Chapter 18.3, Problem 2PPB

A galvanic cell with E°_cell = 0.30 V can be constructed using an iron electrode in a 1.0 M Fe(NO₃)₂ solution, and either a tin electrode in a 1.0 M Sn(NO₃)₂ solution, or a chromium electrode in a 1.0 M Cr(NO₃)₃ solution—even though Sn²⁺/Sn and Cr³⁺/Cr have different standard reduction potentials. Explain and give the overall balanced reaction for each cell.

Interpretation Introduction

Interpretation:

To explain the balanced cell reaction of two galvanic cells constructed through the combination of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Sn}$  in $\text{Sn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Cr}$ in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_3$. But the overall standard potential of both the cell was $\text{0}\text{.30V}\text{.}$

Concept introduction:

Galvanic cell is an experimental set up used to generated electricity through spontaneous redox reaction. The cell works on the principle of oxidation (anode) and reduction (cathode), taking place simultaneously in a separately location with the transfer of electron between them through an external wire. The standard potential of the electrodes (E^o) were measured in relative with standard hydrogen electrode (SHE) at standard temperature and pressure with electrolyte concentration of 1M and it was tabulated.

In this case two galvanic cells were constructed through the combination of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Sn}$ in $\text{Sn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Cr}$ in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_3$. Even though $\text{Sn}$ $\left({\text{E}}^{\text{o}}\text{= +0}\text{.13 V}\right)$ and $\text{Cr}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.74 V}\right)$ have different standard electrode potential (E^o), the overall standard potential of both the cells were determined as same $\left({\text{E}}_{\text{cell}}\text{= 0}\text{.30V}\right)\text{.}$

Answer to Problem 2PPB

Cell-1:

First let us discuss the cell made up of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Sn}$ in $\text{Sn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$. Since the standard electrode of $\text{Sn}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.14 V}\right)$ is fairly higher than standard electrode potential of $\text{Fe}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.44 V}\right)\text{,}$ Fe undergoes oxidation with liberation of two electrons, which was accepted by ${\text{Sn}}^{\text{2+}}$ to get reduce into $\text{Sn}$. The half-cell reaction for the combination-1 was given below.

$\begin{array}{l}\text{Anode (Oxidation)}{\text{Fe}}_{\text{(s)}}\stackrel{}{\to }{\text{ Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\\ {\text{Cathode (Reduction) Sn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Sn}}_{\text{(s)}}\\ \text{Overall}\text{Reaction}{\text{Fe}}_{\text{(s)}}{\text{+ Sn}}^{\text{2+}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{Sn}}_{\text{(s)}}\text{+}{\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}\end{array}$

Then E_cell of the reaction was calculated as follows

$\begin{array}{l}{\text{E}}^{\text{o}}{}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{{\text{Sn}}^{\text{2+}}\text{/Sn}}{\text{-E}}^{\text{o}}{}_{{\text{Fe}}^{\text{2+}}\text{/Fe}}\\ \text{=}(\text{-0}\text{.14}\text{V)-(-0}\text{.44}\text{V)}\\ \text{=}0.30\text{V}\text{.}\end{array}$

Cell-2:

The combination of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Cr}$ in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_3$_­will be considered as cell-2. The standard electrode of $\text{Fe}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.44 V}\right)\text{,}$ is higher than standard electrode potential of $\text{Cr}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.74 V}\right)\text{,}$ so $\text{Cr}$ acts as anode by undergoing oxidation with liberation of three electrons, which was accepted by Fe ²⁺ to get reduce into Fe. The half-cell reaction for the cell-2 was given below.

$\begin{array}{l}\text{Anode (Oxidation)}{\text{Cr}}_{\text{(s)}}\stackrel{}{\to }{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+ 3e}}^{\text{-}}\\ {\text{Cathode (Reduction) Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{ Fe}}_{\text{(s)}}\end{array}$

Charges in the above reaction are unbalanced. In order to balance the charges, oxidation and reduction half need to be multiplied by a whole number as given below.

$\begin{array}{l}\text{Anode (Oxidation)}2{\text{Cr}}_{\text{(s)}}\stackrel{}{\to }2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+ 6e}}^{\text{-}}\\ {\text{Cathode (Reduction) 3Fe}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{6e}}^{\text{-}}\stackrel{}{\to }{\text{ 3Fe}}_{\text{(s)}}\\ \text{Overall}\text{Reaction}2{\text{Cr}}_{\text{(s)}}\text{+}3{\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}\stackrel{}{\to }2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}\text{+}3{\text{Fe}}_{\text{(s)}}\end{array}$

Then E_cell of the above reaction can be calculated as follows

$\begin{array}{l}{\text{E}}^{\text{o}}{}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{{\text{Fe}}^{\text{2+}}\text{/Fe}}{\text{-E}}^{\text{o}}{}_{{\text{Cr}}^{\text{3+}}\text{/Cr}}\\ \text{=}(\text{-0}\text{.44}\text{V)-(-0}\text{.74}\text{V)}\\ \text{=}0.30\text{V}\text{.}\end{array}$

Explanation of Solution

Cell-1:

First let us discuss the cell made up of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Sn}$ in $\text{Sn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$. Since the standard electrode of $\text{Sn}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.14 V}\right)$ is fairly higher than standard electrode potential of $\text{Fe}$ $\left({\text{E}}^{\text{o}}\text{= -0}\text{.44 V}\right)\text{,}$ Fe undergoes oxidation with liberation of two electrons, which was accepted by Sn²⁺ to get reduce into $\text{Sn}$. The half-cell reaction for the combination-1 was given below.

$\begin{array}{l}\text{Anode (Oxidation)}{\text{Fe}}_{\text{(s)}}\stackrel{}{\to }{\text{ Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\\ {\text{Cathode (Reduction) Sn}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{Sn}}_{\text{(s)}}\\ \text{Overall}\text{Reaction}{\text{Fe}}_{\text{(s)}}{\text{+ Sn}}^{\text{2+}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{Sn}}_{\text{(s)}}\text{+}{\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}\end{array}$

Then E_cell of the reaction was calculated as follows

$\begin{array}{l}{\text{E}}^{\text{o}}{}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{{\text{Sn}}^{\text{2+}}\text{/Sn}}{\text{-E}}^{\text{o}}{}_{{\text{Fe}}^{\text{2+}}\text{/Fe}}\\ \text{=}(\text{-0}\text{.14}\text{V)-(-0}\text{.44}\text{V)}\\ \text{=}0.30\text{V}\text{.}\end{array}$

Cell-2:

The combination of $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Cr}$ in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_3$_­will be considered as cell-2. The standard electrode of $\text{Fe}$ (E^o = -0.44 V) is higher than standard electrode potential of $\text{Cr}$ (E^o = -0.74 V), so $\text{Cr}$ acts as anode by undergoing oxidation with liberation of three electrons, which was accepted by Fe ²⁺ to get reduce into Fe. The half-cell reaction for the cell-2 was given below.

$\begin{array}{l}\text{Anode (Oxidation)}{\text{Cr}}_{\text{(s)}}\stackrel{}{\to }{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+ 3e}}^{\text{-}}\\ {\text{Cathode (Reduction) Fe}}^{\text{2+}}{}_{\text{(aq)}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{ Fe}}_{\text{(s)}}\end{array}$

Charges in the above reaction are unbalanced. In order to balance the charges, oxidation and reduction half need to be multiplied by a whole number as given below.

$\begin{array}{l}\text{Anode (Oxidation)}2{\text{Cr}}_{\text{(s)}}\stackrel{}{\to }2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}{\text{+ 6e}}^{\text{-}}\\ {\text{Cathode (Reduction) 3Fe}}^{\text{2+}}{}_{\text{(aq)}}\text{+}{\text{6e}}^{\text{-}}\stackrel{}{\to }{\text{ 3Fe}}_{\text{(s)}}\\ \text{Overall}\text{Reaction}2{\text{Cr}}_{\text{(s)}}\text{+}3{\text{Fe}}^{\text{2+}}{}_{\text{(aq)}}\stackrel{}{\to }2{\text{Cr}}^{\text{3+}}{}_{\text{(aq)}}\text{+}3{\text{Fe}}_{\text{(s)}}\end{array}$

Then E_cell of the above reaction can be calculated as follows

$\begin{array}{l}{\text{E}}^{\text{o}}{}_{\text{cell}}{\text{= E}}^{\text{o}}{}_{{\text{Fe}}^{\text{2+}}\text{/Fe}}{\text{-E}}^{\text{o}}{}_{{\text{Cr}}^{\text{3+}}\text{/Cr}}\\ \text{=}(\text{-0}\text{.44}\text{V)-(-0}\text{.74}\text{V)}\\ \text{=}0.30\text{V}\text{.}\end{array}$

In the cell-1, Fe act as anode due to low standard reduction potential than Sn, but in case of cell-2 Fe act as cathode, since the standard reduction potential of Cr is lower than Fe. Even though Sn and Cr have different standard reduction potential, coupling with iron resulted in the same standard cell potential.

Conclusion

Balanced equation of the two cells with same standard potential made up of two combination, such as $\text{Fe}$  in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Sn}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and $\text{Fe}$ in $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ with $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ were presented and the reason for attaining the same E^o_cell was well explained above.  This is the best example for the role of same metal as anode and cathode.