5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ By seeing this reasonement, it seems that you imply that $\Delta G$ is not a state function, since it depends by the "path" that the reaction establishes by $K_c$. In reality, the actual work depends on the given potential (which does not depend by the path), defined by the initial state of the system; that's why there is not any differential term inside $E$, because the electrons are travelling a definite "jump", no matter how big or small is $K_c$ (any reaction with equal $E$ and $n$ so has an equal $\Delta G$!) $\endgroup$ – alandella Aug 31 '13 at 10:29
  • $\begingroup$ @AndreaL. But then the potential which the electrons cross the cell will keep decreasing as the reaction proceeds and hence the work obtained at different values of concentrations of the reacting species should be different. Did we account for that difference when we calculated the original $\Delta G$ initially? $\endgroup$ – Satwik Pasani Aug 31 '13 at 15:14
  • 1
    $\begingroup$ As stated by the following answer, Work is not a state function, so it does follow the integral proposed to calculate the total work; in my comment I said that the maximum work does depend only to the initial value of the concentration ($[X]\to E$ by Nernst Eq.), hence the initial potential/max-work does not count the decreasing factor by $K_c$. If you want to obtain the total work, then following of the path described by the integtal is necessary. $\endgroup$ – alandella Aug 31 '13 at 19:09
  • $\begingroup$ @AndreaL. Thanks. This helped quite a bit. I didn't need to solve the integral, just to verify that it was correct. $\endgroup$ – Satwik Pasani Sep 1 '13 at 10:37
4
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ If we use this cell to say light a bulb by using superconducting wires, then the net energy dissipated in the ideal case would be the value of the integral, right? Then the maximum obtainable work is then the integral itself. The initial $\Delta G$ was only the work that you would have obtained if those concentrations would have remained unchanged until all the reactants are converted to products and that too would be $\Delta G \times n$ since $\Delta G$ is per mole, is it correct? And any way to evaluate that integral? $\endgroup$ – Satwik Pasani Aug 31 '13 at 15:22
  • 1
    $\begingroup$ @Ben Norris, I apologize for the following question, but is it correct to state your mentioned integral as:$$L_{\mathrm{tot}}=\int_{Q_i}^{K_{eq}}-n\mathcal{F}E\;\mathrm{d}Q\quad ?$$If so it would be easy to calculate given the first equation. $\endgroup$ – alandella Aug 31 '13 at 20:16
  • $\begingroup$ @AndreaL. Seems that way to me. $\endgroup$ – Ben Norris Aug 31 '13 at 22:38
  • $\begingroup$ @Ben_Norris I don't think you can integrate over Q. Electrical work is charge transferred times electrical potential, and the integration variable has to be charge. It is the same with work to extend a spring. The integration variable has to be the extension of the spring, not something else that also changes while the spring is extended. $\endgroup$ – Karsten Theis Apr 7 at 14:55
0
$\begingroup$

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

enter image description here

Lucky for us, it does not change much until the very end. 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 which charge has been transferred from oxidation to reduction half reaction. 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:

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

$$\ce{Ag+ ->Ag+}; \ \ \ \Delta G = R T \ln \frac{\pu{0.178e53}}{1}$$

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

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This is just an abridged version of the answer by Ben. It does not add any more insight to the question. $\endgroup$ – Martin - マーチン Jun 22 '16 at 9:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.