For the cell reaction
`2Fe_((aq))^(3+)+2I_((aq))^(-) to 2Fe_((aq))^(2+)+I_(2(aq))`
`E_(cell)^(Theta)=0.24V` at 298K. The standard Gibbs energy `(Delta_(r)G^(Theta))` of the cell reaction is :
[Given that Faraday constant F=96500 C `mol^(-1)`]

Options

A. 23.16 kJ `mol^(-1)`
B. `-46.32` kJ `mol^(-1)`
C. `-23.16` kJ `mol^(-1)`
D. `46.32" kJ "mol^(-1)`

Answer

B

Explanation

`2Fe_((aq))^(3+)+2I_((aq))^(-) to 2Fe_((aq))^(2+)+I_(2(aq))`
`DeltaG^(@)=-nFE^(@)" "` (where,n=2)
`=-2xx96500xx(0.24)`
`=-46320J`
`=-46.32` kJ `mol^(-1)`