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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 `1.5M Fe^(2+)` and `0.015 M Fe^(3+)` solution at `25^(@)C` is `(E_(Fe^(3+)//Fe^(2+))^(@) = 0.770V)`

A

0.652 V

B

0.88 V

C

0.710 V

D

0.850 V

Text Solution

Verified by Experts

The correct Answer is:
A
`E_(cell)=0.77-(0.059)/(1)log((1.5)/(0.015))=0.652`V
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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 .

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(i) On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti^(4+) species may be used to oxidise Fe(II) to Fe(III) {:(Ti^(4+) + e^(-) to Ti^(3+), E^(@) = +0.01V), (Fe^(3+) + e^(-) to Fe^(2+), E^(@)= +0.77V):} (ii) Based on the data arrange Fe^(3+), Mn^(2+) " and " Cr^(2+) in the increasing order of stability of +2 oxidation state. (Give a brief reason) E_(Cr^(3+)//Cr^(2+))^(@) = -0.4V E_(Mn^(3+)//Mn^(2+))^(@) = +1.5V E_(Fe^(3+)//Fe^(2+))^(@) = +0.8V

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