Question

The standard emf of a galvanic cell involving cell reaction with $n = 2$ is found to be $0.295V$at $25^\circ C$. The equilibrium constant of the reaction would be
(Given F $= 96500Cmo{l^{ - 1}}$, R $= 8.314J{K^{ - 1}}mo{l^{ - 1}}$)

Answer 1

Under standard conditions, the relation between reaction quotient, temperature, cell potential of the electrochemical cell and standard cell potential can be determined with the help of the Nernst equation. The expression for the Nernst equation can be written as:
${E_{cell}} = {E^0} - [\dfrac{RT}{nF}]\ln Q$
Here, ${E_{cell}}$ is the cell potential of the cell, F is Faraday’s constant, Q is reaction quotient, R is the universal gas constant, T is temperature.
The equilibrium constant can be determined with the help of the Nernst equation.

Complete step by step answer:
Let us write the expression used to determine the equilibrium constant. The expression for the Nernst equation is ${E_{cell}} = {E^0} - [\dfrac{RT}{nF}]\ln Q$
At equilibrium we have ${E_{cell}} = 0$
In the question, the temperature is $25^\circ C$
Thus, the expression for the equilibrium constant will be $0 = {E^0}_{cell} - [\dfrac{RT}{nF}]\ln {K_C}$
Here ln can be converted into the log, i.e. $0 = {E^0}_{cell} - [2.303\dfrac{RT}{nF}]\log {K_C}$
${K_C}$ is the equilibrium constant
n is the number of moles of electron involved in the reaction
${E^0}_{cell}$ is the cell potential under standard conditions
We have, T $= 25^\circ C$
When we convert it into Kelvin; the value of T will be 298K.
In this expression $\dfrac{{(2.303)(8.314)(298K)}}{{n(96500)}} = \dfrac{{0.0591}}{n}$
The values of R, T and F are given.
We have, T $= 25^\circ C$
When we convert it into Kelvin; the value of T will be 298K.
We are given the values in the question. i.e. ${E^0}_{cell} = \dfrac{{0.0591}}{n}\log {K_c}$
$\Rightarrow \log {K_c} = \dfrac{{{E^0}_{cell} \times n}}{{0.0591}}$
Substituting the values, we get
$\Rightarrow \log {K_c} = \dfrac{{0.295 \times 2}}{{0.0591}}$
$\Rightarrow \log {K_c} = 10$
$\Rightarrow {K_c} = 1 \times {10^{10}}$
In the last, we can conclude that the value of the equilibrium constant is $1 \times {10^{10}}$

Note: According to the Nernst equation at 298 K, the overall potential of the cell is dependent on reaction quotient. The value of Gibbs free energy is zero, as the reactant and product reach the equilibrium point. The reaction quotient and equilibrium constant are considered to be the same, so we have replaced the reaction quotient to determine equilibrium constant.