Question

At ${25^ \circ }C$ ${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_p} = - 1.25 \times {10^{ - 3}}V{K^{ - 1}}$ And ${E^ \circ } = 1.36V$ .For the cell $X|{X^{ + 2}}||{Y^{ + 2}}|Y.$ Calculate standard entropy change for cell reaction in $\left( {J/K} \right)$ .
$A.$ $- 241.45$
$B.$ $400$
$C.$ $- 3.34 \times {10^2}$
$D.$ $200$

Answer 1

EMF measurement of electrochemical cells is one of the convenient methods to determine the value of physical quantities such as equilibrium constant, ionization constant, $pH$ of a solution and the value of Thermodynamics properties of ions. Entropy change, Enthalpy change, and free Gibbs energy change. $$$$

Complete answer:
In the given question it is given that,
${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_p} = - 1.25 \times {10^{ - 3}}V{K^{ - 1}}$ And ${E^ \circ } = 1.36V$
Before solving the question first we have to find the relation between entropy change $\Delta S$ and ${\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_p}$
Since, the free Gibbs energy changes accompanying a cell reaction can be determined from the emf of a cell.
$\Rightarrow$ $\Delta G = - nf{E^ \circ }$ $\left( 1 \right)$
The change in enthalpy and entropy of the cell reaction can also be obtained if the temperature depends on the emf of the cell. For this we make use of Gibbs- Helmholtz equation,
$\Rightarrow$ $\Delta G = \Delta H + T\left( {\dfrac{{\partial \left( {\Delta G} \right)}}{{\partial T}}} \right)$ $\left( 2 \right)$
Now, From equation $\Delta G = - nf{E^ \circ }$differentiate with respect to Temperature $\left( T \right)$ at constant pressure $\left( p \right)$ , we get

$\Rightarrow$ $\dfrac{{\partial \left( {\Delta G} \right)}}{{\partial T}} = - nf{\left( {\dfrac{{\partial {E^ \circ }}}{{\partial T}}} \right)_p}$ $\left( 3 \right)$
Substituting equation $\left( 1 \right)$ and $\left( 3 \right)$ in equation $\left( 2 \right)$, we get
$\Rightarrow$ $- nf{E^ \circ } = \Delta H - nf{\left( {\dfrac{{\partial E}}{{\partial T}}} \right)_p}$
By rearranging we get,
$\Rightarrow$ $\Delta H = - nf\left[ {{E^ \circ } - T{{\left( {\dfrac{{\partial E}}{{\partial T}}} \right)}_p}} \right]$ $\left( 4 \right)$
For entropy change we , know
$\Delta G = \Delta H - T\Delta S$
$\Rightarrow$ $\Delta S = \dfrac{{\Delta H - \Delta G}}{T}$
Substituting the value of $\Delta H$ and $\Delta G$ from equation $\left( 1 \right)$ and $\left( 4 \right)$ , we get
$\Rightarrow$ $\Delta S = - nf{\left( {\dfrac{{\partial {E^ \circ }}}{{\partial T}}} \right)_p}$
Now, we can solve the equation using above equation
From the given cell, we have $n = 2$

Electrode	Reduction reaction
Right	${Y^{2 + }} + 2{e^ - } \to Y$
Left	${X^{2 + }} + 2{e^ - } \to X$

Hence, $\Delta S = 2 \times 96500\left( { - 1.25 \times {{10}^{ - 3}}} \right)$
$\Rightarrow$ $\Delta S = - 241.45J/K$

Thus, the option $\left( A \right)$ is correct.

Note:
Oxidation takes place at anode and reduction at cathode in a cell of either type (Galvanic cell and electrolytic cell). The main difference of the sign of the electrode. Table given below summarizes the sign convention for electrodes.

Anode	Cathode
Cell	Sign
Galvanic	Negative
Electrolytic	Positive