Question

For the reaction: \[2F{e^{3 + }}(aq) + 2{I^ - }(aq) \to 2F{e^{2 + }}(aq) + {I_2}(s)\] . The magnitude of the standard molar free energy change, \[\Delta r{G^o}_m = - \_\_\_\_\_\_kJ\](Round off the nearest integer).
\[[{E^o}_{\dfrac{{F{e^{2 + }}}}{{Fe(s)}}} = - 0.440V;\,{E^o}_{\dfrac{{F{e^{3 + }}}}{{Fe(s)}}} = - 0.036V]\]
\[[{E^{^O}}_{\dfrac{{{I_2}}}{{2I}}} = - 0.539V;\,F = 96500C]\]

Answer 1

Hint: In Gibbs free energy both entropy and enthalpy are combined and it is represented by \[\Delta G\]. When \[\Delta G\] is positive, the reaction will be non-spontaneous and when \[\Delta G\]is negative, the reaction will be spontaneous.

Formula used: The standard molar free energy or Gibbs’s free energy is calculated by the addition of entropy and enthalpy of a chemical reaction under constant pressure and temperature. It is represented as follows:
\[G = H - TS\] or can also be represented as,
\[G = U + PV - TS\] where,
\[U\] represents internal energy(SI unit: Joule) , \[P\] represents pressure(SI unit: pascal) , \[V\]represents volume (SI unit: \[{m^3}\] ) , \[S\]represents entropy (SI unit: Joule/Kelvin) ,\[H\] represents enthalpy (SI unit: Joule), \[T\] is the temperature in kelvin.
To calculate the standard Gibb’s free energy, we will apply the formula:
\[\Delta {G^o} = - nF{E^o}_{cell}\]
Where \[n\] is the total number of transferred electrons, \[F\]is the amount of electricity flowing and \[{E^o}_{cell}\] is the standard cell potential.

Complete Step by Step Solution:
It is given that \[{E^o}_{\dfrac{{F{e^{2 + }}}}{{Fe(s)}}} = - 0.440V\] , \[{E^o}_{\dfrac{{F{e^{3 + }}}}{{Fe(s)}}} = - 0.036V\]and \[{E^{^O}}_{\dfrac{{{I_2}}}{{2I}}} = - 0.539V\]
Where \[{E^{^O}}\]represents the internal energy that is a state variable, we can say that
\[2F{e^{3 + }}(aq) + 2{I^ - }(aq) \to 2F{e^{2 + }}(aq) + {I_2}(s)\]
\[1 \times {E^o}_{\dfrac{{F{e^{ + 3}}}}{{F{e^{ + 2}}}}} + 2 \times {E^o}_{\dfrac{{F{e^{ + 2}}}}{{Fe}}} = 3 \times {E^o}_{\dfrac{{F{e^{ + 3}}}}{{F{e^{ + 2}}}}}\]
\[ = 3 \times ( - 0.036) - 2 \times ( - 0.44)\]
\[ = 0.772V\]
\[{E^o}_{cell} = {E^o}_{\dfrac{{F{e^{ + 3}}}}{{F{e^{ + 2}}}}} + {E^o}_{\dfrac{{{I^ - }}}{{{I_2}}}}\]
\[ = 0.772 - 0.539 = 0.233V\]
\[\Delta {G^o} = - nF{E^o}_{cell}\]
\[ = + 2 \times 96500 \times 0.233\]
\[ = 44969J = 44.9KJ = 45KJ\]
Therefore, the magnitude of the molar energy-free change will be equal to -45KJ.

Note: Important points of Gibbs’s free energy that are to be considered are as follows:
1) If the free energy of the reactants is greater than the products, then the reaction will tend to take place spontaneously.
2) If the free energy of the products is greater than the reactants, then the chemical reaction will not take place.
3) In a spontaneous reaction, Gibbs free energy will always decrease and it will never increase.