Gibbs Free Energy of reaction

Gibbs free energy, denoted as ( G ), is a thermodynamic potential that measures the maximum reversible work that can be performed by a thermodynamic system at constant temperature and pressure. The change in Gibbs free energy (( $\Delta G$)) is used to predict the feasibility of reactions. For a reaction at constant temperature and pressure, if ( Delta G < 0 ), the reaction is spontaneous; if ( Delta G > 0 ), it is non-spontaneous.

This Story also Contains

1. Electrode Potential and EMF of Cells -
2. Feasibility and Gibbs Free Energy of Reaction
3. Equilibrium Constant
4. Some Solved Examples
5. Summary

Electrode Potential and EMF of Cells -

It is the potential difference between the two terminals of the cell when no current is drawn from it. It is measured with the help of potentiometer or vacuum tube voltmeter.

Calculation of the EMF of the Cell
Mathematically, it may be expressed as
$\begin{aligned} & \mathrm{E}_{\text {cell }} \text { or } \mathrm{EMF}=\left[\mathrm{E}_{\text {red }}(\text { cathode })-\mathrm{E}_{\text {red }}(\text { anode })\right] \\ & \mathrm{E}_{\text {ell }}^{\circ} \text { or } \mathrm{EMF}^{\circ} \\ & =\left[\mathrm{E}_{\text {red }}^{\circ}(\text { cathode })-\mathrm{E}_{\text {red }}^{\circ}(\text { anode })\right]\end{aligned}$

• Characteristics of cell and cell potential.
• For cell reaction to occur the E_cell should be positive. This can happen only if E_red (cathode) > E_red(anode).
• E^o cell must be positive for a spontaneous reaction.
• It measures free energy change for maximum convertibility of heat into useful work.
• It causes the flow of current from the electrode of the higher E^o value to the lower E^o value.

Difference between EMF and Cell Potential

Feasibility and Gibbs Free Energy of Reaction

Let n faraday charge be involved in a cell generating an emf E, then the magnitude of the work done by the cell will be calculated as:
|Work| = Charge × Potential = $(n F) \times(E)$

Now, the maximum work that can be extracted from the cell is equal to the decrease in Gibb's free energy.

So, it can be said that
$-\Delta \mathrm{G}=\mathrm{nFE}$

The negative sign is incorporated to include the spontaneity relation between E and Gibb's Free Energy

Similarly, the maximum obtainable work from the cell at standard conditions will be:

$\mathrm{W}_{\max }=\mathrm{nFE} E_{\text {cell }}^0 \quad$ where $\mathrm{E}_{\text {cell }}^0=$ standard emf of standard cell potential
$-\Delta \mathrm{G}^{\circ}=\mathrm{nFE}_{\text {cell }}^0$

Thus, for a spontaneous cell reaction,

E > 0 and $\Delta$ G < 0

Equilibrium Constant

If the circuit in Daniell cell is closed then we note that the reaction
$\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$
takes place and as time passes, the concentration of Zn²⁺ keeps on increasing while the concentration of Cu²⁺ keeps on decreasing. At the same time, the voltage of the cell as read on the voltmeter keeps on decreasing. After some time, we shall note that there is no change in the concentration of Cu²⁺ and Zn²⁺ ions and at the same time, voltmeter gives zero reading. This indicates that equilibrium has been attained. In this situation the Nernst equation may be written as:

$
\begin{aligned}
& \mathrm{E}_{(\text {cell })}=0=\mathrm{E}_{(\text {cell })}^{\ominus}-\frac{2.303 \mathrm{RT}}{2 \mathrm{~F}} \log \frac{\left[\mathrm{Zn}^2\right]}{\left[\mathrm{Cu}^{2+}\right]} \\
& \text { or } \mathrm{E}_{(\text {cell })}^0=\frac{2.303 \mathrm{RT}}{2 \mathrm{~F}} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}
\end{aligned}
$

But at equilibrium,
$\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}=\mathrm{K}_{\mathrm{c}}$ for the above reaction and at $T=298 \mathrm{~K}$ the above equation can be written as
$
\begin{aligned}
& \mathrm{E}_{\text {(cell) }}^{\ominus}=\frac{0.059 \mathrm{~V}}{2} \log \mathrm{K}_{\mathrm{C}}=1.1 \mathrm{~V} \quad\left(\mathrm{E}_{\text {cell }}^{\Theta}=1.1 \mathrm{~V}\right) \\
& \log \mathrm{K}_{\mathrm{C}}=\frac{(1.1 \mathrm{~V} \times 2)}{0.059 \mathrm{~V}}=37.288 \simeq 37.3 \\
& \mathrm{~K}_{\mathrm{C}}=2 \times 10^{37} \text { at } 298 \mathrm{~K}
\end{aligned}
$

