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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)
@ rishikesh del g = – nf enot cell n =2 F= 96485 e not cell = 1.015 v del g = -2 * 96485 * (-1.015 ) = 195864.55delg = 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 ..
If the EMF of the cell at 20°C and 30°C are 0.6973 and 0.6998 volt respectively, calculate its temperature coefficient.
Dear student,Please find the solution to your problem. ΔG = – nFEo n = 2 F = 96485 C Eo = 1.015 VΔG = – 2 x 96485 x (–1.015 ) = 195864.55ΔG = ΔH – TΔS 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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