Chapter 18, Problem 18.23QP

(a)

Interpretation Introduction

Interpretation: The questions based on the voltaic cell 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.

To determine: The half reactions for anode and cathode for the given reaction between $\text{Cd}\left(s\right)$ and ${\text{MnO}}_4^{-}\left(aq\right)$ that produces $\text{Cd}{\left(\text{OH}\right)}_2\left(s\right)$ and ${\text{MnO}}_2\left(s\right)$ respectively in basic aqueous electrolyte solution.

Answer to Problem 18.23QP

Solution

The half reactions for anode and cathode for the given reaction is given as,

$\begin{array}{c}\text{Cd}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{e}}^{-}\\ {\text{MnO}}_4^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(s\right)+4{\text{OH}}^{-}\end{array}$

Explanation of Solution

Explanation

As given above that the net ionic reaction is the combination of two half reactions that are oxidation half and reduction half.

Oxidation always occurs at anode and reduction always occurs at cathode in such type of electrochemical cell. Reduction is an electron gain process and oxidation is an electron loss process.

The reaction of $\text{Cd}\left(s\right)$ to produce $\text{Cd}{\left(\text{OH}\right)}_2\left(s\right)$ is shown as,

$\text{Cd}\left(s\right)\to \text{Cd}{\left(\text{OH}\right)}_2\left(s\right)$

This reaction is also represented as,

$\text{Cd}\left(s\right)\to {\text{Cd}}^{2+}\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

From the above reaction it is clear that cadmium metal loses its electrons and forms cadmium cation. Therefore, the above reaction is an oxidation reaction and occurs at anode. Hence, the balanced reaction is represented as,

$\text{Cd}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{e}}^{-}$ (1)

The reaction of ${\text{MnO}}_4^{-}\left(aq\right)$ to produce ${\text{MnO}}_2\left(s\right)$ is represented as,

${\text{MnO}}_4^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right)$

The balanced reaction is shown as,

${\text{MnO}}_4^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(s\right)+4{\text{OH}}^{-}$ (2)

This reaction is an electron gain reaction. Therefore, it is an reduction reaction. Hence, it occurs at cathode.

The equation (1) and (2) are the half-reactions for the anode and cathode.

(b)

Interpretation Introduction

To determine: The net ionic balanced equation for describing the cell reaction.

Answer to Problem 18.23QP

Solution

The net ionic balanced equation for the given reaction is given as,

${\text{2MnO}}_4^{-}\left(aq\right)+\text{3Cd}\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{MnO}}_2\left(s\right)+3\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

Explanation of Solution

Explanation

To get the net ionic cell reaction, the individual half-reactions are to be added. Hence, the equations, given in part (a) are represented as,

$\begin{array}{c}\text{Cd}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{e}}^{-}\\ {\text{MnO}}_4^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(s\right)+4{\text{OH}}^{-}\end{array}$

In both the reactions different number of electrons is involved therefore, equation (1) is multiplied by the coefficient three and the equation (2) is multiplied by the coefficient two. The reactions are as follows,

$\begin{array}{c}\text{3Cd}\left(s\right)+6{\text{OH}}^{-}\left(aq\right)\to 3\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+6{\text{e}}^{-}\\ {\text{2MnO}}_4^{-}\left(aq\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)+6{\text{e}}^{-}\to 2{\text{MnO}}_2\left(s\right)+8{\text{OH}}^{-}\end{array}$

On the addition of both the equations, electrons are cancelled with each other. Hence, the net balanced ionic reaction is represented as,

${\text{2MnO}}_4^{-}\left(aq\right)+\text{3Cd}\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{MnO}}_2\left(s\right)+3\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

(c)

Interpretation Introduction

To determine: The cell diagram for the above reaction.

Answer to Problem 18.23QP

Solution

The cell diagram for the above reaction is given as,

$\text{Cd}\left(s\right)|\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)\Vert {\text{MnO}}_4^{-}\left(aq\right)|{\text{MnO}}_2\left(s\right)|\text{Pt}\left(s\right)$

Explanation of Solution

Explanation

The cell diagram is a representation of overall reaction in the form of individual half reactions that are cathodic half and anodic half reaction. In the accepted convention for cell representation the anodic half reaction is putted to the left and cathodic half is putted to the right.

Both the halves are separated by the two parallel vertical lines which are the representation of salt bridge. Here, to support solid ${\text{MnO}}_{\text{2}}$ platinum is act as supporting electrode. The cell diagram for the net ionic balanced reaction is represented as,

$\text{Cd}\left(s\right)|\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)\Vert {\text{MnO}}_4^{-}\left(aq\right)|{\text{MnO}}_2\left(s\right)|\text{Pt}\left(s\right)$

Conclusion

The net ionic equations for the given reactions are given as,

1. a) The half reactions for anode and cathode for the given reaction is given as,

   $\begin{array}{c}\text{Cd}\left(s\right)+2{\text{OH}}^{-}\left(aq\right)\to \text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{e}}^{-}\\ {\text{MnO}}_4^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+3{\text{e}}^{-}\to {\text{MnO}}_2\left(s\right)+4{\text{OH}}^{-}\end{array}$

2. b) The net ionic balanced equation for the given reaction is given as,

   ${\text{2MnO}}_4^{-}\left(aq\right)+\text{3Cd}\left(s\right)+4{\text{H}}_{\text{2}}\text{O}\left(l\right)\to 2{\text{MnO}}_2\left(s\right)+3\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)+2{\text{OH}}^{-}\left(aq\right)$

3. c) The cell diagram for the above reaction is given as,

   $\text{Cd}\left(s\right)|\text{Cd}{\left(\text{OH}\right)}_2\left(aq\right)\Vert {\text{MnO}}_4^{-}\left(aq\right)|{\text{MnO}}_2\left(s\right)|\text{Pt}\left(s\right)$