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Question: For the reaction: $2Fe^{3+}$ aq + $2I^-$aq $\rightarrow$ $2Fe^{2+}$ aq + $I_2s$ The magnitude of th...

For the reaction: 2Fe3+2Fe^{3+} aq + 2I−2I^-aq →\rightarrow 2Fe2+2Fe^{2+} aq + I2sI_2s

The magnitude of the standard molar free energy change, ΔrGmo\Delta_r G_m^o = - ______________ kJ ( Round off to the Nearest Integer).

A

45

B

44

C

46

D

43

Answer

45

Explanation

Solution

  1. Identify cathode (Fe3+/Fe2+Fe^{3+}/Fe^{2+}) and anode (I2/I−I_2/I^-) half-reactions.
  2. Determine the number of electrons transferred (n=2n=2).
  3. Calculate the standard potential for Fe3+/Fe2+Fe^{3+}/Fe^{2+} using given potentials relative to Fe(s). EFe3+/Fe2+o=3×Eo(Fe3+/Fe)−2×Eo(Fe2+/Fe)3−2=3×(−0.036 V)−2×(−0.440 V)1=0.772 VE_{Fe^{3+}/Fe^{2+}}^o = \frac{3 \times E^o(Fe^{3+}/Fe) - 2 \times E^o(Fe^{2+}/Fe)}{3 - 2} = \frac{3 \times (-0.036 \text{ V}) - 2 \times (-0.440 \text{ V})}{1} = 0.772 \text{ V}
  4. Calculate the standard cell potential (Ecello=Ecathodeo−EanodeoE_{cell}^o = E_{cathode}^o - E_{anode}^o). Ecello=0.772 V−0.539 V=0.233 VE_{cell}^o = 0.772 \text{ V} - 0.539 \text{ V} = 0.233 \text{ V}
  5. Calculate the standard Gibbs free energy change (ΔrGmo=−nFEcello\Delta_r G_m^o = -nFE_{cell}^o). ΔrGmo=−(2)×(96500 C)×(0.233 V)=−44989 J=−44.989 kJ\Delta_r G_m^o = -(2) \times (96500 \text{ C}) \times (0.233 \text{ V}) = -44989 \text{ J} = -44.989 \text{ kJ}
  6. Take the magnitude and round to the nearest integer. ∣−44.989 kJ∣≈45 kJ|-44.989 \text{ kJ}| \approx 45 \text{ kJ}