Question

In the galvanic cell one mole electron moves from anode to cathode at $500K$ causing the potential difference of $20V$ . What is $\Delta {S_{universe}}$ during the phenomenon ?
A. $ - 2460J{K^{ - 1}}$
B.$ - 3860J{K^{ - 1}}$
C.$ + 2460J{K^{ - 1}}$
D.$ + 3860J{K^{ - 1}}$

Answer 1

Hint:$\Delta S$ is the entropy change . It is a measure of the randomness or disorder of the system . Greater the randomness , the higher is the entropy of the system.

Complete step by step answer: We have to calculate the value of $\Delta {S_{universe}}$ in this question . To calculate the entropy change in the universe , we use the following formula :
$\Delta S = nF{(\dfrac{{\delta E}}{{\delta T}})_P}$
where , n= number of moles of electrons involved in the process
F= one faraday of electricity
E=potential difference
T = temperature
It is given that one mole of electrons move from anode to cathode , so the value of n is $n = 1$
Further it is given that the reaction takes place at $500K$ so , $T = 500K$
The potential difference which is caused is given by , $E = 20V$
and we take the value of F as $96500C$
So , on substituting the above values in the formula , we get
$\Delta S = 1 \times 96500 \times \dfrac{{20}}{{500}} = 3860J{K^{ - 1}}$
Since the value of entropy is positive , that means it is accompanied by increase in randomness and hence increase of entropy . Hence, the process is spontaneous .
Hence option D is correct .


Note: By looking at the value of entropy change we can determine the spontaneity of the reaction , if the value of $\Delta S$ is positive , the process is spontaneous . If its value is negative , the direct process is non - spontaneous ( the reverse process may be spontaneous ) and if the value of $\Delta S$ is zero then the process is said to be in equilibrium .