Question

The maximum work that can be obtained from the cell
$Al|A{{l}^{3+}}\left( 0.1M \right)||\,F{{e}^{2+}}\left( 0.2M \right)|Fe$ is
[given :$E_{A{{l}^{3+}}/Al}^{0}=-1.66V\,\,and\,E_{F{{e}^{2+}}/Fe}^{0} = -0.44V\,$
a.) $605.8KJ$
b.) $505.8KJ$
c.) $705.8KJ$
d.) $905.8KJ$

Answer 1

As the electrode potentials are given in standard condition, we need to find the work obtained by cell in standard conditions.
- The change in free Gibbs energy is a measure of maximum work done by the system done by the cell.
- Charge on one mole of electrons is one faraday.
- Find the standard cell potential from the standard reduction potential of electrodes given in the question.

Complete Solution :
The maximum work done by the cell which is equal to change to Gibbs free energy is given by
$\Delta {{G}^{0}} = -nFE_{cell}^{o}$
Where $n$ is number of moles of electrons changed between electrodes, $F$ is one faraday which is equals to $96548/\left ( V\,mole\,\,{{e}^{-}} \right)$ , $E_{cell}^{o}$ is standard cell potential

Let us write the electrode reactions for finding $n$,
Anode- $2Al\to 2A{{l}^{3+}}+6{{e}^{-}}$ (oxidation)
Cathode- $3F{{e}^{2+}}+6{{e}^{-}}\to 3Fe$ (reduction)
From the equations $n=6$

Also,
$E_{cell}^{o}=E_{Cathode}^{0}-E_{anode}^{0}$
$\Rightarrow E_{cell}^{o} = -0.44+1.66$
$\Rightarrow E_{cell}^{o} = -0.44+1.66$
$\Rightarrow E_{cell}^{o} = 1.22V$

Putting the values
$\Delta {{G}^{0}} = -6\times 1.22\times 96548$
$\Delta {{G}^{0}} = -706.2KJ$
Negative sign indicates work done by the system.
So, the correct answer is “Option C”.

Note: The total amount of energy produced by an electrochemical cell, and thus the amount of energy available to do electrical work, depends on both the cell potential and the total number of electrons that are transferred from the cathode to the anode during the course of a reaction.
- The work done in a cell can also be understood as work done to transfer some charge across a potential barrier which is the developing cell potential in the cell.
- A spontaneous reaction is characterized by positive cell potential.