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`.

Options

A. 0.653 V
B. 0.889 V
C. 0.683 V
D. 2.771 V

Answer

A

Explanation

`Fe^(3+)+e^(-)toFe^(2+)`
`E_(red)=E_(Fe^(2+)//Fe^(2+))^(@)-(0.0591)/(2)"log"([Fe^(2+)])/([Fe^(3+)])`
`=0.771-0.0591"log"(2.0)/(0.02)`
`=0.771-0.591xxlog100`
`=0.771-0.0591xx2=0.653V`.