MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 12th pass

                        

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)

4 years ago

Answers : (3)

Umakant biswal
5359 Points
							
@ 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 ..
4 years ago
Swapnil admane
15 Points
							
If the EMF of the cell at 20°C and 30°C are 0.6973 and 0.6998 volt respectively, calculate its temperature coefficient.
7 months ago
Kushagra Madhukar
askIITians Faculty
605 Points
							
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
one month ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 141 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details