Question

Calculate the equilibrium constant and free energy change of the given following reaction for Daniell cell at 298 K temperature.
$\text{Zn(s)}+\text{C}{{\text{u}}^{\text{2+}}}\text{(aq)}\rightleftharpoons \text{Z}{{\text{n}}^{\text{2+}}}\text{(aq)}+\text{Cu(s)}$
$\text{Cell potential}=1.1\text{ volt, F}=96500\text{ coulomb}$

Answer 1

The Nernst equation is used to determine the equilibrium constant of a reaction taking place in an electrochemical cell and the equilibrium constant is related to Gibbs free energy change by the following equation:
$\Delta \text{G}=-2.303\text{RTlog}{{\text{K}}_{\text{eq}}}$

Complete answer:
Electrochemical cells convert chemical energy to electrical energy or vice-versa through a chemical redox reaction taking place in it. There are two electrodes dipped in a common electrolytic solution. At one of the electrodes, oxidation reaction takes place and at another electrode, reduction part of reaction takes place. The former electrode is known as the anode while the latter one is known as cathode.
The reactions taking place at each electrode are equilibrium reactions. The equilibrium constant is used to study an equilibrium reaction in terms of the concentration of reactants and products. It can be defined as a ratio of the concentration of reactant and product raised to their stoichiometric coefficient in a balanced chemical reaction.
The Nernst gave a relationship to calculate the equilibrium constant for an electrochemical reaction at non-standard conditions. It is given as:
${{\text{E}}_{\text{cell}}}=\text{E}_{\text{cell}}^{\text{o}}-\dfrac{\text{2}\text{.303RT}}{\text{nF}}{{\log }_{10}}{{\text{K}}_{\text{eq}}}$
Where ${{\text{E}}_{\text{cell}}}$ and $\text{E}_{\text{cell}}^{\text{o}}$ are cell potential at a given temperature and standard condition respectively.
n is the number of electrons transferred in a balanced cell reaction.
R is gas constant $\left( 8.314\text{ J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}} \right)$.
T is the absolute temperature.
F is Faraday constant $\left( 1\text{ F}=96500\text{ C mo}{{\text{l}}^{-1}} \right)$.
${{\text{K}}_{\text{eq}}}$ is the equilibrium constant.
At equilibrium condition,

& \text{E}_{\text{cell}}^{\text{o}}=0 \\\ & \therefore {{\text{E}}_{\text{cell}}}=\dfrac{2.303\text{RT}}{\text{nF}}{{\log }_{10}}{{\text{K}}_{\text{eq}}} \\\ \end{aligned}$$ For the given Daniell cell, first, write the two half-cell reactions and balance them to find the value of ‘n’. At cathode: $$\text{C}{{\text{u}}^{\text{2+}}}\text{(aq)}+2{{\text{e}}^{-}}\to \text{Cu(s)}$$ At anode: $$\text{Zn(s)}\to \text{Z}{{\text{n}}^{2+}}\text{(aq)}+2{{\text{e}}^{-}}$$ Thus, we can see that transfer of 2 electrons is taking place in the overall reaction. So, $\text{n}=2$. Now solve for ${{\text{K}}_{\text{eq}}}$ : $$\begin{aligned} & 1.1\text{ V}=\dfrac{\text{2}\text{.303}\times \text{8}\text{.314 J mo}{{\text{l}}^{-1}}{{\text{K}}^{-1}}\times \text{298 K}}{2\times 96500\text{ C mo}{{\text{l}}^{-1}}}{{\log }_{10}}{{\text{K}}_{\text{eq}}} \\\ & \Rightarrow {{\log }_{10}}{{\text{K}}_{\text{eq}}}=37.21 \\\ & \Rightarrow {{\text{K}}_{\text{eq}}}=1.62\times {{10}^{37}} \\\ \end{aligned}$$ The change in free energy is related to the equilibrium constant as follows: $$\begin{aligned} & \Delta \text{G}=-2.303\text{RT}{{\log }_{10}}{{\text{K}}_{\text{eq}}} \\\ & \Rightarrow \Delta \text{G}=-2.303\times 8.314\text{ J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}\times \text{298 K}{{\log }_{10}}\left( 1.62\times {{10}^{37}} \right) \\\ & \Rightarrow \Delta \text{G}=-212.31\text{ kJ} \\\ \end{aligned}$$ Hence, the equilibrium constant and free energy change for a given Daniell cell is $1.62\times {{10}^{37}}$ and $-212.31\text{ kJ}$ respectively. **Note:** The negative value of free energy change indicates that the reaction is spontaneous while a positive value indicates that the reaction is nonspontaneous. Its value becomes zero at equilibrium. And free energy change is also related to cell potential as: $$\begin{aligned} & \Delta {{\text{G}}^{\text{o}}}=-\text{nFE}_{\text{cell}}^{\text{o}}\text{ (at standard conditions)} \\\ & \Delta \text{G}=-\text{nF}{{\text{E}}_{\text{cell}}}\text{ } \\\ \end{aligned}$$ That is why at equilibrium, standard cell potential becomes zero.