Inorganic Chemistry

What is ∆S of a cell if Ecell=0.045V at 298K ang temperature coefficient of cell is 3×10^-4VK^-1

Karan Kumawat

7 Years agoGrade

1 Answer

Khimraj

7 Years ago

According to Gibbs helmholtz equation ,

$\bold{\Delta{G}=\Delta{H}+T(\frac{\delta(\Delta{G})}{\delta{T}})_P}$

Here ∆G is Gibbs free energy , ∆H is heat ,T is temperature and is rate of change of Gibbs free energy at constant pressure.

But we know, ∆G = -nEF
∴ $\bold{(\frac{\delta(\Delta{G})}{\delta{T}})_P} = -nF\bold{(\frac{\delta{E}}{\delta{T}})_P}$
Put it in above equation,
Then, ∆G = ∆H - nFT $\bold{(\frac{\delta{E}}{\delta{T}})_P}$
Now, compare this equation with , ∆G = ∆H - T∆S,
Here ∆S is the entropy change.
Hence, ∆S = - nF $\bold{(\frac{\delta{E}}{\delta{T}})_P}$
,here is temperature coefficient and ∆S is entropy change.