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. 4 1 2 3
B. 2 4 4 1
C. 1 2 3 4
D. 3 4 1 2

Explanation

`Fe^(+3)+e^(-)to Fe^(+2) E^(@)=+0.77`
`Fe^(+2)+2e^(-)to Fe E^(@)=-0.44`
`overline(Fe^(+3)+3e^(-)toFe) E^(@)-?`
`E^(@)=(0.77xx1xx2(-0.44))/3=(-0.11)/3=-0.04V`
So, `P to 3` and hence Answer is (D).