# Why transition metal chloride complex have lower reduction potential than metal aqua ions?

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?

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

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.

• 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. Mar 23 at 13:12
• $\pu{1E-30M}$ is effectively nothing. Reduction of aquo and chlorido complex have only the product in common. Mar 23 at 16:17