(a) The cell in which, the following reaction occurs:
$2{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+2{\mathrm{I}}^{-}\left(\mathrm{aq}\right)⟶2{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{I}}_2\left(\mathrm{s}\right)$ 
has ${\mathrm{E}}_{\text{cell }}^{\circ }=0.236\mathrm{V}\text{ at }298\mathrm{K}.$ Calculate the standard Gibbs energy of the cell reaction. (Given: 1 F = 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: 1 F = 96,500 C mol⁻¹)

(a) Given,
n =2
${\mathrm{E}}_{\text{cell }}^0=0.236\mathrm{V}$
T= 298K
Formula: $\begin{array}{l}{\mathrm{\Delta G}}^0=-\mathrm{nFE}\\ =-2\times 96500\times 0.236\\ =-45547{\mathrm{Jmol}}^{-1}=-45.548{\mathrm{kJmol}}^{-1}\end{array}$
(b) Given; i = 0.5A 
t = 2 hours =2×60×60 s =7200s Q=it
Q = 0.5× 7200= 3600C
As 96500C= 6.023 ×10²³
Therefore, $\frac{3600\mathrm{C}=6.023\times {10}^{23}\times 3600}{96500}$ number of electrons
$=2.25\times {10}^{23}$ number of electrons
$2.25\times {10}^{23}$number of electrons will flow through the wire.