For the cell reaction:
`2Fe^(3+)(aq)+2l^(-)(aq)to2Fe^(2+)(aq)+l_(2)(aq)`
`E_(cell)^(ɵ)=0.24V` at `298K`. The rstandard gibbs energy `(triangle,G^(ɵ))` of the cell reaction is
[Given that faraday constnat `F=96400Cmol^(-1)]`

For the cell reaction: 2Fe^(3+)(aq)+2l^(-)(aq)to2Fe^(2+)(aq)+l_(2)(aq) E_(cell)^(ɵ)=0.24V at 298K . The standard gibbs energy (triangle,G^(ɵ)) of the cell reaction is [Given that faraday constnat F=96500Cmol^(-1)]

For the cell reaction 2Fe^(3+)(aq) + 2l^(-1) (aq) to 2Fe^(2+) (aq) + I_2(aq) E_("cell")^(Theta) = 0.24 V at 298 K. The standard Gibbs energy (Delta_r G^Theta) of the cell reaction is [Given that Faraday constnat F = 96500 C mol^(-1) ]

(a) The cell in which the following reactions occurs: 2Fe^(3+)(aq)=2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s) has E_(cell)^(@)=0.236V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F=96,500" C "mol^(-1) ) (b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F=96,500" C "mol^(-1) )