Heat given off from an electrochemical cell compared to mixing reactants

Question

My confusion regarding the difference between the heat given off in a reaction that takes place in an electrochemical cell compared to in a beaker/appropriate container when the reactants are simple mixed, arose from the following problem question:

For the reaction:

\begin{align} \ce{HgCl2~(s) + H2~(g,1~bar) &-> 2Hg~(l) + 2H+~(aq) + 2Cl- }\\ \Delta G^\circ &=-51.64~\mathrm{kJmol^{-1}},\\ \Delta S^\circ &=-61.6~\mathrm{JK^{-1}mol^{-1}},\\ \Delta H^\circ &=-70~\mathrm{kJmol^{-1}}.\\ \end{align}

Part (a) of the question then asks: "Determine how much heat is given off at $298~\mathrm{K}$ per mole of $\ce{HgCl2}$ if the cell is operated reversibly"

The correct approach is to use $T\,\mathrm{d}S=\mathrm{d}q$ giving a value of: $18.36~\mathrm{kJmol^{-1}}$

Then part (b) asks: "Now determine the heat given off if the reaction is brought about by simply mixing the reactants"

I realised that the electrical work done in the cell would now be given off as heat so I just added the value for the Gibbs free energy to the value calculated in (a) and got: $70~\mathrm{kJmol^{-1}}$. This answer is correct but my tutor noted that this value is the value for the enthalpy change which would have been a shortcut to the answer. I didn't spot this and I don't understand why this is the case so wouldn't spot it again, should another question like this come up. Please can you clear up why this value is the enthalpy change.

I have some issues with the above working (it is all correct but I don't fully understand it).

Firstly, in part (a) I had to use the Clausius inequality (except it was an equality since the process was carried out reversibly) to find $\mathrm{d}q$. However, why must I use this equation because $\Delta H=q_p$ - the process was carried out at a constant pressure?

Secondly, if the heat given off in part (b) is $70~\mathrm{kJmol^{-1}}$ doesn't this violate the Clausius inequality? I must be wrong but it seems that this must be the case since $\mathrm{d}S \geq \frac{\mathrm{d}Q}{T}$ and using the value for the entropy change ($\Delta S^\circ=-61.6~\mathrm{JK^{-1}mol^{-1}}$) and the new calculated value for the heat given off ($70~\mathrm{kJmol^{-1}}$) then dividing this by $298~\mathrm{K}$ gives a value larger than the value for the entropy change but the Clausius inequality says that the change in entropy is greater than or equal to $\mathrm{d}q/T$. What is going on there?

Answer 1

I think the problem here is related to what each of these terms means, in a fundamental sense.

$\mathrm{d}S$ is the entropy of the system

$\frac{-\mathrm{d}q}{T}$ is the entropy of the surroundings - the heat change of the surroundings divided by the temperature.

$\mathrm{d}H$ is the enthalpy of the system - it is equal to $\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.

In other words, you can't just use $\mathrm{d}q = \Delta H$ for the 1st question because $\mathrm{d}q_p \neq\Delta H$ when you are doing any work besides PV work.

Instead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:

$$\Delta S - \frac{\mathrm{d}q}{T} \geq 0$$

Solving that for dq gives you the heat absorbed by the surroundings, which is equal to temperature times the entropy change of the system.

For a process that is only (possibly) doing PV work, at constant pressure, $\Delta H = \mathrm{d}q_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV.

In your case, you used: \begin{align} \Delta G &= \Delta H - T\,\mathrm{d}S\\ \Delta H &= \Delta G + T\,\mathrm{d}S\\ \end{align}

This works because the equation for $\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.

Edit

To answer your last questions, and questions from the comments:

why is $T\,\mathrm{d}S$ less than $\mathrm{d}q$ in my example? I don't really get this bit

It's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:

\begin{align} T\,\mathrm{d}S &\geq \mathrm{d}q\\ (298~\mathrm{K})\left( -61.6~\mathrm{\frac{J}{K\cdot mol}}\right) &\geq -70~\mathrm{\frac{kJ}{mol}}\\ -18~\mathrm{kJ} &\geq -70~\mathrm{kJ}\\ \end{align}

Which is true, although it is easier to see if you divide by -1:

$$18~\mathrm{kJ} \leq 70~\mathrm{kJ}$$

The entropy divided by the temperature gives $234.9~\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?

Here I assume you meant "enthalpy divided by temperature." The second reaction is not carried out in a reversible manner, so we have no guarantee that $\mathrm{d}S = \mathrm{d}q/T$. All we know is that it will be greater than or equal to $\mathrm{d}q/T$, and if you plug in the numbers, it is.

Remember that all of this is built into the $\Delta G$ equation as well, which is why we use it when doing "regular" chemical reactions at constant T & P.

Answer 2

For first part since system is operated reversibly at constant temperature: $$\newcommand{\d}{{\rm d}} \d S=\frac{\d q_{rev}}{T}\implies \Delta S=\frac{Q_{rev}}{T}$$ This is what you have done. For second part, Yes Gibbs energy is the maximum (in reversible process) non-expansion/additional work (here the electrical work) $$\Delta G^\circ=-\nu FE^\circ=w_{add,max}$$ This value would be same as $\Delta H^\circ$ because from rearrangement of definition of $$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$$$$\Delta H^\circ=\underbrace{\Delta G^\circ}_{\text{non expansion work}}+\underbrace{T\Delta S^\circ}_{\text{work extracted as heat}}$$ Also it can be viewed in another way, since $\Delta H^\circ$ is "the energy supplied as heat at constant pressure".Another thing to keep in mind is $\d H=q_p$ only under constant pressure and no additional work, both to be satisfied. And for the Classius Inequality, there's some calculation error by you: $$\frac{Q}{T}=\frac{-70000}{298}=-234.8\le-61.6(\Delta S^\circ)$$