Question

Using the standard electrode potentials given in table 3.1, predict if the reaction between the following is feasible.

Given reaction is: ${\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)$ and ${\mathrm{Br}}^{-}\left(\mathrm{aq}\right)$

Answer 1

Electromotive force

• The electromotive force or EMF of a cell is the maximum potential distance between two electrodes of a cell.
• It is also known as the net voltage between the oxidation and reduction half-reactions.
• The formula of EMF of a cell is as follows:

${\mathrm{E}}_{\mathrm{cell}}^{°}={\mathrm{E}}_{\mathrm{cathode}}^{°}-{\mathrm{E}}_{\mathrm{anode}}^{°}$

• If the EMF of a cell is positive, then the reaction is feasible.

Half cell reaction

• The overall reaction of the cell is as follows:

$\mathrm{Fe}{}^{3+}\left(\mathrm{aq}\right)+2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{Br}}_2\left(\mathrm{aq}\right)$

• The half-cell reaction is as follows:

$$\begin{gathered} \mathrm{Cathode}:2{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)\to 2{\mathrm{Fe}}^{2+}\left(\mathrm{s}\right)+2{\mathrm{e}}^{-},{\mathrm{E}}_{\mathrm{cathode}}^{°}=0.77\mathrm{V} \\ \mathrm{Anode}:2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to {\mathrm{Br}}_2\left(\mathrm{aq}\right),{\mathrm{E}}_{\mathrm{Anode}}^{°}=-1.09\mathrm{V} \end{gathered}$$

Feasibility of the reaction

• The EMF of a cell is as follows:

$\begin{array}{rcl}{\mathrm{E}}_{\mathrm{cell}}^{°}& =& {\mathrm{E}}_{\mathrm{cathode}}^{°}-{\mathrm{E}}_{\mathrm{anode}}^{°}\\ & =& 0.77\mathrm{V}-1.09\mathrm{V}\\ & =& -0.32\mathrm{V}\end{array}$

The EMF of a cell is negative hence the reaction is not feasible.