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Question: The \( {E^\circ }_{F{e^{2 + }}/Fe} = - 0.44V \) and \( {E^\circ }_{C{r^{3 + }}/Cr} = - 0.74V \) free...

The E∘Fe2+/Fe=−0.44V{E^\circ }_{F{e^{2 + }}/Fe} = - 0.44V and E∘Cr3+/Cr=−0.74V{E^\circ }_{C{r^{3 + }}/Cr} = - 0.74V free energy is (F=96500C)(F = 96500C)
A. 57,900
B. -57,900
C. -173,700
D. 173,700

Explanation

Solution

Hint : In order to the question, first we have to determine the cathode and anode for the reaction and then find the standard E∘{E^\circ } value for the formula to find the Standard free energy change.

Complete Step By Step Answer:
In the question, we have the standard E∘{E^\circ } value or Equilibrium Constant of both Iron and Chromium E∘Fe2+/Fe=−0.44V{E^\circ }_{F{e^{2 + }}/Fe} = - 0.44V and E∘Cr3+/Cr=−0.74V{E^\circ }_{C{r^{3 + }}/Cr} = - 0.74V . Here, the standard E∘{E^\circ } value of Iron is at the higher side, so Iron is acted as a Cathode here and Chromium is acted as an Anode in the given reaction.
Now, we have to calculate the standard E∘{E^\circ } value of the cell, we have the formula: E∘cell=E∘c−E∘a{E^\circ }_{cell} = {E^\circ }_c - {E^\circ }_a
here, E∘c{E^\circ }_c is the standard E∘{E^\circ } value of Cathode
E∘a{E^\circ }_a is the standard E∘{E^\circ } value of Anode.
And, we already know that Iron and Chromium act as Cathode and Anode respectively. ∴E∘cell=−0.44−(−0.74)\therefore {E^\circ }_{cell} = - 0.44 - ( - 0.74) =0.30 Volt= 0.30\,Volt .
Now, in the given reaction 3Fe2+(aq)+2Cr(s)⇒2Cr3+(aq)+3Fe(s)3F{e^{2 + }}(aq) + 2Cr(s) \Rightarrow 2C{r^{3 + }}(aq) + 3Fe(s) , 3Fe2+3F{e^{2 + }} is converted into 3Fe3Fe and 2Cr2Cr is converted into 2Cr3+2C{r^{3 + }} . So, here we can see that the n factor is 6 (n=6).
Hence, we have the formula to find the Standard free energy change for the given reaction:-
The standard free energy of a substance represents the free energy change associated with the formation of the substance.
∴ΔG=−nFE∘cell\therefore \Delta G = - nF{E^\circ }_{cell}
here, ΔG\Delta G is the standard free energy change
nn is the n-factor of the reaction
FF is given (F=96500C)
E∘cell{E^\circ }_{cell} , we calculated above
∵ΔG=−6×96500×0.30\because \Delta G = - 6 \times 96500 \times 0.30 =−173,700= - 173,700
So, we now have the Standard free energy change is -173,700.
Hence, the correct option is (C.) -173,500.

Note :
Or in other words, Calculate the change in standard free energy for a particular reaction. The standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under standard conditions.