Question

The cell in which the following reaction occurs:
$2F{e^{3 + }}_{(aq)} + 2{I^ - }_{(aq)} \to 2F{e^{2 + }}_{(aq)} + {I_2}_{(s)}$ ${E^0}_{(cell)} = 0.236V$ at $298K$
Calculate the standard Gibbs energy and equilibrium constant of the cell reaction.

Answer 1

Hint: We have studied Gibbs energy, cell potential and Nernst equation in electrochemistry. We can solve this problem with the help of formulas. In the electrochemical cell, a redox reaction takes place which is responsible for electrical energy and by the cell reaction, we find its potential, energy etc.

Complete step by step answer:
Nernst equation is used for determination of potential of electrolytic cells. By the Nernst equation we can also derive Gibbs energy and equilibrium constant. Nernst equation for cell is:
\[{E_{cell}} = {E^0} - \left[ {RT/nF} \right]\ln Q\] ,where ${E_{cell}}$ is the potential of the cell, ${E^0}$ is potential at standard conditions, $R$is universal gas constant, $n$ is number of electrons transferred in the redox reaction, $T$ is temperature, $F$ is faraday constant and $Q$ is reaction quotient.
When the electrons move through the solution, they do some work. Formula for work done is:
${W_{red}} = nF{E_{red}}$ . The maximum work done in the process is equal to the change in the Gibbs free energy and the Gibbs free energy is an indication of the spontaneity. Hence,
${W_{red}} = nF{E_{red}}$ $ = - \Delta G$.

At the standard condition change in free energy is called standard Gibbs free energy i.e.$\Delta {G^0} = - nF{E^0}_{red}$.
Relationship between equilibrium constant and the Gibbs free energy is:
$\Delta G = \Delta {G^0} + RT\ln K$ ,where $K$is equilibrium constant.
In the question, it is given that ${E^0}_{(cell)} = 0.236V$ at $298K$. And the cell reaction is,
$2F{e^{3 + }}_{(aq)} + 2{I^ - }_{(aq)} \to 2F{e^{2 + }}_{(aq)} + {I_2}_{(s)}$

Therefore, standard Gibbs energy $\Delta {G^0} = - nF{E^0}_{cell}$$ \Rightarrow - 2 \times 0.236 \times 96487 = - 45.54kJ/mol$
We know that, $\log {K_c} = \dfrac{{nF{E^0}_{cell}}}{{2.303RT}}$
$ \Rightarrow \dfrac{{2 \times 96487 \times 0.326}}{{2.303 \times 8.31 \times 298}} = 7.9854$
${K_c} = 9.62 \times {10^7}$
The equilibrium constant for the above cell reaction is ${K_c} = 9.62 \times {10^7}$.

Note: We cannot use Nernst equation for measuring the cell potential when the current is flowing through electrodes. The Gibbs free energy normally measures $kJ/mol$ but we can convert the unit according to the need of the question.