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 = 4, Q = 1, R = 2, S = 3
B. P = 2, Q = 3, R = 4, S = 1
C. P = 1, Q = 2, R = 3, S = 4
D. P = 3, Q = 4, R = 1, S = 2

Answer

D

Explanation

`n_(1)E_(1)^(@)+n_(2)E_(2)^(@)=n_(3)E_(3)^(@)`
`1xx0.77+2(-0.44)=3xxE_(3)^(@)`
`E_(3)^(@)=-0.11/3~~-0.04 V`
`{:([Q],2H_(2)O rarr O_(2)+4H^(+)+4e^(-),E^(@)=-1.23 V),(,O_(2)+2H_(2)O+4e^(-) rarr 4OH^(-),E^(@)=+0.40 V),(,ulbar(4H_(2)O hArr 4H^(+)+4OH^(-)),):}`

`n_(1)E_(1)^(@)+n_(2)E_(2)^(@)=n_(3)E_(3)^(@)`
`1xxE_(1)^(@)+1xx0.52=2xx0.34`
`E_(1)^(@)=+0.16 V`
`{:(Cu^(2+)+e^(-) rarr Cu^(+),E^(@)=+0.16 V),(Cu rarr Cu^(+)+e^(-),E^(@)=+0.52 V),(ulbar(Cu^(2+)+Cu hArr 2Cu^(+)),):}`
`E^(@)=E_("reduced species")^(@)-E_("oxidised species")^(@)=0.16-0.52=-0.36 V`
However, in the given option `(-0.18)` is printed.

`n_(1)E_(1)^(@)+n_(2)E_(2)^(@)=n_(3)E_(3)^(@)`
`1xxE_(1)^(@)+2(-0.91)=3xx(-0.74)`
`E_(1)^(@)=-0.4 V`