CBSE Class 12 Chemistry 2016 Delhi Set 1 Solved Paper

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Question : 16
Total: 26
Calculate e.m.f. of the following cell at 298K :
2Cr( s)+3Fe2+(0.1M)2Cr3+(0.01M) +3Fe( s)
Given : E°(Cr3+Cr)=0.74VE
(Fe2+Fe)=0.44V3
Solution:  
E°celi =Ec0Ea0
=(0.44)(0.74)V
=0.30V
Ecell =E°cell
0.059
n
log
[Cr3+]2
[Fe2+]3

Ecell =E°cell
0.059
6
log
[0.01]2
[0.1]3

=0.30(
0.059
6
)

=0.3098V
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