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The reduction potential of a half-cell c...

The reduction potential of a half-cell consisting of a Pt electrode immersed in `1.5 M Fe^(2+)` and ` 0.015 M Fe^(3+)` solutin at `25^@ C` is `(E_(Fe^(3+)//Fe^(2+))^@ = 0.770 V)` is .

A

`0.652V`

B

`0.88V`

C

`0.710V`

D

`0.850V`

Text Solution

Verified by Experts

The correct Answer is:
A
`Fe^(3+) rarr Fe^(2+)`
`E_(cell) = 0.770 V -(0.0591)/(1)log.(1.5)/(0.015)`
`= 0.770 -0.059 xx 2 = 0.652 V`
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