Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Chemistry
Given, E° Cr 3+/Cr =-0.74 V;E°MnO-4/Mn2+=1.51 V E° Cr2 O 2-7/Cr3+ =-1.33 V;E°Cl/Cl-=1.36 V Based on the data given above strongest oxidising agent will be
Q. Given,
E
C
r
3
+
/
C
r
∘
=
−
0.74
V
;
E
M
n
O
4
−
/
M
n
2
+
∘
=
1.51
V
E
C
r
2
O
7
2
−
/
C
r
3
+
∘
=
−
1.33
V
;
E
Cl
/
C
l
−
∘
=
1.36
V
Based on the data given above strongest oxidising agent will be
6932
207
JEE Main
JEE Main 2013
Redox Reactions
Report Error
A
Cl
10%
B
C
r
3
+
24%
C
M
n
2
+
25%
D
M
n
O
4
−
41%
Solution:
Higher the standard reduction potential, better is oxidising agent. Among the given
E
M
n
O
4
−
/
M
n
2
+
∘
is highest
M
n
O
4
−
is the strongest oxidising agent.