Calculating the electrode potential from thermochemical data (part 2)

This question is in continuation of this question.

I am taking an example of $$\ce{Cu^2+}$$ in aqueous medium.

\begin{align} \ce{Cu (s) &-> Cu (g)} & \Delta_\text{a}H^\circ &= \pu{339 kJ mol-1} \\ \ce{Cu (g) &-> Cu+ (g) + e-} & \Delta_\text{i}H_1^\circ &= \pu{745 kJ mol-1} \\ \ce{Cu+ (g) &-> Cu^2+ (g) + e-} & \Delta_\text{i}H_2^\circ &= \pu{1960 kJ mol-1} \\ \ce{Cu^2+ (g) &-> Cu^2+ (aq)} & \Delta_\text{hyd}H^\circ &= \pu{-2121 kJ mol-1} \end{align}

Net reaction:

\begin{align} \ce{Cu (s) &-> Cu^2+ (aq) + 2e-} & \Delta_\text{r}H^\circ &= \pu{923 kJ mol-1} \end{align}

From Lange's Handbook of Chemistry- 15E, at page 6.91:

\begin{align} \Delta_\text{r}S^\circ &= S^\circ(\ce{Cu^2+ (aq)}) - S^\circ(\ce{Cu (s))} \\ &= (\pu{-98.4 J K-1 mol-1}) - (\pu{33.15 J K-1 mol-1}) \\ &= \pu{-131.55 J K-1 mol-1} \end{align}

and hence

\begin{align} \Delta_\text{r}G^\circ &= \Delta_\text{r}H^\circ - T\Delta_\text{r}S^\circ \\ &= \pu{923 kJ mol-1} - (\pu{298.15 K})(\pu{-131.55 J K-1 mol-1}) \\ &= \pu{962.2 kJ mol-1} \\ E^\circ &= -\frac{\Delta_\text{r}G^\circ}{nF} \\ &= -\frac{\pu{962.2 kJ mol-1}}{(2)(\pu{96485 C mol-1})} \\ &= \pu{-4.986 V} \end{align}

But the value of $$E^\circ(\ce{Cu|Cu^2+})$$ is $$\pu{-0.34 V}$$.

Where did the calculations go wrong?

• I'd strongly suggest looking at my edit for typography. In particular, it's far easier to enclose the number with the unit inside \pu{...} i.e. use \pu{2 mol} not 2~\pu{mol} (the latter is correct but a bit more tedious). And you should really make use of the align environment for this sort of thing, it makes things look far better. See this or this. Lastly, don't leave out units!! Oct 13 '21 at 13:34
• Ok... But why plimsoll replaced by circle? Oct 13 '21 at 16:02
• No particular reason, to be honest; I guess I just retyped the equations and I'm more used to using the circle so ended up using that "by default". Both symbols are acceptable according to the IUPAC Green Book. Feel free to change that back if you wish. Oct 13 '21 at 16:26

I'm going to drop the standard state symbol throughout this answer for ease of notation.

Your value for $$\Delta_\mathrm{r} H$$ is way too large. Specifically, if you want to do it this way (by counting ionisation energies, etc.), then you must do the same for the reference electrode i.e. the standard hydrogen electrode (SHE), because $$E^\circ$$ is not an absolute value but is rather measured against the SHE. So, you should calculate the full $$\Delta_\mathrm{r}H$$ for the reaction

$$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$

and in this, you'll find that there are extra terms corresponding to the solvation of protons, ionisation energy of hydrogen, and the $$\ce{H-H}$$ bond formation which you've neglected in your analysis.

