Question

The potential of the Daniel cell, Zn\left| {\begin{array}{*{20}{c}} {ZnS{O_4}} \\\ {\left( {1M} \right)} \end{array}} \right|\left| {\begin{array}{*{20}{c}} {CuS{O_4}} \\\ {\left( {1M} \right)} \end{array}} \right|Cu was reported by Buckbee, Surdzial and Metz as ${E^0} = 1.1028 - 0.641 \times {10^{ - 3}}T + 0.72 \times {10^{ - 5}}{T^2},$ where T is the temperature in degree Celsius. Calculate $\Delta {S^0}$ for the cell reaction at $25^\circ {\text{C}}$ .
(A) ${\text{ - 45}}{\text{.32J}}{{\text{K}}^{{\text{ - 1}}}}$
(B) ${\text{ - 34}}{\text{.52J}}{{\text{K}}^{{\text{ - 1}}}}$
(C) ${\text{704}}{\text{.48J}}{{\text{K}}^{{\text{ - 1}}}}$
(D) ${\text{ - 54}}{\text{.23J}}{{\text{K}}^{{\text{ - 1}}}}$

Answer 1

Hint : The cell potential of an electrochemical cell can be utilized to determine the entropy change for the reaction. Since, entropy is related to the Gibbs free energy through the temperature dependence of the Gibbs free energy by the relation: ${\left( {\dfrac{{\partial \Delta G}}{{\partial T}}} \right)_P} = - \Delta S$ , a similar relationship has also been derived which relates the entropy to the cell potential via its temperature dependence:
$\Delta {S^0} = nF\dfrac{{d{E^0}}}{{dT}}$

Complete Step By Step Answer:
Given that the potential cell of the Zn – Cu Daniel cell is ${E^0} = 1.1028 - 0.641 \times {10^{ - 3}}T + 0.72 \times {10^{ - 5}}{T^2},$ and the given temperature in degree Celsius is $25^\circ {\text{C}}$ .
We need to find out the value of the standard entropy change at $25^\circ {\text{C}}$ .
We know that the entropy change for a reaction is related to the electrochemical cell potential by the following expression:
$\Delta S = nF\dfrac{{\Delta E}}{{\Delta T}}$
Here, $\Delta {\text{S}}$ is the change in entropy for the reaction, n is the number of electrons involved in the cell reaction, F is the Faraday’s constant which is equal to 96500C per mol, $\Delta {\text{E}}$ is the electrochemical cell potential and $\Delta {\text{T}}$ is the temperature change.
Or we can write $\Delta S = nF\dfrac{{dE}}{{dT}}$
Under standard conditions, that is at $25^\circ {\text{C}}$ , we can write the relation between the electrochemical cell potential and the entropy change for a reaction as,
$\Delta {S^0} = nF\dfrac{{d{E^0}}}{{dT}}$
Now, we have been given the expression of cell potential as: ${E^0} = 1.1028 - 0.641 \times {10^{ - 3}}T + 0.72 \times {10^{ - 5}}{T^2}$
Now differentiate the cell potential ${{\text{E}}^{\text{0}}}$ with respect to the temperature T. So we will now have,
$\dfrac{{d{E^0}}}{{dT}} = - 0.641 \times {10^{ - 3}} + 2 \times 0.72 \times {10^{ - 5}} \times T$
The temperature T is given to be equal to $25^\circ {\text{C}}$ .
Since the temperature is given in degrees Celsius, we will convert it into Kelvin. So, now we will have: T ${\text{ = }}\left( {{\text{25 + 273}}} \right){\text{K = 298K}}$ . So now we can replace T by 298 K in the above equation.
$\dfrac{{d{E^0}}}{{dT}} = - 0.641 \times {10^{ - 3}} + 2 \times 0.72 \times {10^{ - 5}} \times 298$
Now, we need to determine the value of n for this cell reaction. From the cell representation, it can be understood that the reaction is between zinc and ${\text{CuS}}{{\text{O}}_{\text{4}}}$ :
$Zn(s) + C{u^{2 + }}(aq) \to Z{n^{2 + }}(aq) + Cu(s)$ which can be split into the two half reactions:
$Zn(s) \to Z{n^{2 + }}(aq) + 2{e^ - }$
$C{u^{2 + }}(aq) + 2{e^ - } \to Cu(s)$
So, n for the above reaction is 2 as two electrons are involved in the overall reaction. Now put n is equal to 2 in the expression for entropy change.
$\Delta {S^0} = 2 \times 96500 \times \left( { - 0.641 \times {{10}^{ - 3}} + 2 \times 0.72 \times {{10}^{ - 5}} \times 298} \right) \Rightarrow \Delta {S^0} = 704.48J{K^{ - 1}}$
So the correct option is C.

Note :
In this problem, the value of the entropy change is to be found out in the units of Joule per Kelvin, so if we substitute the value of temperature in degree Celsius, the answer will not be correct. So, the temperature must be changed into Kelvin scale at first.
Daniel cell is a form of electrochemical cell. Electrochemical cells are based upon the redox reaction which is spontaneous. In electrochemical cells, electrons flow from anode to cathode in the external circuit and the inner circuit is completed by the flow of ions through the salt bridge.