Express the maximum work from a voltaic cell

Question

The net cell reaction of an electrochemical cell and its standard potential is given below: $$\ce{ Mg + 2Ag+ ->Mg^{2+} + 2Ag} \ \ \ \ \ \ \ \ E^\circ=3.17\:\mathrm{V}$$ The question is to find the maximum work obtainable from this electrochemical cell if the initial concentrations of $\ce{Mg^{2+}}=0.1\ \mathrm{M}$ and of $\ce{Ag+}=1\ \mathrm{M}$.

The solution just uses the Nernst equation to find the potential at this concentrations and uses $\Delta G=-nFE$ to calculate the Gibbs free energy change which is then equated to the maximum work obtainable.

But this is just the work obtainable per mole at this concentration only. As soon as the reaction proceeds, the concentration change and so does the value of $\Delta G$ per mole and therefore the maximum work obtainable is different for different concentrations.

Therefore, to find the maximum work obtainable, shouldn't we first calculate the equilibrium constant of this reaction (I calculated it to be $K_c=1.89\times10^{107}$), then the equilibrium concentrations(which is probably $\ce{Mg^{2+}}=0.6\ \mathrm{M}, \ \ce{Ag+}=0.178\times10^{-53}\ \mathrm{M}$) and then use something like: $$ \int_{\mathrm{in}}^{\mathrm{eq}}\Delta G* \mathrm{d\,M} $$ where $\Delta G$ is per mole and the $dM$ is a small amount in moles that the reaction proceeds at that concentration.

Please help me verify my method and to proceed further.

Answer 1

There is a simpler approach to this problem.

First, since both initial concentrations are not 1 M, you need to determine $E$ for the initial conditions, where $Q$ is the reaction quotient $Q=\frac{[\ce{Mg^{2+}}]}{[\ce{Ag^{+}}]^2}$:

$$E=E^\circ-\dfrac{RT}{nF}\ln{Q}$$

The value of $n$ for this reaction is $n=2$, and assuming $T=298\ \mathrm{K}$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ K})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.1 \text{ M}}{(1.0 \text{ M})^2}}=3.22 \text{ V}$$

You can then go and convert this potential into free energy or maximum work or whatever else you want.

At equilbrium $E=0$ and $Q=K_c$, so we can calculate the equilibrium constant if we want to compare it to $Q$ and determine which direction is favored. We don't need to. Since $E_i>0$, this reaction procedures toward the products.

Now, as the reaction proceeds the concentrations of $\ce{Ag+}$ and $\ce{Mg^{2+}}$ will change. From the law of mass action:

$$-\Delta \ce{[Ag+]}= 2\Delta \ce{[Mg^{2+}]}$$

After some time $\Delta \ce{[Ag+]} = 0.25 \text{ M}$, so $\Delta \ce{[Mg^{2+}]}=0.125\text{ M}$ and $[\ce{Ag+}]=0.75 \text{ M}$ and $[\ce{Mg{2+}}]=0.225 \text{ M}$. The value of $Q$ has changed and so has the value of $E$:

$$E_i=3.17\ \mathrm(V)-\dfrac{(8.314 \dfrac{\text{J}}{\text{K}\cdot \text{mol}})(298 \text{ V})}{(2)(9.648\times10^4 \dfrac{\text{C}}{\text{mol}})}\cdot\ln{\dfrac{0.225 \text{ M}}{(0.75 \text{ M})^2}}=3.18 \text{ V}$$

As the reaction proceeds, the value of $E$ decreases, which means the amount of work that can be done by the reaction also decreases. The maximum value of $E$ occurs at the initial state, and so the maximum work that this cell can do happens at the initial state. Since $E$ decreases as the reaction proceeds, so too does the work the cell can do at any moment.

If you want the total work, then you need to integrate with respect to $Q$, not worrying about the actual concentrations:

$$\int_{Q_i}^{K_c}E\cdot\mathrm{d}Q$$

Answer 2

Here is a plot of the cell potential for your example:

Lucky for us, it does not change much until the very end (at the end, the cell potential is zero - this does not show up on the graph, but I used a smaller interval near the end to show the trend). This means numerical integration would work quite well. Notice that the x-axis is the progress of the reaction (related in a linear way to the electrical charge transferred from oxidation to reduction half reaction via the wire). To integrate, you would have to integrate over that charge because the work is equal to charge times electrical potential.

What is the answer to the sample problem

Work is an extensive quantity, and the problem contains intensive quantities only, so there is some information missing. It is like asking the total work an alkaline battery can do, but leaving out the information of whether the battery is a D cell or a AAA cell.

The best we can do is to say that between initial state and equilibrium state, the average cell potential is 3.15 V. Once some information about the amount of reactants is available (e.g. volume of solution, amount of electrolyte), we can calculate the charge transferred and calculate the total work.

Why do we have to integrate?

We don't because the free energy is a state function. If we had a way to calculate the change in free energy between initial state and equilibrium state, we could use that as the work. Let's say the half-cells both have a volume of 1L, so then almost 1 mol of silver ions get oxidized and 0.5 mol of magnesium ions are formed (on top of the 0.1 mol already there initially). One mole of electrons will travel along the wire, corresponding to a charge of 96,485 coulomb.

Alternatively, we could write down the reaction as a two step process, running the reaction at initial concentrations (imagine that there are 1,000,000 L instead of 1 L, but the same amount reacts), and then two reactions transferring ions to their equilibrium concentrations. We know how to calculate the molar Gibbs energy for both:

$$\ce{2Ag+ + Mg ->2Ag + Mg^2+}; \ \ \ \Delta_r G = \Delta_r G^\circ + R T \ln(Q_\text{initial})$$

$$\ce{Ag+ ->Ag+}; \ \ \ \Delta G = R T \ln \frac{\pu{0.178e-53}}{1}$$

$$\ce{Mg^2+ ->Mg^2+}; \ \ \ \Delta G = R T \ln \frac{\pu{0.6}}{0.1}$$

Then we multiply these molar Gibbs energies by the amount that reacts (for the first one) and by the amount present at equilibrium (for the latter two) and add them up to get the total work.

Answer 3

It seems to me that your understanding of the question is flawed. You would require the integral suggested if you wish to calculate the total work done. The question asked is to calculate the maximum work at any given instant, which is obviously at the the initial state where the concentrations would be maximum and thereby the value of E cell. This would give you maximum value of work done at any particular instant until equilibrium is reached and not the summation of work done during the whole process until the cell dies.