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The reduction electrode potential E of 0...

The reduction electrode potential E of 0.1 M solutoin of `M^(+) "ions" (E_(RP)^(@)=-2.36 V)` is

A

`-4.82 V`

B

`-2.41 v`

C

`+2.41 v`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`E=E_(RP)^(@)+(0.0591)/(n)log[M^(+)]`
Given `E_(RP)^(@)=-2.36 V,[M^(+)]=0.1 M`
`n=1 ("for" M^(+) rarr M)`
`E=E_(RP)^(@)+(0.0591)/(n)log[M^(+)]`
`=-2.36 +0.591 xx(-1)`
`=-2.36 -0.591 =-2.419 V`
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