# How can normal potentials be explained?

For these two metals, the standard reduction potentials are: $$\ce{2e-} + \ce{Cd^{2+}} \rightarrow \ce{Cd}:\mbox{ } E^0 = -0.403 V\\ \ce{2e-} + \ce{Ni^{2+}} \rightarrow \ce{Ni}:\mbox{ } E^0 = -0.25 V$$

But the simple ionization energies are:

$$\ce{Ni} \rightarrow \ce{Ni^{2+}} \mbox{ } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2490 \ce{kJ. mol^{-1}}$$ $$\ce{Cd} \rightarrow \ce{Cd^{2+}} \mbox{ } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2500\ce{kJ. mol^{-1}}$$ The difference in $E^\theta$ is greater than the difference in ionization energy of both elements.
How can this be explained?

• Can you be more specific regarding your query? Why shouldn't they be different? – Nicolau Saker Neto Dec 11 '13 at 19:39
• If I get asked, for which of these two equations, the normal potential will be higher, how should I go about answering that? – TMOTTM Dec 11 '13 at 20:27
• Without looking at result, it's not obviously for me which one will be higher. I have a guess based on [this post][1]. There is a choice of filling electron from d$^8$ to d$^{10}$ vs s$^0$ to s$^2$. The prior one will cause much more electrostatic repulsion (like why 3d$^3$ is not the electronic configuration of Sc). However, this argument will predict Ni$^{2+}$ higher than Zn$^{2+}$. [1]: chemistry.stackexchange.com/questions/6921/… – user26143 Dec 12 '13 at 0:34
• If I go down one period, the smaller $Z/r$ of Cd$^{2+}$ will also make the $E^0$ more positive, which is an opposite trend. It's hard to say how large this effect. From the experimental $E^0$, the downing period expected to be a smaller contribution. Anyway, all these are hand-waving argument, quantitative calculations are needed to justify this "explanation". – user26143 Dec 12 '13 at 0:37
• I now found a theoretical workaround. The standard potential can in principle be calculated from a thermodynamic cycle in which an electron is transfered from a metal to $\frac{1}{2}H_2(g)$. In principle, I believe, all properties along the cycle should be calculable. – TMOTTM Dec 13 '13 at 14:27