Chapter 19, Problem 19.19QP

(a)

Interpretation Introduction

Interpretation: The questions, based on the super iron battery for the given cell reaction with iron (VI) that is “super iron” are to be answered

Concept introduction: Voltaic cell is a type of electrochemical cell which converts the chemical energy to electrical energy. This chemical energy generates by processing the redox reaction. Redox reaction is the combination of two half reactions that are oxidation half and reduction half. Oxidation is a process in which loss of electrons take place, while in reduction gain of electron take place.

In voltaic cell during the process of redox reaction electrons are generated on anode and then transfer to the cathode by the external circuit. In this way we can say that the process of oxidation occurs at the anode and the process of reduction occurs at cathode and hence, electron flows. Platinum behaves as a supportive and an inert electrode. Oxidation number is the charge experienced by the individual atom is combined state.

To determine: The number of transferred electrons for the given cell reaction.

Answer to Problem 19.19QP

Solution

The number of transferred electrons for the given cell reaction is $\underset{\_}{6}$.

Explanation of Solution

Explanation

Given

The given reaction is as follows,

${\text{2K}}_{\text{2}}{\text{FeO}}_{\text{4}}\left(aq\right)+3\text{Zn}\left(s\right)\to {\text{Fe}}_2{\text{O}}_3\left(s\right)+\text{ZnO}\left(s\right)+2{\text{K}}_2{\text{ZnO}}_2\left(aq\right)$

The above given reaction the redox reaction and the combination of two half reactions. The change in oxidation state represents the electron transfer.

To calculate the involved number of electrons the oxidation state of atoms is to be calculated.

The oxidation state of zinc at the left side is zero due to the atomic nature.

At the right side in combined form the oxidation of zinc is calculated by the formula,

$\text{Sum}\text{of}\text{oxidation}\text{state}\text{of}\text{all}\text{the}\text{individual}\text{atom}=0$

For $\text{ZnO}$,

Oxidation state of oxygen is $\left(-2\right)$ and oxidation state of zinc is supposed to be x.

Therefore, the oxidation state of zinc in $\text{ZnO}$ is calculated as,

$\begin{array}{c}x+\left(-2\right)=0\\ x=2\end{array}$

For ${\text{K}}_{\text{2}}{\text{ZnO}}_{\text{2}}$,

The oxidation state of potassium is $\left(+1\right)$. Therefore, the oxidation state of zinc in ${\text{K}}_{\text{2}}{\text{ZnO}}_{\text{2}}$ is calculated as,

$\begin{array}{c}2\left(+1\right)+x+2\left(-2\right)=0\\ x-2=0\\ x=2\end{array}$

At the right side in both the compound the oxidation state of zinc is $\left(+2\right)$. Both the side three zinc atoms are present. Therefore, the electron transfer for zinc is written as,

$\begin{array}{c}3\text{Zn}\to 3{\text{Zn}}^{2+}\\ 3\left(0\right)\to 3\left(2\right)\\ 0\to 6\end{array}$

Hence, six electrons are transferred for zinc.

Now same calculations are involved for iron. The oxidation state of iron is supposed to be x. Therefore, the oxidation state of iron in ${\text{K}}_{\text{2}}{\text{FeO}}_{\text{4}}$ is calculated as,

$\begin{array}{c}2\left(+1\right)+x+4\left(-2\right)=0\\ 2+x+\left(-8\right)=0\\ x=+6\end{array}$

The oxidation state of iron in ${\text{Fe}}_2{\text{O}}_3$ is calculated as,

$\begin{array}{c}2x+3\left(-2\right)=0\\ 2x-6=0\\ x=+3\end{array}$

At the right side the oxidation state of iron is $\left(+3\right)$ and the left side it is $\left(+6\right)$. Both the side two iron atoms are present. Therefore, the electron transfer for iron is written as,

$\begin{array}{c}2{\text{Fe}}^{+6}\to 2{\text{Fe}}^{3+}\\ 2\left(+6\right)\to 2\left(+3\right)\\ 12\to 6\end{array}$

Hence, six electrons are transferred for iron.

In both the case same electrons are involved. Hence, the number of transferred electrons for the given cell reaction is $\underset{\_}{6}$.

(b)

Interpretation Introduction

To determine: The oxidation state of transition metals in the given reaction.

