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The standard reduction potential data at...

The standard reduction potential data at `25^(@)C` is given below .
`E^(@) (Fe^(3+) , Fe^(2+)) = +0.77 V , E^(@) (Fe^(2+) , Fe) = -0.44 V `
`E^(@) (Cu^(2+) , Cu) = +0.34 V , E^(@) (Cu^(+) , Cu) = +0.52` V
`E^(@) [O_(2 (g)) + 4 H^(+) + 4e^(-) to 2 H_(2) O ] = + 1.23 V` ,
`E^(@) [O_(2 (g)) + 2 H_(2) O + 4e^(-) to 4 OH^(-) ] = +0.40 V `
`E^(@) (Cr^(3+) , Cr) = -0.74 V , E^(@) (Cr^(2+) , Cr) = -0.91 V`
Match `E^(@)` of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists :

A

`{:( P , Q,R , S) , ( 4 ,1 , 2 ,3):}`

B

`{:( P , Q,R , S) , ( 2 ,3 , 4 ,1):}`

C

`{:( P , Q,R , S) , ( 1 ,2 , 3 ,4):}`

D

`{:( P , Q,R , S) , ( 4 ,3 , 1 ,2):}`

Text Solution

Verified by Experts

The correct Answer is:
D

`Delta G_(3)^(@) = Delta G_(1)^(@) + Delta G_(2)^(@)`
`3 xx x F = 1 xx 0.77 xx F + 2 (-0.44 ) xx F`
`3 x = 0.77 - 0.88`
` x = - (0.11)/(3) = -0.037 = -0.04 V`

`Delta G_(3)^(@) = Delta G_(1)^(@) + Delta G_(2)^(@)`
`- 3 F xx 0.74 = 1 F xx x + 2 F xx (-0.91 )`
`x = -2.22 + 1.82`
or = `-0.40` V
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Copper from copper sulphate solution can be displaced by …………. The standard reducation potentials of some electrodes are given below: (a) E^(@) (Fe^(2+), Fe) = -0.44 V (b) E^(@) (Zn^(2+), Zn) = -0.76 V (c) E^(@) (Cu^(2+), Cu) = +0.34 V (d) E^(@) (H^(+), 1//2H_(2)) = +0.34 V

What is the standard reduction potential (E^(@)) for Fe^(3+) to Fe ? Given that : Fe^(2+) + 2e^(-) to Fe , E_(Fe^(2+) // Fe)^(@) = -0.47 V Fe^(3+) + e to Fe^(2+) , E_(Fe^(3+)// Fe^(2+))^(@) = +0.77 V

Knowledge Check

  • The standard reduction potential data at 25^(@)C is given below E^(@) (Fe^(3+), Fe^(2+)) = +0.77V , E^(@) (Fe^(2+), Fe) = -0.44V , E^(@) (Cu^(2+),Cu) = +0.34V , E^(@)(Cu^(+),Cu) = +0.52 V , E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V , E^(@) (Cr^(3+), Cr) =- 0.74V , E^(@) (Cr^(2+),Cr) = - 0.91V , Match E^(@) of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: {:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):} Codes:

    A
    `{:(P,Q,R,S),(4,1,2,3):}`
    B
    `{:(P,Q,R,S),(2,3,4,1):}`
    C
    `{:(P,Q,R,S),(1,2,3,4):}`
    D
    `{:(P,Q,R,S),(3,4,1,2):}`

  • The standard reduction potential data at 25^(@)C is given below: E^(@) (Fe^(3+),Fe^(2+))=+0.77V E^(@) (Fe^(2+), Fe)=-0.44V E^(@) (Cu^(2+),Cu)=+0.52V E^(@) [O_(2)(g)+4H^(+)+4e^(-) to 2H_(2)O]=+1.23V E^(@) [O_(2)(g)+2H_(2)O+4e^(-) to 4OH^(-)]=+0.40V E^(@) (Cr^(3+), Cr)=-0.74V , E^(@) (Cr^(2+), Cr)=-0.91V Match E^(@) of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below the lists:

    A
    `{:(,,P,Q,R,S),(,A,4,1,2,3):}`
    B
    `{:(,,P,Q,R,S),(,B,2,3,4,1):}`
    C
    `{:(,,P,Q,R,S),(,C,1,2,3,4):}`
    D
    `{:(,,P,Q,R,S),(,D,3,4,1,2):}`

  • E^(@) of Fe^(2 +) //Fe = - 0.44 V, E^(@) of Cu //Cu^(2+) = -0.34 V . Then in the cell

    A
    `Cu^(2+)` Oxidizes Fe
    B
    `Fe^(2+)` oxidizes Cu
    C
    Cu Reduces `Fe^(2+)`
    D
    Fe reduces `Cu^(2+)`

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    E^(@) for Fe//Fe^(2+) is +0.44 V and E^(@) for Cu//Cu^(2+) is -0.32 V . Then, in the cell,

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    Using the standard electrode potential, find out the pair between which redox reaction is not feasible. E^(@) values : Fe^(3+)//Fe^(2+)=+0.77, I_(2)//I^(-)=+0.54 V Cu^(2+)//Cu= + 0.34 V, Ag^(+)//Ag= +0.80 V

    What is the standard reduction potential (E^(@)) for Fe^(3+) to Fe ? Given that : Fe^(2+) + 2e^(-) to Fe, E_(Fe^(2+)//Fe)^(@) =-0.47V Fe^(3+) + e^(-) to Fe^(2+), E_(Fe^(3+)//Fe^(2+))^(@) = +0.77V

    Standard electrode potentials are Fe^(2+)//Fe, E^(@) = -0.44 V Fe^(3+)//Fe^(2+), E^(@) = -0.77 V If Fe^(3+), Fe^(2+) , and Fe block are kept together, then

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