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Why transition metal chloride complex have lower reduction potential than metal aqua ions? For example:

$$ \begin{align} \ce{Au^3+(aq) + 3 e- &-> Au(s)} &\quad E^\circ &= \pu{+1.52 V}\tag{1}\\ \ce{[AuCl4]-(aq) + 3 e- &-> Au(s) + 4 Cl-} &\quad E^\circ &= \pu{+0.93 V}\tag{2} \end{align} $$

Is it because the chloride complex formation constant is very high (higher than aqua ions) so the reduction of $\ce{Au^3+}$ is less favored?

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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:

$$ \begin{align} \ce{Au^3+(aq) + 3 e- &<=> Au(s)} &\quad E^\circ_\mathrm{cathode} \tag1\\ \ce{[AuCl4]-(aq) + 3 e- &<=> Au(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} \tag2 \end{align} $$

If we substract the equation $(2)$ from the equation $(1)$, we get:

$$\ce{Au^3+(aq) + 4 Cl- (aq) <=> [AuCl4]-(aq)} \quad \quad E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag3$$

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 ($K_f = 1.0 \times 10^{16}$):

$$ \begin{align} \ce{Pt^2+(aq) + 2 e- &<=> Pt(s)} &\quad E^\circ_\mathrm{cathode} = \pu{1.18 V} \tag5\\ \ce{[PtCl4]^2-(aq) + 2 e- &<=> Pt(s) + 4 Cl- (aq)} &\quad E^\circ_\mathrm{anode} = ? \tag6 \end{align} $$

If we substract the equation $(6)$ from the equation $(5)$, we get:

$$\ce{Pt^2+(aq) + 4 Cl- (aq) <=> [PtCl4]^2-(aq)} \quad \quad E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} \tag7$$

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 $E^\circ_\mathrm{cathode} = \pu{1.18 V}$,

$$E^\circ_\mathrm{anode} = 1\pu{1.18 V} - \pu{0.474 V} = \pu{0.706 V}$$

$$\therefore \ \ce{[PtCl4]^2-(aq) + 2 e- <=> Pt(s) + 4 Cl- (aq)} \quad E^\circ = \pu{0.706 V}\text{ (calculated)} $$

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

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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.

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  • $\begingroup$ Thank you. With this part of the last period "the reduction of Au is less favourited" I meant exactly what you have written (my fault, I was not very clear). If chloride complex's Kf is higher, than the concentration of free Au3+ is very low. Thank you again for answering. $\endgroup$
    – Alessia
    Mar 23 at 13:12
  • $\begingroup$ $\pu{1E-30M}$ is effectively nothing. Reduction of aquo and chlorido complex have only the product in common. $\endgroup$
    – Mithoron
    Mar 23 at 16:17

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