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The reducation potential of a Pt(s)|Cl(2...

The reducation potential of a `Pt(s)|Cl_(2)(g)|Cl^(-)(aq.)` electrode is found to be `1.42 V` when the pressure olf `Cl_(2)` is 0.25 arm. The concentration of chloride ion in this half cell is

A

`0.43 M`

B

`0.043 M`

C

`0.34 M`

D

`0.034 M`

Text Solution

Verified by Experts

The correct Answer is:
B
The half cell reaction is
`Cl_(2)(g) + 2e^(-) hArr 2Cl^(-)(aq.)`
and its `E_(Cl_(2)//Cl^(-)` is `1.36 V`
The value of `n`, the number of elctrons transferred, is `2`. The expression for `Q` is `C_(Cl^(-)//P_(Cl_(2)))`
Substituting these data into the Nernst equation gives
`E = E^(@) -(0.0592 v)/(n) log Q`
`1.42 V = 1.36 V - (0.0592 V)/(n) "log" (C_(Cl^(-))^(2))/(0.25)`
`C_(Cl^(-)) = 0.043 M`
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