Chapter 7, Problem 12Q

(a)

Interpretation Introduction

Interpretation:

The given equations have to be balanced by adding electrons as needed.

Concept introduction:

Balancing the equation:

• • There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
• • The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
• • First write the reaction from the given information.
• • Then count the number of atoms of each element in reactants as well as products.
• • Finally obtained values could place it as coefficients of reactants as well as products.
• • Loss of electron and loss of Hydrogen in a compound is oxidation - the compound is oxidized.  Gain of electron, gain of Oxygen in a compound is reduction - the compound is reduced.
• • Oxidation reduction and reduction reaction occur simultaneously in same reaction.

Explanation of Solution

Given equations are:

A complete equation will have same elements present on both sides of the equation.

The above equation can be completed by writing as follows,

$\begin{array}{l}\text{1}\text{. Pb(s)}\text{+}{\text{SO}}_4^{2-}\text{(}\text{aq) }\stackrel{}{\to }{\text{PbSO}}_{\text{4}}\text{(s)}\\ \text{2}{\text{. PbO}}_2(\text{s})+{\text{4H}}^{\text{+}}(\text{aq})+{\text{SO}}_4^{2-}\text{(}\text{aq) }\underset{}{\to }{\text{PbSO}}_{\text{4}}+{\text{2H}}_{\text{2}}\text{O(l)}\end{array}$

Balancing the equation:

A balanced equation will have same elements, electrons and same number of atoms of each side of the reaction. The balanced equation is given below.

$\begin{array}{l}\text{1}\text{. Pb(s)}\text{+}{\text{SO}}_4^{2-}\text{(}\text{aq) }\stackrel{}{\to }{\text{PbSO}}_{\text{4}}\text{(s)}\text{+}{\text{2e}}^{-}\\ \text{2}{\text{. PbO}}_2(\text{s})+{\text{4H}}^{\text{+}}(\text{aq})+{\text{SO}}_4^{2-}\text{(}\text{aq)}\text{+}{\text{2e}}^{-}\underset{}{\to }{\text{PbSO}}_{\text{4}}+{\text{2H}}_{\text{2}}\text{O(l)}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Among the given equations, which is oxidation reaction and reduction reaction have to be identified.

Concept introduction:

Oxidation: The gain of oxygen or the loss of hydrogen or the loss of an electron in a species during a redox reaction is called oxidation.

Reduction: The loss of oxygen or the gain of hydrogen or gain of an electron in a species during a redox reaction is called reduction.

Rules for assigning oxidation number

• • The oxidation number of the atom of an uncombined element is zero.
• • The sum of the oxidation numbers of all atoms in a compound must equal to the total charge on the species.
• • Hydrogen has an oxidation number of +1 unless it is combined with metals, in which case it has an oxidation number of -1.
• • Oxygen has an oxidation number of -2, except in peroxides where it is -1 and in $O{F}_2$ where it is +2.

Explanation of Solution

Oxidation reaction:

Let us consider the following reaction

  $\text{1}\text{. Pb(s)}\text{+}{\text{SO}}_4^{2-}\text{(}\text{aq) }\stackrel{}{\to }{\text{PbSO}}_{\text{4}}\text{(s)}\text{+}{\text{2e}}^{-}$

This reaction is oxidation half reaction, because the reactant lead $Pb)$ is lost two electrons and changed into $PbS{O}_4$. Therefore, the given cell reaction is undergoes for oxidation reaction.

Reduction reaction:

$\text{2}{\text{. PbO}}_2(\text{s})+{\text{4H}}^{\text{+}}(\text{aq})+{\text{SO}}_4^{2-}\text{(}\text{aq)}\text{+}{\text{2e}}^{-}\underset{}{\to }{\text{PbSO}}_{\text{4}}+{\text{2H}}_{\text{2}}\text{O(l)}$

This reaction is a reduction half reaction, because the two electrons are on the reactants side, then the reactant is gaining these two electrons. So, this is a reduction reaction.

(c)

Interpretation Introduction

Interpretation:

Form the given information, which electrode is anode and which is cathode have to be identified.

Concept introduction:

In any electrochemical reaction, an electron goes from one substance to another substance driven by the oxidation-reduction reaction. Electrochemical cells have two conducting electrodes, called the anode and cathode. The anode is defined as the electrode where oxidation occurs. The cathode is the electrode where reduction takes place. Standard electrode potential $\left(E^{°}cell\right)$ can be calculated for both oxidation and reduction reactions. A positive value of $\left(E^{°}cell\right)$ indicates that the reaction proceeds spontaneously in the forward direction whereas, a negative value of $\left(E^{°}cell\right)$ indicates that the reaction proceeds spontaneously in backward direction. The standard cell potential expression is written as follows,

$E^{°}cell=E^{°}{}_{cathode}+E^{°}{}_{anode}$

Explanation of Solution

Let us consider the following reaction

  $\text{1}\text{. Pb(s)}\text{+}{\text{SO}}_4^{2-}\text{(}\text{aq) }\stackrel{}{\to }{\text{PbSO}}_{\text{4}}\text{(s)}\text{+}{\text{2e}}^{-}(\text{left}\text{side}\text{cathode})$

  $\text{2}{\text{. PbO}}_2(\text{s})+{\text{4H}}^{\text{+}}(\text{aq})+{\text{SO}}_4^{2-}\text{(}\text{aq)}\text{+}{\text{2e}}^{-}\underset{}{\to }{\text{PbSO}}_{\text{4}}+{\text{2H}}_{\text{2}}\text{O(l)}$

Lead $(Pb)$ is oxidized is lost for two electrons, so lead is a anode. While lead dioxide (or) lead (IV) oxide is reduced. Thus lead dioxide $Pb{O}_2$ is the cathode.

Note: The electrons on the left side of the half reaction it’s a cathode.