Question

For the cell reaction
$2F{e^{3 + }}(aq) + 2{I^ - }(aq) \to 2F{e^{2 + }}(aq) + {I_2}(aq)$
$E_{cell}^0$= 0.24 V at 298 K. The standard Gibbs energy (${\Delta _r}{G^0}$) of the cell reaction is :
Given that faraday constant, F = 96500 C$mo{l^{ - 1}}$
a.) -46.32$kJmo{l^{ - 1}}$
b.) -23.16$kJmo{l^{ - 1}}$
c.) 46.32$kJmo{l^{ - 1}}$
d.) 23.16$kJmo{l^{ - 1}}$

Answer 1

Hint:. The standard Gibbs free energy of a reaction is the change in Gibbs free energy of a system during formation of 1 mole of products from reactants at standard conditions of temperature and pressure. It is given by -
$\Delta {G^0}$= - nF$E_{cell}^0$
Where ‘n’ is the number of electrons transferred
F is faraday constant

Complete step by step answer:
For such a type of question, let us first write what is given to us and what we need to find out
Given :
The reaction is $2F{e^{3 + }}(aq) + 2{I^ - }(aq) \to 2F{e^{2 + }}(aq) + {I_2}(aq)$
$E_{cell}^0$= 0.24 V
Temperature = 298 K
faraday constant, F = 96500 C$mo{l^{ - 1}}$
To find :
Standard Gibbs energy of the cell reaction
We know that change in Gibbs free energy can be given by -
$\Delta {G^0}$= - nF$E_{cell}^0$
Where ‘n’ is the number of electrons transferred
F is faraday constant
From the above reaction, we see that Fe has accepted one electron and its oxidation state is changed by +1. For two atoms of Fe, there is transfer of two electrons which are donated by iodide.
So, n = 2

Thus,
$\Delta {G^0}$= - nF $E_{cell}^0$
$\Delta {G^0}$= -2$ \times $96500$ \times $0.24
$\Delta {G^0}$= - 46320 $Jmo{l^{ - 1}}$
$\Delta {G^0}$= - 46.32 $kJmo{l^{ - 1}}$
So, the correct answer is “Option A”.

Note: It must be noted that the value of standard Gibbs energy of a cell reaction is positive or greater than 1 if the reaction is non spontaneous, negative or less than 1 in case the reaction is spontaneous and zero if the reaction is at equilibrium.