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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

0.653 V

B

0.889 V

C

0.683 V

D

2.771 V

Text Solution

Verified by Experts

The correct Answer is:
A
`E_("cell")=E_("cell")^@-(RT)/(nF) xx 2.303log""([Fe^(2+)])/([Fe^(3+)])`
`E_("cell")=0.771 -(8.314 xx2.303xx298)/(96500)log""2/(0.02)`
`=0.771 - (0.0591xx 2) = 0.771 - 0.1182`
`E_("cell") = 0.653V.`
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