Question

The cell in which the following reaction occurs:
$2{\text{F}}{{\text{e}}^{3 + }}_{\left( {{\text{aq}}} \right)} + 2{{\text{I}}^ - }_{\left( {{\text{aq}}} \right)} \to 2{\text{F}}{{\text{e}}^{2 + }}_{_{\left( {{\text{aq}}} \right)}} + {{\text{I}}_2}_{\left( {\text{s}} \right)}$ has ${{\text{E}}^ \circ }_{{\text{cell}}} = 0.236{\text{V}}$ at $298{\text{K}}$.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Answer 1

If the reactants and products are in their standard states, the resulting cell potential is called standard cell potential. It is related to the standard Gibbs free energy for the reaction and thus to the equilibrium constant. This equation is known as Nernst equation.

Complete step by step answer:

Initially, a chemical reaction may start at standard condition, but it changes its concentration and thus the cell potential becomes zero. Thus the reaction is at equilibrium.
Nernst equation is used to calculate the Gibbs energy at standard conditions. It is given by:
Gibbs energy, $\Delta {\text{G}} = - {\text{nF}}{{\text{E}}^ \circ }_{{\text{cell}}} \to \left( 1 \right)$
${\text{n}} \to$ number of electrons exchanged in the reaction.
${\text{F}} \to$ Faraday constant $\left( {96500{\text{C}}.{\text{mo}}{{\text{l}}^{ - 1}}} \right)$
${{\text{E}}^ \circ }_{{\text{cell}}} \to$ cell potential
It can be expressed with respect to equilibrium constant.
i.e. $\Delta {\text{G}} = - 2.303{\text{RT}}\log {{\text{K}}_{\text{c}}} \to \left( 2 \right)$
${\text{R}} \to$ gas constant
${\text{T}} \to$ standard temperature
${{\text{K}}_{\text{c}}} \to$ equilibrium constant
When $\left( 1 \right)$ and $\left( 2 \right)$ is combined, we get
${\text{nF}}{{\text{E}}^ \circ }_{{\text{cell}}}{\text{ = 2}}{\text{.303RT}}\log {{\text{K}}_{\text{c}}} \to \left( 3 \right)$
It is given that the cell potential, ${{\text{E}}^ \circ }_{{\text{cell}}} = 0.236{\text{V}}$
Here, the total number of electrons exchanged is two, i.e. ${\text{n}} = 2$
Initially we have to calculate the Gibbs free energy from the formula $\left( 1 \right)$
i.e. $\Delta {\text{G}} = - 2 \times 96500{\text{C}}.{\text{mo}}{{\text{l}}^{ - 1}} \times 0.236{\text{V = - 45548J = - 45}}{\text{.548kJ}}$
Now we have to calculate the equilibrium constant of the cell. Thus we can use the formula $\left( 3 \right)$.
i.e. ${\text{nF}}{{\text{E}}^ \circ }_{{\text{cell}}}{\text{ = 2}}{\text{.303RT}}\log {{\text{K}}_{\text{c}}}$
From the above equation, equilibrium constant can be written as:
$\ln {{\text{K}}_{\text{c}}} = \dfrac{{{\text{nF}}{{\text{E}}^ \circ }_{{\text{cell}}}}}{{{\text{2}}{\text{.303RT}}}}$
Substituting the values, we get
$\ln {{\text{K}}_{\text{c}}} = \dfrac{{45.54 \times {{10}^3}{\text{J}}}}{{2.303 \times 8.314{\text{J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}} \times 298{\text{K}}}}$
On simplifying, we get
$\log {{\text{K}}_{\text{c}}} = \dfrac{{45.54 \times {{10}^3}}}{{5705.84}} = 7.98$
${{\text{K}}_{\text{c}}} =$ antilog $\left( {7.98} \right)$
${{\text{K}}_{\text{c}}} = 9.57 \times {10^7}$
Hence the Gibbs energy, $\Delta {\text{G}} = - 45.54{\text{kJ}}$
Equilibrium constant, ${{\text{K}}_{\text{c}}} = 9.57 \times {10^7}$

Note:
This equation was developed on the basis of Faraday’s law of electrolysis. It explains that the quantities of substances liberated at electrodes depend upon the following three factors:
Quantity of current passed
Time duration of passing current at uniform rate
Charge on ions being deposited.