3.7: Thermodynamics from Electrochemistry

Electrochemistry for indirect measurement of free energy

Many reactions (particularly biochemical) consist of multiple coupled half-reactions. It is often not possible to measure all of these half-reactions against one known half-cell to determine their reduction potentials. Many reduction potentials are arrived at indirectly using the fact that \(\Delta G\) is a state function and thus so is E^o because \(\Delta_r G = -n_e FE\). This means we can calculate an E^o by following any path that overall leads to the same electrochemical change. As a generic example consider three half-cell potentials \(\ce{E_1 = E^{o} (A->B)}\) , \(\ce{E_2 = E^{o} (B->C)}\) and \(\ce{E_3 = E^{o} (A->C)}\). We know that:

\[\ce{\Delta G^{o}_1 (A->B) + \Delta G^{o}_2 (B->C)= \Delta G^{o}_3 (A->C)} \nonumber\]

Using \(\Delta_r G = -n_e FE\) to replace the free energies gives us:

\[-n_{e1} FE^{o}_1+-n_{e2} FE^{o}_2= -n_{e3} FE^{o}_3 \nonumber\]

Thus as long as we know two of the three cell potentials and all n_e's (moles e- in each reduction), we can solve for the third E^o. We could also solve for an unknown number of electrons transferred knowing all the potentials and two of the n_e's. The other possibility is to do a complete cycle so that \(\Delta G^o_{overall} = 0\), which means that \(E^o_{overall} = 0\). Then you can solve for the E^o of one reduction in the cycle knowing all the others.

Once E^o is known you can calculate equilibrium constants and \(\Delta G\) from:

\[\Delta G^o = -n_e FE^o\text{,}\quad\Delta G^o = -RTlnK\quad\text{and}\quad E^o = \frac{RT}{n_e F}lnK\]

Example \(\PageIndex{1}\)

Given the following half-reactions calculate the \(E^o(\ce{Hg2^{2+} +2e- -> 2Hg(l)})\).

\[\ce{Hg^{2+} +2e- -> Hg(l)}\quad E^o = 0.8535\]

\[\ce{2Hg^{2+} +2e- -> Hg2^{2+}}\quad E^o = 0.911\]

Solution

By inspection two times the first reaction minus the second reaction gives us the desired overall reaction:

\[2(\ce{Hg^{2+} +2e- -> Hg(l)})\quad\text{RXN 1}\\[4pt]
\underline{-(\ce{2Hg^{2+} +2e- -> Hg2^{2+}})\quad\text{RXN 2}}\\[4pt]
\ce{Hg2^{2+} +2e- -> 2Hg(l)}\quad\text{RXN 3} \nonumber\]

 

The multiplication of RXN 1 by 2 means that there are n_e1 = 4, while number electrons transferred in RXN 2 is unchanged at n_e2 =2. Calling the desired reaction #3, we can write:

\[n_{e1}FE^o_1 - n_{e2}FE^o_2 = n_{e3}FE^o_3 \implies 2E^o_3 = 4E^o_1 - 2E^o_2\]

So:

\[E^o_3 = \frac{4(0.8535 V) - 2(0.911V)}{2} = 0.796 V\]

Getting to enthalpy and entropy

Electrochemical measurements of cell potentials versus temperature can be used to determine reaction free energies, entropies and enthalpies. This is because \(\Delta_r G = -n_e FE\) under the current galvanic cell conditions. Thus measurement of E, directly gives us the Gibbs free energy. From general thermodynamics we know:

\[ \left( \dfrac{\partial \Delta G}{\partial T} \right)_p = - \Delta S \label{EQ:dGdT} \]

Substituting in -n_eFE for \(\Delta G\) yields:

\[ nF \left( \dfrac{\partial E}{\partial T} \right)_p = \Delta S \label{EQ:dEdT} \]

So if we measure the slope of the potential at a given temperature we have the \(\Delta S\) of the reaction. Because \(\Delta G = \Delta H - T\Delta S\) we also get \(\Delta H\):

\[\Delta G + T\Delta S= \Delta H \]

If we are at standard conditions for all the reactants we get the standard Gibbs, entropy and enthalpy of reaction.

Example

Consider the following data for the Daniel cell (Buckbeei, Surdzial, & Metz, 1969) which is defined by the following reaction

\[Zn(s) + Cu^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Cu(s) \nonumber \]

T (°C)	0	10	20	25	30	40
E (V)	1.1028	1.0971	1.0929	1.0913	1.0901	1.0887

From a fit of the data to a quadratic function, the temperature dependence of

\[\left( \dfrac{\partial E^o}{\partial T} \right)_p \nonumber \]

is easily established.

Figure \(\PageIndex{1}\): Temperature dependence of the cell potential for a Daniel cell.

The quadratic fit to the data results in

\[\left( \dfrac{\partial E^o}{\partial T} \right)_p = 3.8576 \times 10^{-6} \dfrac{V}{°C^2}(T) - 6.3810 \times 10^{-4} \dfrac{V}{°C} \nonumber \]

So, at 25 °C,

\[\left( \dfrac{\partial E^o}{\partial T} \right)_p = -5.4166 \times 10^{-4} V/K \nonumber \]

noting that \(K\) can be substituted for \(°C\) since in difference they have the same magnitude. So the entropy change calculated using equation \ref{EQ:dEdT} is

\[ \Delta S = nF \left( \dfrac{\partial E^o}{\partial T} \right)_p = (2\,mol)(96485\,C/mol) (-5.4166 \times 10^{-4} V/K) \nonumber \]

Because

\[ 1\,C \times 1\,V = 1\,J \nonumber \]

The standard entropy change for the Daniel cell reaction at 25 °C is

\[ \Delta S = -104.5\, J/(mol\,K). \nonumber \]

It is the negative entropy change that leads to an increase in standard cell potential at lower temperatures.