Instead, a more straightforward way to calculate $$\Delta_\mathrm{r}H$$ is to use formation enthalpies:

$$\Delta_\mathrm{r}H = \Delta_\mathrm{f}H(\ce{Cu^2+(aq)}) - \Delta_\mathrm{f}H(\ce{Cu(s)})$$

The second of these terms is obviously zero. Looking up the first term in Table 2.8 of the "Resource Section" in Atkins's Physical Chemistry (9th ed.) I get a value of $$\pu{+64.77 kJ/mol}$$. Plugging this in:

\begin{align} E^\circ &= -\frac{\Delta_\text{r}H - T\Delta_\text{r}S}{nF} \\ &= -\frac{\pu{64.77 kJ mol-1} - (\pu{298.15 K})(\pu{-131.55 J K-1 mol-1})}{(2)(\pu{96485 C mol-1})} \\ &= \pu{-0.539 V} \end{align}

which is not quite perfect, but at least it's much closer.

The reason why this value works is because the enthalpies of formation for ions are already normalised against $$\ce{H+}$$: that is to say, by definition, $$\Delta_\mathrm{f}H(\ce{H+ (aq)}) = 0$$, and so the enthalpy for the SHE half-reaction is zero. (The electrons can be safely ignored, because they would cancel out on both sides of the equation.) In Atkins's own words (p 71, Section 2.8):

The standard enthalpy of formation of ions in solution poses a special problem because it is impossible to prepare a solution of cations alone or of anions alone. This problem is solved by defining one ion, conventionally the hydrogen ion, to have zero standard enthalpy of formation at all temperatures:

$$\Delta_\mathrm{f}H^\circ(\ce{H+, aq}) = 0$$

Thus, if the enthalpy of formation of $$\ce{HBr (aq)}$$ is found to be $$\pu{-122 kJ mol−1}$$, then the whole of that value is ascribed to the formation of $$\ce{Br-(aq)}$$, and we write $$\Delta_\mathrm{f}H^\circ(\ce{Br-, aq}) = \pu{-122 kJ mol-1}$$. That value may then be combined with, for instance, the enthalpy formation of $$\ce{AgBr(aq)}$$ to determine the value of $$\Delta_\mathrm{f}H^\circ(\ce{Ag+, aq})$$, and so on. In essence, this definition adjusts the actual values of the enthalpies of formation of ions by a fixed amount, which is chosen so that the standard value for one of them, $$\ce{H+(aq)}$$, has the value zero.

It's worth pointing out that the standard molar entropies that you quoted are also already adjusted in this way. That is to say, the standard molar entropy of $$\ce{H+}$$ is already set to $$0$$, by definition.

• Suppose, I want to calculate it by first method: (i) Enthalpy for reverse of solvation of proton= +1150, multiply by 2 gives= +2300 kJ/mol; (ii) reverse of ionisation energy of H atom= -1312, multiply by 2 gives= -2624 kJ/mol; (iii) reverse of bond enthalpy of H2 molecule= -436 kJ/mol; (iv) addition of these gives= -760 kJ/mol; (v) Now adding my DH=+923 kJ/mol, I get= +163 kJ/mol against your +64.77 kJ/mol? Oct 13 '21 at 16:16
• I'm not really sure how much agreement we should expect to get in the final number, to be honest. When you add up many terms on the order of 1000 kJ/mol, even if each of them have an error of 1% i.e. 10 kJ/mol, the final result may easily end up being not perfect. Generally, I'm thinking more in terms of broad strokes here: your original calculation and the correct one differs by a factor of 9 or so. I'd say that even +163 is a much better improvement over the original value of +923. Oct 13 '21 at 16:31
• I'd expect the solvation enthalpies to be the most inaccurate of all the terms in the sum. Oct 13 '21 at 16:33
• I am getting -1.05 V, which is not at all in agreement with accepted value of -0.34 V. I think, I missed something. Oct 13 '21 at 17:15
• Hence my point about the numbers that you're subtracting, in my previous comment. If $X$ and $Y$ are both large numbers of similar magnitude, then $X - Y$ is a small number. Even if the relative error in $X$ and $Y$ is small, the absolute error is still gonna be large compared to their difference, which translates into a large relative error for $X - Y$. Oct 13 '21 at 19:29