Question

Calculate ${{E}}_{{{cell}}}^{{0}}$, ${{\Delta }}{{{G}}^{{0}}}$ and equilibrium constant for the reaction
${{2C}}{{{u}}^{{ + }}}{{(aq)}} \to {{C}}{{{u}}^{{{2 + }}}}{{(aq) + Cu(s)}}$.
${{{E}}^{{0}}}_{{{C}}{{{u}}^{{ + }}}{{/Cu}}}{{ = 0}}{{.52V}}$ and ${{{E}}^{{0}}}_{{{C}}{{{u}}^{{ + }}}{{,C}}{{{u}}^{{{2 + }}}}}{{ = 0}}{{.16V}}$

Answer 1

${{{E}}^{{o}}}_{{{cell}}}$ is also known as cell voltage or cell potential between two half-cells. The greater the ${{{E}}^{{o}}}_{{{cell}}}$ of a reaction, greater is the force by which electrons are driven through the system. We shall calculate the ${{{E}}^{{o}}}_{{{cell}}}$ using the values of electrode potential of cathode and anode. Then, we shall use the Nernst equation to find the equilibrium constant and thus the value of ${{\Delta }}{{{G}}^{{0}}}$

Formula used: ${{{E}}_{{{cell}}}}{{ = E}}_{{{cell}}}^{{o}} - \dfrac{{{{0}}{{.0591}}}}{{{n}}}{{logQ}}$
Where Q is the reaction quotient.
${{\Delta }}{{{G}}^{{o}}}{{ = }} - {{nF}}{{{E}}^{{o}}}_{{{cell}}}$
where ${{\Delta }}{{{G}}^{{o}}}$ is the Gibbs free energy, n is the number of moles of electrons exchanged. F is the Faraday’s constant and is equal to 96500 coulomb approximately. ${{E}}_{{{cell}}}^{{0}}$ is the cell potential.

Complete Step by step solution:
The reaction that takes place is as follows,
${{2C}}{{{u}}^{{ + }}}{{(aq)}} \to {{C}}{{{u}}^{{{2 + }}}}{{(aq) + Cu(s)}}$
In the above reaction, 1 mole of copper(I) ion is undergoing oxidation by increasing the oxidation state of itself and converting into copper(II) ion and another mole of copper(I) ion is undergoing reduction by decreasing the oxidation state of itself and converting into neutral copper atom. The net exchange of electrons taking place in this reaction is 1. That is why, the value of n in this reaction is 1.
It is given that,
${{{E}}^{{0}}}_{{{C}}{{{u}}^{{ + }}}{{/Cu}}}{{ = 0}}{{.52V}}$
${{{E}}^{{0}}}_{{{C}}{{{u}}^{{ + }}}{{,C}}{{{u}}^{{{2 + }}}}}{{ = 0}}{{.16V}}$
Now,
${{{E}}^{{o}}}_{{{cell}}}$ = ${{{E}}^{{0}}}_{{{C}}{{{u}}^{{ + }}}{{/Cu}}}$ - ${{{E}}^{{0}}}_{{{C}}{{{u}}^{{ + }}}{{,C}}{{{u}}^{{{2 + }}}}}$ = $0.52 - 0.16$ = ${{0}}{{.36}}$ V
Therefore, the value of ${{E}}_{{{cell}}}^{{0}}$ is ${{0}}{{.36}}$ V.
According to the Nernst equation,
As the reaction is in equilibrium, Q will change to ${{{K}}_{{c}}}$ and ${{{E}}_{{{cell}}}}$ will be zero.
${{0 = E}}_{{{cell}}}^{{o}}{{ - }}\dfrac{{{{0}}{{.0591}}}}{{{n}}}{{log}}{{{K}}_{{c}}}$
${{E}}_{{{cell}}}^{{o}}{{ = 0}}{{.0591log}}{{{K}}_{{c}}}$
We have found that ${{E}}_{{{cell}}}^{{0}}$ is ${{0}}{{.36}}$ V.
${{0}}{{.36 = 0}}{{.0591log}}{{{K}}_{{c}}}$
${{log}}{{{K}}_{{c}}}{{ = }}\dfrac{{{{0}}{{.36}}}}{{{{0}}{{.0681}}}}{{ = 6}}{{.09}}$
Hence as per the log table,
${{{K}}_{{c}}}{{ = 1}}{{.2 \times 1}}{{{0}}^{{6}}}$
Therefore, the value of equilibrium constant is $1.2 \times {10^6}$.
In a galvanic cell, Gibbs free energy is related to the potential of cell by the following equation,
${{\Delta }}{{{G}}^{{o}}}{{ = }} - {{nF}}{{{E}}^{{o}}}_{{{cell}}}$
Hence substituting the values,
${{\Delta }}{{{G}}^{{o}}} = {{ - 1}} \times 96500 \times 0.36{{ = - 34740 J = - 34}}{{.74 kJ}}$
Therefore, the value of ${{\Delta }}{{{G}}^{{o}}}$ for the reaction is ${{ - 34}}{{.74 kJ}}$.

Note:
In Thermodynamics, Gibbs free energy is the maximum amount of the reversible work at constant pressure and temperature. If it is negative, this means that the reaction is spontaneous, i.e. it doesn’t require external energy to occur.