1.

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)]

Answer»

23.16 KJ `mol^(-1)`
`-46.32` kJ `mol^(-1)`
`-23.16` kJ `mol^(-1)`
`46.32" kJ "mol^(-1)`

SOLUTION :`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)`


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