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)

Question

Umakant biswal · Answer

@ 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 · Answer

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 · Answer

Dear student, Please find the solution to your problem. Δ G = – nFE o n = 2 F = 96485 C E o = 1.015 V Δ G = – 2 x 96485 x (–1.015 ) = 195864.55 Δ G = Δ H – T Δ S Δ S = dE/dT = – 4.02 x 10 -4 Δ H = Δ G + T ΔS T = 293 k Hence, Δ H = 195864.55 + 293*(–4.02 x 10 -4 ) = 195864.432 Jmol-1 = 195.86 kJmol-1 Thanks and regards, Kushagra

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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, E_cell = 1.015V, T = 293K)

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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)

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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 = 1.015V, T = 293K)