(a) The cell in which the following reaction occurs: 2Fe^(3+) (aq) + 2I^(-) (aq) to 2Fe^(2+) (aq) + I_(2) (s) has E_("cell")^(@) 0.236 V 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 mol^(-1)]

(a) The cell in which the following reaction occurs: 2Fe^(3+) (aq) +

1.	(a) The cell in which the following reaction occurs: 2Fe^(3+) (aq) + 2I^(-) (aq) to 2Fe^(2+) (aq) + I_(2) (s) has E_("cell")^(@) 0.236 V 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 mol^(-1)]
Answer» Solution :(a) `DeltaG^(@) =-nFE_("cell")^(@)` `DeltaG^(@) =-2 xx 96500 C MOL^(-1) xx 0.236 V = 45548 J mol^(-1)` `=45.548 kJ mol^(-1)` (B) Given, l = 0.5 A, t = 2hours `=2 xx 60 xx 60 = 7200 s` QUANTITY of electricity `= I xx t = 0.5 xx 7200 = 3600 C` 96500 C electricity is given by `=6.02 xx 10^(23)` electrons. 3600 C electricity is given by `=(6.02 xx 10^(23))/96500 xx 3600 = 0.2245 xx 10^(23)` electrons. `=2.245 xx 10^(22)` electrons.