Home
Class 12
CHEMISTRY
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`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    P BAHADUR|Exercise Exercise (6) INTEGER ANSWER TYPE PROBLEAMS|2 Videos

  • ELECTROCHEMISTRY

    P BAHADUR|Exercise Exercise (7) COMPREHENSION BASED OBJECTIVE PROBLEMS|3 Videos

  • ELECTROCHEMISTRY

    P BAHADUR|Exercise Exercise (3B) OBJECTIVE PROBLEMS: (More than One OptIons are Correct)|1 Videos

  • DILUTE SOLUTION AND COLLIGATIVE PROPERTIES

    P BAHADUR|Exercise Exercise 9 Advanced Numerical|12 Videos

  • MOCK TEST PAPER

    P BAHADUR|Exercise Exercise|92 Videos

Similar Questions

Explore conceptually related problems

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

Given E_(Cr^(3+)//Cr)^(@)= 0.72V , E_(Fe^(2+)//Fe)^(@)=-0.42V . The potential for the cell Cr|Cr^(3+)(0.1M)||Fe^(2+) (0.01M) | Fe is :

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`

    • Similar Questions

    Explore conceptually related problems

    Given E_(Fe^(3+)//Fe)^(o)=-0.036V,E_(Fe^(2+)//Fe)^(o)=-439V the value of standard electrode potential for the change, Fe_((aq))^(3+)+e^(-)toFe^(2+)(aq) will be

    Given, {:(E_(Fe^(3+) //Fe)^o + 3e E^o =- 0.036 V),(E_(Fe^(3+) //Fe)^o =- 0.439 V):} The value of standard electrode ptoential for the charge,

    Given E_(Cr^(3+)//Cr)^(@)=-0.72 V, E_(Fe ^(2+)//Fe)^(@)=-0.42V. The potential for the cell Cr|Cr^(3l) (0.1M)||Fr ^(2l)(0.01M)|Fe is

    Given E_(Cr^(3+)//Cr)^(@)=-0.72V , and E_(Fe^(2+)//Fe)^(@)=-0.42V The potential for the cell. Cr|Cr^(3+)(0.1M)||Fe^(2+)( 0. 01M)Fe is

    The standard reduction for the following reactions are : Fe^(3+) + 3e^(-) rarr Fe with E^(@) = - 0.036 V Fe^(2+) + 2e^(-) rarr Fe with E^(@) = - 0.44 V What would be the standard electrode potential for the reaction Fe^(3+) + e^(-) rarr Fe^(2+) ?

    Home
    Profile