ag}$ with ${\text{H}}_{\text{2}}\text{S}$ is,

$2\text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Ag}}_{\text{2}}\text{S}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)$

In the reactant side oxidation number of pure element that is $\text{Ag}$ is $\underset{\_}{0}$ and  the common oxidation number of hydrogen is $\underset{\_}{+1}$.

In the product side oxidation number of $\text{Ag}$ in ${\text{Ag}}_{\text{2}}\text{S}$ is $\underset{\_}{+1}$, oxidation state of sulfur is $\underset{\_}{-2}$ and oxidation number of hydrogen in molecular hydrogen is $\underset{\_}{0}$.

The number of electrons transferred per mole of silver is $\underset{\_}{1mole}$.s

Explanation of Solution

Explanation

When $\text{Ag}$ reacts with ${\text{H}}_{\text{2}}\text{S}$ then tarnishing of $\text{Ag}$ takes place that leads to the formation of ${\text{Ag}}_{\text{2}}\text{S}$ as a product. The reaction that takes place is,

$\text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Ag}}_{\text{2}}\text{S}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)$

To balance the above reaction the coefficient $2$ is placed in front of silver atom on the reactant side. The balanced equation thus obtained is,

$2\text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Ag}}_{\text{2}}\text{S}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)$

In the reactant side, the oxidation number of pure element that is $\text{Ag}$ is $\underset{\_}{0}$.

The oxidation state hydrogen is usually $+1$.

The oxidation number of $\text{S}$ in ${\text{H}}_{\text{2}}\text{S}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on H}}_{\text{2}}\text{S}=\left[\begin{array}{c}\left(\text{Number of}\text{S}\text{atoms}\times \text{oxidation number of S}\right)\\ +\left(\text{Number of H atoms}\times \text{oxidation number of H}\right)\end{array}\right]$

The charge on ${\text{H}}_{\text{2}}\text{S}$ is $0$.

Substitute the value of the number of atoms and their oxidation number in the above formula to calculate the oxidation number of sulfur.

$\begin{array}{c}{\text{Charge on H}}_{\text{2}}\text{S}=\left[\left(2\times \left(1\right)\right)+\left(x\right)\right]\\ 0=\left[\left(2\right)+x\right]\\ x=\underset{\_}{-2}\end{array}$

Thus, the oxidation number of $\text{S}$ in ${\text{H}}_{\text{2}}\text{S}$ is $\underset{\_}{-2}$.

In the product side, the oxidation number of hydrogen in molecular hydrogen is $\underset{\_}{0}$.

The oxidation number of sulfur in ${\text{Ag}}_{\text{2}}\text{S}$ is $-2$.

The oxidation number of $\text{Ag}$ in ${\text{Ag}}_{\text{2}}\text{S}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on Ag}}_{\text{2}}\text{S}=\left[\begin{array}{c}\left(\text{Number of}\text{S}\text{atoms}\times \text{oxidation number of S}\right)\\ +\left(\text{Number of Ag}\text{atoms}\times \text{oxidation number of Ag}\right)\end{array}\right]$

The charge on ${\text{Ag}}_{\text{2}}\text{S}$ is $0$.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of sulfur.

$\begin{array}{c}{\text{Charge on Ag}}_{\text{2}}\text{S}=\left[\left(2\times \left(x\right)\right)+\left(-2\right)\right]\\ 0=\left[\left(-2\right)+2x\right]\\ x=\underset{\_}{+1}\end{array}$

Thus, the oxidation number of $\text{Ag}$ in ${\text{Ag}}_{\text{2}}\text{S}$ is $\underset{\_}{+1}$.

Therefore, in the reactant side oxidation number of pure element that is $\text{Ag}$ is $\underset{\_}{0}$ and the common oxidation number of hydrogen is $\underset{\_}{+1}$.

In the product side oxidation number of $\text{Ag}$ in ${\text{Ag}}_{\text{2}}\text{S}$ is $\underset{\_}{+1}$, oxidation state of sulfur is $\underset{\_}{-2}$ and oxidation number of hydrogen in molecular hydrogen is $\underset{\_}{0}$.

