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) e^(-) + Fe^(3+) rarr Fe^(+2)`
`DeltaG_(1)^(ul(@)) =- 1 xx F xx 0.77`
`2e^(-) +Fe^(2+) rarr Fe`
`DeltaG_(2)^(ul(@))=- 2 xx F xx (-0.44)`
Required equation
`Fe^(3+) +3e^(-) rarr Fe`
`DeltaG_(3)^(ul(@)) =- 3 xx F xx E^(@)`
`DeltaG_(3)^(ul(@)) = DeltaG_(2)^(ul(@)) +DeltaG_(1)^(ul(@))`
`- 3 xx F xx E^(ul(@)) =- 1 xx F xx 0.77 - 2 xx F xx (-0.44)`
`E^(ul(@)) =- (0.11)/(3) =- 0.04V`
`(Q) O_(@)(g) +4H^(+) +4e^(-) rarr 2H_(2)O`
`E_(1)^(ul(@)) = 1.23` ...(i)
`O_(2)(g) +2H_(2)O +4e^(-) rarr 40H^(-) E_(2)^(ul(@)) = 0.40V` ...(ii)
Required equation
`4H_(2)O hArr 4H^(+) + 4OH^(-) (E_(2)^(ul(@))-E_(1)^(ul(@))) =- 0.83V`
`(R) {:(Cu^(+2)+e^(-),rarr,Cu^(+)),(E_(1)^(ul(@))=0.34V,,),(Cu,rarr,Cu^(+)+e^(-)),(E_(2)^(ul(@))=-0.52,,):}`
`bar(Cu^(+2) +Cu rarr 2Cu^(+1)`
`-0.18 V`
`(S) Cr^(+3) +3e^(-) rarr Cr`
`DeltaG_(1)^(ul(@)) =- 3 xx F xx (-0.74)`
`Cr^(+2) +2e^(-) rarr Cr`
`DeltaG_(2)^(ul(@)) = - 2xx F xx (-0.91)`
Required equation
`Cr^(3+) +e^(-) rarr Cr^(+2)`
`DeltaG_(3)^(ul(@)) =- 1 xx F xx (E^(@))`
`DeltaG_(1)^(ul(@)) = DeltaG_(2)^(ul(@)) +DeltaG_(3)^(ul(@))`
`- 3 xx F xx (-0.74) =- 2 xx F xx (-0.91) - 1 xx F xx E^(@)`
`E^(@) = 1.82 - 2.22 =- 0.4V`