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The temperature coefficient of emf , i.e., dE/dT = -4.02×10 -4 volt deg -1 for a cell. The enthalpy change for the cell is (n=2, E cell = 1.015 V , T = 293 K)

The temperature coefficient of emf, i.e., dE/dT = -4.02×10-4 volt deg-1 for a cell. The enthalpy change for the cell is (n=2, Ecell = 1.015V, T = 293K)

Grade:12th pass

3 Answers

Umakant biswal
5359 Points
4 years ago
@ rishikesh 
del g = – nf enot cell 
n =2 
F= 96485 
e not cell = 1.015 v 
del g = -2 * 96485 * (-1.015 ) = 195864.55
delg = del h- t del s 
del s = de/dt = -4.02 * 10 ^ -4 
del h = del g + t del s 
t = 293 k 
= 195864.55+293*(-4.02*10^-4) 
HOPE IT CLEARS YOUR DOUBT 
ALL THE BEST ..
Swapnil admane
15 Points
one year ago
If the EMF of the cell at 20°C and 30°C are 0.6973 and 0.6998 volt respectively, calculate its temperature coefficient.
Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the solution to your problem.
 
ΔG = – nFEo 
n = 2 
F = 96485 C 
Eo = 1.015 V
ΔG = – 2 x 96485 x (–1.015 ) = 195864.55
ΔG = ΔH –  TΔ
ΔS = dE/dT = – 4.02 x 10-4 
ΔH = ΔG + TΔS 
T = 293 k 
Hence, ΔH = 195864.55 + 293*(–4.02 x 10-4
                 = 195864.432 Jmol-1
                 = 195.86 kJmol-1
 
Thanks and regards,
Kushagra

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