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Given E(Cr^(3+)//Cr)^(@) = -0.72 V, E(Fe...

Given `E_(Cr^(3+)//Cr)^(@) = -0.72 V, E_(Fe^(2+)//Fe)^(@) = -0.439 V`. The value of standard electrode potential for the change,
`Fe_(aq.)^(3+) + e^(-) rarr Fe_(aq.)^(2+)` will be:

A

`-0.072 V`

B

`0.385 V`

C

`0.770 V`

D

`-0.270 V`

Text Solution

Verified by Experts

The correct Answer is:
A
As `E_(Cr^(3+)//Cr)^(@) = -0.72 V` and
`E_(Fe^(2+)//Fe)^(@) = -0.42 V`
`2Cr + 3Fe^(2+) rarr 3Fe + 2Cr^(3+)`
Six electrons `(n = 6)` are used in redox change.
`E_(cell) = E_(cell)^(@) - (0.0591)/(n)log'([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
`= (-0.42 + 0.72) - (0.0591)/(6)log'((0.01)^(2))/((0.01)^(3))`
`= 0.30 - (0.0591)/(6)log.((0.1)^(2))/((0.01)^(3))`
`= 0.30 - (0.0591)/(6)log10^(4)`
`E_(cell) = 0.2606 V`
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Knowledge Check

  • Given : E_(Fe^(3+)//Fe)^@=-0.036 V E_(Fe^(2+)//Fe)^@=-0.439 V. The value of standard electrode potential for the change, Fe_((aq))^(3+)+e^(-)toFe_((aq))^(2+) will be

    A
    -0.072 V
    B
    0.385 V
    C
    0.770 V
    D
    -0.270 V

  • E_(Fe^(3+)//Fe)^(@)=-0.036V,E_(Fe^(2+)//Fe)^(@)=-0.0439V . The value of standard electrode potential for the change, Fe^(3+)(aq)+e^(-)toFe^(2+)(aq) will be

    A
    `-0.072V`
    B
    `-0.385V`
    C
    `-0.020V`
    D
    `-0.270V`

  • E_(M^(3+)//(M))^(@) = -0.036V , E_(M^(2+)//M)^(@)= -0.439V . The value of standard electrode potential for the change, M^(3+)(aq) + e^(-)rightarrow M^(2+) (aq) will be :

    A
    `(-0.072V)`
    B
    `(0.385V)`
    C
    `(0.770V)`
    D
    `0.270V`