$\begin{aligned} & \text { In general, } \\ & \mathrm{E}_{(\text {cell })}^{\ominus}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \log \mathrm{K}_{\mathrm{C}}\end{aligned}$

Alternatively

$\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln \mathrm{O}$

At equilibrium, $\Delta \mathrm{G}=0$ and $\mathrm{Q}=\mathrm{K}$, so.

$0=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \ln \mathrm{K}$

$\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{K}$

$\Delta \mathrm{E}^{\mathrm{o}}=\frac{2.303 \mathrm{RT}}{\mathrm{nF}} \ln \mathrm{K}$

$\ln \mathrm{K}=\frac{\mathrm{nE} \mathrm{E}^{\mathrm{o}}}{0.059}$

Thus, the above equation gives a relationship between the equilibrium constant of the reaction and the standard potential of the cell in which that reaction takes place. Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding E^o value of the cell.

Recommended topic video on (Gibbs Free Energy of reaction )

Some Solved Examples

Example.1

1. Calculate the standard cell potential (in V) of the cell in which following reaction takes place:

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

Given that

$\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0=\mathrm{xV} \\ & \mathrm{E}_{\mathrm{Fe}^2+\mid \mathrm{Fe}}^0=y \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}}^0=\mathrm{zV}\end{aligned}$

Solution

$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})$

Given,

$\begin{aligned} & \mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0=\mathrm{xV} \\ & \mathrm{E}_{\mathrm{Fe}^2+\mid \mathrm{Fe}}^0=y \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}}^0=\mathrm{zV}\end{aligned}$

The given equation can be represented in the form of a cell as

$\mathrm{Fe}^{2+}\left|\mathrm{Fe}^{+3} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$

Standard EMF of given cell reaction

$\mathrm{E}_{\mathrm{cell}}^0=\mathrm{E}_{\mathrm{Ag}^{+} \mid \mathrm{Ag}}^0-\mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}^2+}^0$

It is evident that in order to find the above cell potential, we need to find the electrode potential of the $\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}$ couple.

In order to calculate the value of $\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0$, we need to use the given values of $\mathrm{E}_{\mathrm{Fe}^2+\mid \mathrm{Fe}}^0$ and $\mathrm{E}_{\mathrm{Fe}^3+\mid \mathrm{Fe}}^0$.

Now,

$\mathrm{Fe}^{+2}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \quad \mathrm{E}^0=\mathrm{y}, \Delta \mathrm{G}^0=-2 \mathrm{Fy}$

$\mathrm{Fe}^{+3}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \mathrm{E}^0=\mathrm{z}, \Delta \mathrm{G}^0=-3 \mathrm{Fz}$

Now, subtracting (ii) from (iii), we have

$\mathrm{Fe}^{+3}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{+2} ; \Delta \mathrm{G}^0=-3 \mathrm{Fz}+2 \mathrm{Fy}$

Thus we can write

$-1 \times \mathrm{F} \times \mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=-3 \mathrm{Fz}+2 \mathrm{Fy}$

$\mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=3 \mathrm{z}-2 \mathrm{y}$

Putting this in equation $(\mathrm{i})$, we have

$E_{\text {cell }}^0=x-(3 z-2 y)=x+2 y-3 z$

Hence, the answer is the option (3).

Example.2

2. The standard Gibbs energy for the given cell reaction in $\mathrm{K} \mathrm{J} \mathrm{mol}^{-1}$ at 298 K is:

$\begin{aligned} & Z n(s)+C u^{2+}(a q) \rightarrow Z n^{2+}(a q)+C u(s) \\ & E^0=2 V \text { at } 298 K\end{aligned}$

(Faraday's constant, $\stackrel{F}{F}=96000 \mathrm{C} \mathrm{mol}^{-1}$)

1) (correct)-384

2)384

3)192

4)-192

Solution

$\begin{gathered}\Delta G^{\circ}=-n F E_{\text {cell }}^{\circ}=-2 \times 96500 \times 2.0 \\ \Delta G^{\circ}=-386 \times 10^3 \\ \Delta G^{\circ}=-386 \mathrm{Kj} / \mathrm{mol} \\ \Delta G^{\circ} \simeq-384 \mathrm{Kj} / \mathrm{mol}\end{gathered}$

Hence, the answer is the option (1).

