Please give detailed explanation.

Solution:
In this reaction Cr undergoes reduction from +6 to +3.
Cr2O72- to Cr+3 (Acts as cathode) 
I undergoes oxidation from -1 to 0
Ito I2 (Acts as anode) 
We know, 
Ecell = E°cathode - E°anode (in terms of standard reduction potential) 
  0.79= 1.33 - E°anode
anode= 0.54 V 
I2/I- = 0.54 V

Option A is correct

 

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