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Calculate the reduction potential of a h...

Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in `2.0M Fe^(2+)` and `0.02M Fe^(3+)` solution. Given `E_(Fe^(3+)//Fe^(2+))^(@) = 0.771 V`.

A

0.653V

B

0.889V

C

0.683V

D

2.771V

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(cell)=E_(cell)^(0)-(RT)/(nF)xx2.303"log"([Fe^(2+)])/([Fe^(3+)])`
`E_(cell)=0.771-(8.314xx2.303xx298)/(96500)"log"(2)/(0.02)`
`=0.771-(0.0591xx2)=0.771-0.1182`
`E_(cell)=0.653V`.
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