Example.3

3. What is the value of the equilibrium constant for a reaction when $\Delta G^0=-15.38 k j ?$

1) (correct)$10^{2.695}$

2)$e^{2.695}$

3)$e^{1.738}$

4)$10^{1.738}$

Solution

Standard Gibbs Energy - $\Delta_r G^0=-R T \ln k$

K= equilibrium constant of the reaction

$\Delta G^0=-R T \ln K$

$-15.38=-8.314 \times 298 \times 2.303 \log K$

$K=10^{2.695}$

Hence, the answer is the option (1).

Example.4

4. What is the value of equilibrium constant for a reaction when $\Delta G^0=-21.28 k j ?$ ?

1)$e^{3.73}$

2) (correct)$10^{3.73}$

3)$10^{2.59}$

4)$e^{2.59}$

Solution

Standard Gibbs Energy -

$\Delta_r G^0=-R T \ln k$

K= equilibrium constant of the reaction

$\begin{aligned} & \Delta G^0=-R T \ln K \\ & -21.28 \times 10^3=-8.314 \times 298 \times 2.303 \log k \\ & K=10^{3.73}\end{aligned}$

Hence, the answer is the option (2).

Example.5

5. Given that the standard potentials (E^o) of Cu²⁺|Cu and Cu⁺|Cu are 0.34V and 0.522V respectively, the E^o of Cu²⁺|Cu⁺ is:

1)-0.182V

2) (correct)+0.158V

3)-0.158V

4)0.182V

Solution

Given,

$\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}, \mathrm{E}^0=0.34 \mathrm{~V}$

$\mathrm{Cu}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}, \mathrm{E}^0=0.522 \mathrm{~V}$

We need to find out the electrode potential of the reaction

$\mathrm{Cu}^{2+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}^{+}, \mathrm{E}^0=?$

Reaction (3) can be obtained by carrying out the operation (1) - (2)

Thus, we can write

$\Delta \mathrm{G}_2^0=\Delta \mathrm{G}_1^0-\Delta \mathrm{G}_n^0$

$\Rightarrow-\mathrm{FE}_3^0=-2 \mathrm{~F}(0.34)-(-\mathrm{F}(0.522))$

$\Rightarrow \mathrm{E}_3^0=2(0.34)-(0.522)$

$\Rightarrow \mathrm{E}_3^0=0.158 \mathrm{~V}$

Hence, the answer is the option(2).

Example.6

6. For the reaction $\mathrm{Mg}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{Ag}$

$E^0=3.17 \mathrm{~V}$ , find the value of $\Delta G^0$(in kJ).

1) (correct)-612

2)-603

3)-598

4)-591

Solution

Feasibility and Gibbs Free Energy of Reaction -

Let n faraday charge be taken out of a cell of emf E, then work done by the cell will be calculated as:
Work = Charge × Potential
Work done by the cell is equal to the decrease in the free energy.

$-\Delta \mathrm{G}=\mathrm{nFE}$

Similarly, the maximum obtainable work from the cell at standard conditions will be:

$\begin{aligned} & \mathrm{W}_{\max }=\mathrm{nFE}_{\text {cell }}^0 \quad \text { where } \mathrm{E}_{\text {cell }}^0=\text { standard emf of standard cell potential } \\ & -\Delta \mathrm{G}^{\circ}=\mathrm{nFE}_{\mathrm{cell}}^0 \\ & \Delta G^0=-n F E^0 \\ & =-2 \times 96487 \times 3.17=-611.7 k j \approx-612 k j\end{aligned}
Hence, the answer is the option (1).

Summary

Gibbs free energy is used to analyze phase transitions (like melting and boiling). The condition for phase equilibrium (e.g., between solid and liquid) is when the Gibbs free energy of the phases is equal. Gibbs's free energy also helps calculate other thermodynamic properties. For example, the relationship between ( Delta G ), enthalpy (Delta H ), and entropy (Delta S ) is given as [ Delta G = Delta H - T Delta S ].