The standard reduction potentials are as follows: Cr³⁺(aq) + 3 e^- → Cr(s) ; E° = -0.74 V
Fe²⁺(aq) + 2 e^- → Fe(s) ; E° = -0.41 V
Calculate the standard Gibbs free energy change for the following reaction.
2 Cr(s) + 3 Fe²⁺ → 3 Fe(s) + 2 Cr³⁺(aq)

Question

Quizplus · Accepted Answer

The standard Gibbs free energy change ($\Delta G^\circ$) for a reaction can be calculated using the equation $\Delta G^\circ = -nFE^\circ$, where $n$ is the number of moles of electrons transferred, $F$ is the Faraday constant (96485 C/mol), and $E^\circ$ is the standard cell potential. The standard cell potential ($E^\circ$) is the difference between the reduction potentials of the cathode and anode. For the given reaction, the reduction potential for $Fe^{2+}$ to $Fe$ is -0.41 V (anode, so we take it as positive since it will be reversed in the reaction), and for $Cr^{3+}$ to $Cr$ is -0.74 V (cathode). Thus, $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.74 - (-0.41) = -0.33 V$. The total number of electrons transferred in the balanced reaction is 6 (2 Cr atoms each gaining 3 electrons, and 3 Fe atoms each losing 2 electrons). Plugging these values into the formula gives $\Delta G^\circ = -6 \times 96485 \times -0.33 = -191,055.9 J$ or approximately -191 kJ.

The Standard Reduction Potentials Are as Follows: Cr3+(aq)+ 3 E-

The Standard Reduction Potentials Are as Follows: Cr³⁺(aq)+ 3 E^-