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:

Options

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):}`

Answer

D

Explanation

`(P) DeltaG_(fe^(3+)//Fe)^(@) = DeltaG_(Fe^(3+)//Fe^(2+))^(@) + DeltaG_(Fe^(2+)//Fe)^(@)`
`rArr -3 xx FE_((Fe^(+3)//Fe))^(@)`
`=- 1 xx FE_((Fe^(+3)//Fe^(+2)))^(@) + (-2FE_(Fe^(+2)//Fe)^(@))`
`rArr E_(Fe^(+3)//Fe)^(@) = - 0.04V`
`(Q) O_(2)(g)+2H_(2)O +4e^(-) rarr 4bar(O)H E^(@) = 0.40V` ........(i)
`2H_(2)O rarr O_(2)(g)+4H^(+) +4e^(-), E^(@) =- 1.23 V` ......(ii)
So, `4H_(2)O hArr 4H^(+) +4 bar(O)H` ........(iii)
`E^(@)` for IIIrd reduction `= 0.40 - 1.23 =- 0.83 V`
`(R) DeltaG_((Cu^(2+)//Cu))^(@)`
`DeltaG_((Cu^(2+)//Cu^(+)))^(@) + DeltaG_((Cu^(2+)//Cu))^(@)`
`-2 xx FE_(Cu^(+2)//Cu)^(@) =- 1 xx FE_(Cu^(+2)//Cu^(+))^(@)+(-1xxFxxE_(Cu^(+)//Cu)^(@))`
`rArr E_(Cu^(+2)//Cu)^(@) =- 0.18V`
(S) `DeltaG_(Cr^(+3)//Cr^(+2))^(@) = DeltaG_(Cr^(+3)//Cr)^(@) +DeltaG_(Cr//Cr^(+2))^(@)`
`-1 xx F xx E_(Cr^(+3)//Cr^(+2))^(@) =- 3 xx F xx E_(Cr^(+3)//Cr)^(@)`
`+(-2xx F xx E_(Cr//Cr^(+2))^(@))`
`rArr E_(Cr^(+3)//Cr^(+2))^(@) =- 0.4V`