 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
```
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)

```
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 ) = 195864.55delg = 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 ..
```
4 years 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.
```
11 months ago 616 Points
```							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
```
6 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Physical Chemistry

View all Questions »  ### 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