Use the following standard electrode potentials, calculate `DeltaG^(@)` in kJ`//`mol for the indicated reaction :
`5Ce^(4+)(aq)+Mn^(2+)(aq)+4H_(2)O(l)to5Ce^(3+)(aq)+MnO_(4)^(-)(aq)+8H^(+)(aq)`
`MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l), E^(@)=+1.51 V`
`Ce^(4+)(aq)+e^(=)toCe^(3+)(aq)" "E^(@)=+1.61 V`

Options

A. `-9.65`
B. `-24.3`
C. `-48.25`
D. `-35.2`

Answer

C

Explanation

`E_(cell)^(@)=E_(RP(RHS))^(@)-E_(RP(LHS))^(@)`
`=1.61-1.51rArr0.10V`
`DeltaG^(@)=-nFE^(@)rArr-5xx96500xx0.10J`
`DeltaG^(@)=-48.25kJ`