wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Which of the following transformations can be carried out by using HI as a reducing agent, under acidic conditions?
[Given:I2(s)2IE=0.54V]
(i)Cu+Cu E=0.52V(ii)Cr3+Cr2+ E= 0.41V(iii)Fe3+Fe2+ E=0.77V(iv)Fe2+Fe E= 0.44V

A
(i) and (iii)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(ii) and (iv)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
only (iii)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
only (ii)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C only (iii)
Here only (iii) E value is more than 0.54V so HI can be used to reduce Fe3+

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Electrolysis and Electrolytes_Tackle
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon