This difference of redox potentials is a consequence of Nernst's law. The second equation is nothing else as the first one, where the concentration of the $\ce{Au^{3+}}$ in a solution of $\ce{AuCl4^{-}}$ is extremely low. This sort of comparison could be used to calculate the equilibrium constant of the equilibrium $$\ce{Au^{3+} + 4 Cl- <-> AuCl4^{-}}$$ Here, Nernst's law can be written : $$\ce{Au -> Au^{3+} + 3 e-}$$ $$\ce{E = 0.93 V = 1.52 V + \frac{0.059 V}{3} . log [Au^3+]}$$ so that the concentration $\ce{[Au^{3+}]}$ in $\ce{AuCl4^-}$ is given by $$\mathrm{log}\ce{[Au^{3+}] = \frac{3}{0.059 V}(0.93 V - 1.52 V) = - 30}$$ This means that in a $1$ molar solution of $\ce{AuCl4^-}$, the concentration of the ion $\ce{Au^{3+}}$ is only $\pu{1E-30M}$. But the redox reaction happening in $\ce{AuCl^{4-}}$ is the same redox reaction that occurs between $\ce{Au}$ and $\ce{Au^{3+}}$. Just the concentration of $\ce{Au^{3+}}$ is extremely low in $\ce{AuCl4^-}$ solutions.

This difference of redox potentials is a consequence of Nernst's law. The second equation is nothing else as the first one, where the concentration of the $\ce{Au^{3+}}$ in a solution of $\ce{AuCl4^{-}}$ is extremely low. This sort of comparison could be used to calculate the equilibrium constant of the equilibrium $$\ce{Au^{3+} + 4 Cl- <-> AuCl4^{-}}$$ Here, Nernst's law can be written : $$\ce{Au -> Au^{3+} + 3 e-}$$ $$\ce{E = 0.93 V = 1.52 V + \frac{0.059 V}{3} . log [Au^3+]}$$ so that the concentration $\ce{[Au^{3+}]}$ in $\ce{AuCl4^-}$ is given by $$\mathrm{log}\ce{[Au^{3+}] = \frac{3}{0.059 V}(0.93 V - 1.52 V) = - 30}$$ This means that in a $1$ molar solution of $\ce{AuCl4^-}$, the concentration of the ion $\ce{Au^{3+}}$ is only $\pu{1E-30M}$. But the redox reaction happening in $\ce{AuCl4^{-}}$ is the same redox reaction that occurs between $\ce{Au}$ and $\ce{Au^{3+}}$. Just the concentration of $\ce{Au^{3+}}$ is extremely low in $\ce{AuCl4^-}$ solutions.