Question

Calculate the enthalpy change at 250C for the reaction occurring in the cell

Zn ( 1.0 M ZnCl2, AgCl (s) ( Ag

The EMF of the cell at 250C is 0.805 V and ((E/(T)p is -3.98X10-4 volt/degree. Faraday = 96500 Columns.

Answer 1

The reaction is:

Zn + 2 AgCl = ZnCl2 + 2 Ag

EMF is linked to ΔG:

ΔG = -2*F*EMF = ΔH - T*ΔS

Accordingly,

d(EMF)/dT = ΔS/2F = -3.98*10^-4 volt/degree

ΔS = 2*96500*(-3.98*10^-4) = 76.814 J/degree

At 25 ⁰C = 298 K:

ΔH = -2*F*EMF + T*ΔS = -2*96500*0.805 + 298*76.814 = -132474.428 J

enthalpy change at 25 ⁰C is -132474.428 J