CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

Passage of one ampere current through 0.1 M Ni(NO3)2 solution using Ni electrodes bring in the concerntration of solution to _________in 60 seconds:

  1. 0.1M
  2. 0.05M
  3. 0.2M
  4. 0.025M

Open in App
Solution

Step 1:
Current, I = 1A
Time, t = 1 × 60 = 60 s (1 min = 60 second)
Charge = I × t
= 1 × 60 = 60 Coulombs
Step 2:
Charge on Ni,
No3 has –1 charge always so that
Ni +2(NO3) =0
Ni + 2(–1) = 0
Ni = +2
Charge required to deposited 1 mol of Ni = nF
= 2 × 96487 Coulombs
= 192974 Coulombs

Step 3:

1 mol of Ni = 58.7 g (use periodic table to get this value)
192974 Coulombs will generate = 58.7 g of Ni
60 Coulombs will generate =58.7g×60C192974C
= 0.00031 g
Hence, 0.0031 g of nickel will be deposited at the cathode.

flag
Suggest Corrections
thumbs-up
15
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Introduction
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon