Chapter 17, Problem 17.21QA

Interpretation Introduction

To determine:

a)  The number of electrons transferred in the cell reaction.

b)  The oxidation states of the transition metals in the reaction.

c)  Draw the cell.

Answer to Problem 17.21QA

Solution:

a)  $6$ electrons are transferred in the cell reaction

b)  

Species	Transition metal	Oxidation state
$K_{2}Fe{O}_4 \left(aq\right)$	$Fe$	$+6$
$Zn\left(s\right)$	$Zn$	$0$
$F{e}_2{O}_3\left(s\right)$	$Fe$	$+3$
$ZnO \left(s\right)$	$Zn$	$+2$
$K_{2}Zn{O}_2\left(aq\right)$	$Zn$	$+2$

c) ${Zn\left(s\right)\left| ZnO\left(s\right), K_{2}Zn{O}_2\left(aq\right)\right|| K}_{2}Fe{O}_4\left(aq\right)| F{e}_2{O}_3\left(s\right)$

Explanation of Solution

1) Concept:

In a voltaic cell, anode is the electrode at which oxidation takes place while cathode is the electrode at which reduction takes place. So, we can write two half-reactions based on the given cell reaction. The number of electrons transferred in the cell reaction can be determined by the number of electrons transferred in the half reactions.

The oxidation states of the transition metals can be determined by knowing the oxidation states of other elements in each species.

In a cell diagram, the anode and the species involved in the oxidation half reaction are written on the left and the cathode and the species involved in the reduction half reaction are written on the right. Single lines are used to separate the phases and a double line to represent the porous bridge separating the two compartments of the cell.

2) Given:

i) Cell reaction: $2 K_{2}Fe{O}_4\left(aq\right)+3 Zn\left(s\right)\to F{e}_2{O}_3\left(s\right)+ZnO\left(s\right)+2 K_{2}Zn{O}_2\left(aq\right)$

3) Calculations:

a) $2 K_{2}Fe{O}_4\left(aq\right)+3 Zn\left(s\right)\to F{e}_2{O}_3\left(s\right)+ZnO\left(s\right)+2 K_{2}Zn{O}_2\left(aq\right)$

Considering potassium as a spectator ion, the net ionic equation is written as

$2 Fe{O}_4^{-2}\left(aq\right)+3 Zn\left(s\right)\to F{e}_2{O}_3\left(s\right)+ZnO\left(s\right)+2 Zn{O}_2^{-2}\left(aq\right)$

Separating the oxidation and reduction half reactions, we get the following:

$2 Fe{O}_4^{-2}\left(aq\right)\to F{e}_2{O}_3\left(s\right)$  is the reduction half reaction.

$3 Zn\left(s\right)\to ZnO\left(s\right)+2 Zn{O}_2^{-2}\left(aq\right)$ is the oxidation half reaction.

Balancing the oxidation half reaction:

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

$3 Zn\left(s\right)+5 H_{2}O \left(l\right)\to ZnO\left(s\right)+2 Zn{O}_2^{-2}\left(aq\right)+10H^{+}\left(aq\right)+6e^{-}$

Balancing the reduction half reaction:

$2 Fe{O}_4^{-2}\left(aq\right)\to F{e}_2{O}_3\left(s\right)$

$2 Fe{O}_4^{-2}\left(aq\right)+10H^{+}\left(aq\right)+6e^{-}\to F{e}_2{O}_3\left(s\right)+5 H_{2}O \left(l\right)$

In each half reaction, there are 6 electrons transferred. So the total number of electrons transferred in the cell reaction is 6.

b) $2 K_{2}Fe{O}_4\left(aq\right)+3 Zn\left(aq\right)\to F{e}_2{O}_3\left(s\right)+ZnO\left(s\right)+2 K_{2}Zn{O}_2\left(aq\right)$

In this cell reaction, $Fe$ and $Zn$ are transition metals. Metal in its pure form has oxidation state 0, so the oxidation state of $Zn$ in $Zn\left(s\right)$ is 0.  $K$ is first group element; therefore, its oxidation state in combined form is $+1$. Oxidation state of $O$ in combined form is $-2$.

Let us assume “x” as oxidation state of $Fe$ in $K_{2}Fe{O}_4$. The sum of oxidation state of K, Fe, and O is zero, so we can get

$\left[2\times \left(+1\right)\right]+x+\left[4\times \left(-2\right)\right]=0$

$+2+x-8=0$

$x-6=0$

$x=+6$

Let us assume that the oxidation state of $Fe in F{e}_2{O}_3$ is x.

$2x+\left[3\times \left(-2\right)\right]=0$

$2x-6=0$

$2x= +6$

$x=+3$

Oxidation state of $Zn$ in $ZnO$: As the oxidation state of $O$ atom in combined state is $-2$, so the oxidation state of $Zn is+2$.

Let us assume that the oxidation state of $Zn i$ n ${ K}_{2}Zn{O}_2$ is x.

$[2\times \left(+1\right)+x+[2\times \left(-2\right)]=0$

$+2+x-4=0$

$x-2=0$

$x=+2$

Thus, the oxidation states of the transition metals are

Species	Transition metal	Oxidation state
$K_{2}Fe{O}_4 \left(aq\right)$	$Fe$	$+6$
$Zn\left(s\right)$	$Zn$	$0$
$F{e}_2{O}_3\left(s\right)$	$Fe$	$+3$
$ZnO \left(s\right)$	$Zn$	$+2$
$K_{2}Zn{O}_2\left(aq\right)$	$Zn$	$+2$

c)

Cell diagram:

Applying the rules for writing the cell diagram, (anode on the left, cathode on the right, bridge in the middle), we get the following:

$Fe$ is reduced from $K_{2}Fe{O}_4\left(aq\right)$ to $F{e}_2{O}_3\left(s\right)$ at cathode while $Zn$ is oxidized from $Zn\left(s\right)$$ZnO\left(s\right)$ and $K_{2}Zn{O}_2\left(aq\right)$ to at anode. So, cell diagram is

${Zn\left(s\right)\left| ZnO\left(s\right), K_{2}Zn{O}_2\left(aq\right)\right|| K}_{2}Fe{O}_4\left(aq\right)| F{e}_2{O}_3\left(s\right)$

Conclusion:

From the given cell reaction, the two half reactions are written and the number of electrons transferred in cell reaction are determined. Applying the rules for assigning oxidation states, we got the oxidation states of transition metals. The cell is drawn using the balanced cell reaction and the rules for writing the cell diagram.