Why transition metal chloride complex have lower reduction potential than metal aqua ions?
The  anion  $\ce{AuCl4−}$ is  the  most  known  complex  of  gold(III). It is very stable complex and have very high formation constant ($K_f$). That is the reason the metal chloride complex have a lower reduction potential than metal aqua ion. As pointed out in the other answer, this is a consequence of Nernst's law. Let's look at the two redox equations we are dealing with:
$$
\begin{align}
\ce{Au^3+(aq) + 3 e- &amp;&lt;=&gt; Au(s)} &amp;\quad E^\circ_\mathrm{cathode} \tag1\\
\ce{[AuCl4]-(aq) + 3 e- &amp;&lt;=&gt; Au(s) + 4 Cl- (aq)} &amp;\quad E^\circ_\mathrm{anode} \tag2
\end{align}
$$
If we substract the equation $(2)$  from the equation $(1)$, we get:
$$\ce{Au^3+(aq) +  4 Cl- (aq) &lt;=&gt; [AuCl4]-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag3$$
The equation $(3)$ is the reaction for the formation of $\ce{AuCl4−}$ with formation constant $K_f$. If you applied the Nernst equation for the equation $(3)$, then at $\pu{25 ^\circ C}$:
$$E^\circ_\mathrm{cell} = \frac{RT}{nF} \ln Q = \frac{RT}{nF} \ln K_f = \frac{0.0592}{n} \log K_f \tag4$$
This last solution of the equation $(4)$ (where $n = 3$) is because, when at equilibrium $Q = K$. If we know the $K_f$, we could have find the $E^\circ_\mathrm{cell}$, and hence $E^\circ_\mathrm{anode}$ (which is for $\ce{AuCl4−}$). Unfortunately, I couldn't find $K_f$ for $\ce{AuCl4−}$ ion formation. Thus, I'd show how big $K_f$ for $\ce{AuCl4−}$ by applying the given values of $E^\circ_\mathrm{cathode} = \pu{+1.52 V}$ and $E^\circ_\mathrm{anode} = \pu{+0.93 V}$ on the equation $(4)$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f \ \Rightarrow \log K_f = \frac{n}{0.0592} E^\circ_\mathrm{cell} = \frac{3}{0.0592} (1.52 - 0.93) = 29.9$$
$$\therefore \ K_f = 10^{29.9} = 7.91 \times 10^{29}$$
It is also in similar way you can find the $E^\circ$ for redox half-reaction of $\ce{PtCl4^2−}$ since we know its formation constant ($K_f = 1.0 \times 10^{16}$):
$$
\begin{align}
\ce{Pt^2+(aq) + 2 e- &amp;&lt;=&gt; Pt(s)} &amp;\quad E^\circ_\mathrm{cathode} = \pu{1.18 V} \tag5\\
\ce{[PtCl4]^2-(aq) + 2 e- &amp;&lt;=&gt; Pt(s) + 4 Cl- (aq)} &amp;\quad E^\circ_\mathrm{anode} = ? \tag6
\end{align}
$$
If we substract the equation $(6)$  from the equation $(5)$, we get:
$$\ce{Pt^2+(aq) +  4 Cl- (aq) &lt;=&gt; [PtCl4]^2-(aq)} \quad \quad  E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag7$$
From equation $(4)$ where $n = 2$:
$$E^\circ_\mathrm{cell} = \frac{0.0592}{n} \log K_f = \frac{0.0592}{2} \log( 1.0 \times 10^{16})  = 0.474$$
Since, $E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.474 V}$, and $E^\circ_\mathrm{cathode} = \pu{1.18 V}$,
$$E^\circ_\mathrm{anode} = 1\pu{1.18 V} - \pu{0.474 V} = \pu{0.706 V}$$
$$\therefore \ \ce{[PtCl4]^2-(aq) + 2 e- &lt;=&gt; Pt(s) + 4 Cl- (aq)} \quad E^\circ = \pu{0.706 V}\text{ (calculated)} $$
This value is in good agreement with the experimental value, which is given as $\pu{0.755 V}$ in electrpchemical series of CRC Handbook of Chemistry and Physics.
Source of formation constant of platinum complex ion