The reaction corresponding to the oxidation of $\text{Ag}$ to ${\text{Ag}}^{\text{+}}$ ion is,

$2\text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Ag}}_{\text{2}}\text{S}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)$

In the above reaction oxidation of silver is represented as,

$\text{Ag}\to {\text{Ag}}^{+}$

This reaction shows that in the process of oxidation, the $\text{Ag}$ atom loses one electron which indicates that only $\underset{\_}{1mole}$ of electrons will be transferred per mole of silver.

Hence, the number of electrons transferred per mole of silver is $\underset{\_}{1mole}$.

(b)

Interpretation Introduction

To determine: The balanced equation for the reaction of ${\text{Ag}}_{\text{2}}\text{S}$ with $\text{Al}$ metal, sodium bicarbonate and water.

Answer to Problem 4.127AP

The balanced equation for the reaction of ${\text{Ag}}_{\text{2}}\text{S}$ with $\text{Al}$ metal, sodium bicarbonate and water is,

$3{\text{Ag}}_{\text{2}}\text{S}\left(s\right)+4\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6\text{Ag}\left(s\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

Explanation of Solution

Explanation

The unbalanced equation of ${\text{Ag}}_{\text{2}}\text{S}$ with $\text{Al}$ metal, sodium bicarbonate and water is,

${\text{Ag}}_{\text{2}}\text{S}\left(s\right)+\text{Al}\left(s\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)+{\text{H}}_{\text{2}}\left(g\right)+\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

The number of oxygen atoms on both the sides are balanced by adding the coefficient $12$ in front of the water molecule on the reactant side and coefficient $4$ is placed in front of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$. Thus equation thus obtained is,

${\text{Ag}}_{\text{2}}\text{S}\left(s\right)+\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)+{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

Now the number of hydrogen atoms is balanced by adding the coefficient $3$ in front of ${\text{H}}_{\text{2}}\text{S}$ and ${\text{H}}_{\text{2}}$ on the product side of the reaction. The equation thus obtained is,

${\text{Ag}}_{\text{2}}\text{S}\left(s\right)+\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ag}\left(s\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

To balance the $\text{Al}$ atoms, the coefficient $4$ is placed in front of $\text{Al}$ on the reactant side of the equation. The equation thus obtained is,

${\text{Ag}}_{\text{2}}\text{S}\left(s\right)+4\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ag}\left(s\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

The sulfur atoms are balanced by adding the coefficient $3$ in front of ${\text{Ag}}_{\text{2}}\text{S}$ on the reactant side of the equation. The equation thus obtained is,

${\text{3Ag}}_{\text{2}}\text{S}\left(s\right)+4\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \text{Ag}\left(s\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

Lastly, the silver atoms are balanced by adding the coefficient $6$ in front of $\text{Ag}$ on the product side. The balanced chemical equation obtained is,

${\text{3Ag}}_{\text{2}}\text{S}\left(s\right)+4\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6\text{Ag}\left(s\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

Conclusion

1. a) The balanced equation for the tarnishing of $\text{Ag}$ with ${\text{H}}_{\text{2}}\text{S}$ is,

   $2\text{Ag}\left(s\right)+{\text{H}}_{\text{2}}\text{S}\left(g\right)\to {\text{Ag}}_{\text{2}}\text{S}\left(s\right)+{\text{H}}_{\text{2}}\left(g\right)$

2. b) The balanced equation for the reaction of ${\text{Ag}}_{\text{2}}\text{S}$ with $\text{Al}$ metal, sodium bicarbonate and water is,

   $3{\text{Ag}}_{\text{2}}\text{S}\left(s\right)+4\text{Al}\left(s\right)+12{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 6\text{Ag}\left(s\right)+3{\text{H}}_{\text{2}}\text{S}\left(g\right)+3{\text{H}}_{\text{2}}\left(g\right)+4\text{Al}{\left(\text{OH}\right)}_{\text{3}}$