Answer to Problem 19.19QP

Solution

The oxidation state of transition metal $\text{Fe}$ is $\underset{\_}{\left(+3\right)}$ and $\underset{\_}{\left(+6\right)}$.

The oxidation state of transition metal $\text{Zn}$ is $\underset{\_}{\left(+2\right)}$.

Explanation of Solution

Explanation

The transition metals are also known as d-block elements. In the given reaction two transition metals are involved which are $\text{Fe}$ and $\text{Zn}$. The oxidation states of both the transition metals are already calculated in the part (a).

Therefore, the oxidation state of iron in ${\text{K}}_{\text{2}}{\text{FeO}}_{\text{4}}$ is $\underset{\_}{\left(+6\right)}$ and in ${\text{Fe}}_2{\text{O}}_3$ is $\underset{\_}{\left(+3\right)}$.

The oxidation state of zinc in both $\text{ZnO}$ and ${\text{K}}_{\text{2}}{\text{ZnO}}_{\text{2}}$ is $\underset{\_}{\left(+2\right)}$.

(c)

Interpretation Introduction

To determine: The cell diagram.

Answer to Problem 19.19QP

Solution

The cell diagram is represented as,

$\text{Zn}\left(s\right)|\text{ZnO}\left(s\right),{\text{ZnO}}_2^{-}\left(aq\right)\Vert {\text{FeO}}_4^{-}\left(aq\right),{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)|\text{Pt}\left(s\right)$

Explanation of Solution

Explanation

From the calculated values it is clear that the zinc undergoes oxidation. Therefore, this reaction occurs at anode. While iron undergoes reduction. Therefore, it occurs at cathode. To represent the given reaction in cathodic and anorthic form the individual atoms are represented as,

$\begin{array}{c}{\text{2K}}_{\text{2}}{\text{FeO}}_{\text{4}}\left(aq\right)\rightleftharpoons 4{\text{K}}^{\text{+}}\left(aq\right)+2{\text{FeO}}_4^{2-}\left(aq\right)\\ 2{\text{K}}_{\text{2}}{\text{ZnO}}_{\text{2}}\left(aq\right)\rightleftharpoons 4{\text{K}}^{\text{+}}\left(aq\right)+2{\text{ZnO}}_2^{2-}\left(aq\right)\\ \text{ZnO}\left(s\right)\rightleftharpoons {\text{Zn}}^{2+}\left(aq\right)+{\text{O}}^{2-}\left(aq\right)\end{array}$

Now, the given equation is represented as,

$\left[\begin{array}{l}4{\text{K}}^{\text{+}}\left(aq\right)+2{\text{FeO}}_4^{2-}\left(aq\right)\\ +3\text{Zn}\left(s\right)\end{array}\right]\to \left[\begin{array}{l}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{ZnO}\left(s\right)+\\ 4{\text{K}}^{+}\left(aq\right)+2{\text{ZnO}}_2^{-}\left(aq\right)\end{array}\right]$

Both the side $4{\text{K}}^{+}$ ions are present. Therefore, it cancels with each other. Hence, the above reaction is represented as,

$2{\text{FeO}}_4^{2-}\left(aq\right)+3\text{Zn}\left(s\right)\to {\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+\text{ZnO}\left(s\right)+2{\text{ZnO}}_2^{-}\left(aq\right)$

Reduction occurs at anode and oxidation occurs at anode. Here, platinum is used as an inert electrode. Hence, for the above reaction the cell diagram is written as,

$\text{Zn}\left(s\right)|\text{ZnO}\left(s\right),{\text{ZnO}}_2^{-}\left(aq\right)\Vert {\text{FeO}}_4^{-}\left(aq\right),{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)|\text{Pt}\left(s\right)$

Conclusion

1. a) The number of transferred electrons for the given cell reaction is $\underset{\_}{6}$.
2. b) The oxidation state of transition metal $\text{Fe}$ is $\underset{\_}{\left(+3\right)}$ and $\underset{\_}{\left(+6\right)}$ and the oxidation state of transition metal $\text{Zn}$ is $\underset{\_}{\left(+2\right)}$.
3. c) The cell diagram is represented as,

   $\text{Zn}\left(s\right)|\text{ZnO}\left(s\right),{\text{ZnO}}_2^{-}\left(aq\right)\Vert {\text{FeO}}_4^{-}\left(aq\right),{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)|\text{Pt}\left(s\right)$