Cell Potential, Gibbs Energy & Equilibrium Constant

From Cell Potential to Thermodynamics

A galvanic cell does electrical work. The maximum electrical work a cell can do is directly linked to the Gibbs energy change of its reaction. This is one of the most beautiful connections in chemistry — it ties the electrochemistry we've built to the thermodynamics of spontaneity.

The relation is:

$$\boxed{\Delta_r G^\circ = -nFE^\circ_{cell}}$$

• $\Delta_r G^\circ$ = standard Gibbs energy change of the cell reaction (in J mol$^{-1}$).
• $n$ = moles of electrons transferred.
• $F$ = Faraday constant = 96487 C mol$^{-1}$ (≈ 96500).
• $E^\circ_{cell}$ = standard cell potential (V).

The minus sign captures the logic: a positive $E^\circ_{cell}$ (spontaneous cell) gives a negative $\Delta_r G^\circ$ (spontaneous reaction). The two ways of saying "spontaneous" agree.

Maximum work: the maximum electrical work obtainable is $w_{max} = -\Delta_r G^\circ = nFE^\circ_{cell}$.

Linking Potential to the Equilibrium Constant

Combining $\Delta_r G^\circ = -nFE^\circ_{cell}$ with the thermodynamic relation $\Delta_r G^\circ = -RT\ln K_c$ gives a direct bridge between the standard cell potential and the equilibrium constant:

$$nFE^\circ_{cell} = RT\ln K_c \quad\Rightarrow\quad E^\circ_{cell} = \frac{RT}{nF}\ln K_c$$

At 298 K, using the 0.059 shortcut:

$$\boxed{E^\circ_{cell} = \frac{0.0591}{n}\log K_c}$$

This makes intuitive sense via the Nernst equation: at equilibrium $E_{cell}=0$ and $Q=K_c$, so $0=E^\circ_{cell}-\dfrac{0.0591}{n}\log K_c$, which rearranges to exactly this.

Reading the Signs

The three quantities $E^\circ_{cell}$, $\Delta_r G^\circ$ and $K_c$ are three views of the same thing — how far and which way a reaction goes:

$E^\circ_{cell}$	$\Delta_r G^\circ$	$K_c$	Reaction
Positive	Negative	> 1	Spontaneous (forward)
Zero	Zero	= 1	At equilibrium
Negative	Positive	< 1	Non-spontaneous (reverse favoured)

[JEE Tip] Watch your units. With $F = 96487$ C mol$^{-1}$ and $E^\circ$ in volts, $\Delta_r G^\circ$ comes out in joules per mole. Divide by 1000 for kJ mol$^{-1}$. A frequent error is mixing kJ and J when comparing with $\Delta G = -RT\ln K$.

Key Point: $\Delta_r G^\circ = -nFE^\circ_{cell}$ converts a voltmeter reading into a thermodynamic spontaneity statement. It is the most important single equation connecting the two halves of physical chemistry.

Solved Examples

Example 1: Gibbs energy from cell potential

Calculate $\Delta_r G^\circ$ for the Daniell cell ($E^\circ_{cell}=1.10$ V, $n=2$). ($F=96487$ C mol$^{-1}$)

Solution: $\Delta_r G^\circ = -nFE^\circ_{cell} = -2\times96487\times1.10 = -212{,}271$ J $\approx -212.3$ kJ mol$^{-1}$.

Answer: $-212.3$ kJ mol$^{-1}$ (negative → spontaneous).

Example 2: Equilibrium constant from E°

For a cell with $E^\circ_{cell}=0.295$ V and $n=2$, find $\log K_c$ and $K_c$.

Solution: $E^\circ_{cell}=\dfrac{0.0591}{n}\log K_c \Rightarrow \log K_c = \dfrac{nE^\circ_{cell}}{0.0591}=\dfrac{2\times0.295}{0.0591}\approx 9.98 \approx 10$. So $K_c \approx 10^{10}$.

Answer: $\log K_c \approx 10$, $K_c \approx 1\times10^{10}$.

Example 3: Maximum work

What is the maximum electrical work obtainable from the Daniell cell per mole of reaction? ($E^\circ=1.10$ V, $n=2$)

Solution: $w_{max}=nFE^\circ_{cell}=2\times96487\times1.10=212{,}271$ J $\approx 212.3$ kJ mol$^{-1}$.

Answer: $212.3$ kJ mol$^{-1}$ of work per mole.

Example 4: Sign check

A cell reaction has $E^\circ_{cell}=-0.25$ V. Is it spontaneous, and what is the sign of $\Delta_r G^\circ$?

Solution: $\Delta_r G^\circ=-nFE^\circ_{cell}$. With $E^\circ_{cell}$ negative, $\Delta_r G^\circ$ is positive → the reaction is non-spontaneous as written (the reverse is favoured).

Example 5: Kc for the Daniell cell

Estimate $K_c$ for the Daniell cell ($E^\circ_{cell}=1.10$ V, $n=2$).

Solution: $\log K_c=\dfrac{nE^\circ}{0.0591}=\dfrac{2\times1.10}{0.0591}\approx 37.2$. So $K_c\approx 10^{37.2}\approx 2\times10^{37}$.

Answer: $K_c\approx 10^{37}$ — enormous, confirming the reaction goes essentially to completion.

Example 6: ΔG° from electrode potentials

For Mg + 2Ag$^+$ → Mg$^{2+}$ + 2Ag, $E^\circ_{cell}=3.16$ V, $n=2$. Find $\Delta_r G^\circ$.

Solution: $\Delta_r G^\circ=-nFE^\circ=-2\times96487\times3.16=-609{,}880$ J $\approx -609.9$ kJ mol$^{-1}$.

Answer: $-609.9$ kJ mol$^{-1}$.

Example 7: Finding E° from ΔG°

A reaction transferring 2 electrons has $\Delta_r G^\circ=-193$ kJ mol$^{-1}$. Find $E^\circ_{cell}$. ($F=96500$)

Solution: $E^\circ_{cell}=-\dfrac{\Delta_r G^\circ}{nF}=-\dfrac{-193000}{2\times96500}=1.0$ V.

Answer: $1.0$ V.

Example 8: Relating Kc and spontaneity

If $K_c=1$ for a cell reaction, what is $E^\circ_{cell}$?

Solution: $E^\circ_{cell}=\dfrac{0.0591}{n}\log K_c=\dfrac{0.0591}{n}\log 1=0$ V. With $K_c=1$ the system is balanced, so $E^\circ_{cell}=0$ and $\Delta_r G^\circ=0$.

Example 9: Two-electron cell Kc

A cell has $E^\circ_{cell}=0.118$ V and $n=2$. Find $K_c$.

Solution: $\log K_c=\dfrac{2\times0.118}{0.0591}\approx 3.99\approx 4.0$. $K_c\approx 10^4$.

Answer: $K_c\approx 1\times10^4$.

Example 10: ΔG° in kJ from a 1-electron cell

For a cell with $E^\circ_{cell}=0.80$ V and $n=1$, find $\Delta_r G^\circ$ in kJ. ($F=96500$)

Solution: $\Delta_r G^\circ=-1\times96500\times0.80=-77{,}200$ J $=-77.2$ kJ.

Answer: $-77.2$ kJ mol$^{-1}$.