# Cell Potential and Free Energy

> AP Chemistry · Unit 9: Applications of Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u9-cell-potential-and-free-energy/

This subtopic connects electrochemistry to core thermodynamics, explaining how standard cell potential relates to Gibbs free energy change and reaction spontaneity. You will learn to calculate these values and predict reaction favorability.

**Prerequisites:** [Standard reduction potentials for half-cells](https://www.owlsprep.com/study/ap-chemistry-u9-standard-reduction-potentials/); [Gibbs free energy and reaction spontaneity](https://www.owlsprep.com/study/ap-chemistry-u8-gibbs-free-energy/); [Equilibrium constant basics](https://www.owlsprep.com/study/ap-chemistry-u7-equilibrium-constants/)

## Learning objectives

- Relate standard cell potential to Gibbs free energy change for redox reactions
- Calculate ΔG° from E°cell and vice versa
- Predict reaction spontaneity using E°cell, ΔG°, and equilibrium constant K
- Derive and use the relationship between E°cell and K

## The Core Relationship Between E°cell and ΔG°

**ΔG° and E°cell Relationship** — The standard Gibbs free energy change of a redox reaction equals the negative product of moles of electrons transferred ($n$), Faraday's constant ($F$), and standard cell potential ($E^\circ_{\text{cell}}$).

*Notation:* $\Delta G^\circ = -nFE^\circ_{\text{cell}}$

*Example:* $E^\circ_{\text{cell}} = +1.10$ V gives $\Delta G^\circ < 0$, so the reaction is spontaneous.

Each variable has a clear meaning: $n$ is the total number of moles of electrons transferred in the balanced full reaction, always a positive integer. Faraday's constant is approximated as $9.65 \times 10^4$ C/mol for AP Chemistry calculations. The negative sign creates the consistent spontaneity rule: positive $E^\circ_{\text{cell}}$ gives negative $\Delta G^\circ$ (spontaneous), while negative $E^\circ_{\text{cell}}$ gives positive $\Delta G^\circ$ (non-spontaneous).

**Worked example:** Calculate the standard Gibbs free energy change for the reaction: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$. Given $E^\circ_{\text{red}}(\text{Zn}^{2+}/\text{Zn}) = -0.76$ V and $E^\circ_{\text{red}}(\text{Cu}^{2+}/\text{Cu}) = +0.34$ V.

1. First calculate $E^\circ_{\text{cell}}$. Zn is oxidized (anode), Cu is reduced (cathode):
2. $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10 \ \text{V}$$
3. Identify $n$: 2 moles of electrons are transferred in the balanced reaction, so $n=2$.
4. Substitute into the core equation, noting that 1 V·C = 1 J:
5. $$\Delta G^\circ = -(2)(96500 \ \text{C/mol})(1.10 \ \text{V}) = -212300 \ \text{J/mol}$$
6. Convert to kJ and confirm spontaneity:
7. $$\Delta G^\circ = -212 \ \text{kJ/mol}$$
8. $\Delta G^\circ$ is negative, so the reaction is spontaneous under standard conditions.

## Relating E°cell to the Equilibrium Constant K

We can combine the relationship $\Delta G^\circ = -RT \ln K$ with $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ to get a direct connection between $E^\circ_{\text{cell}}$ and the equilibrium constant $K$. This lets us predict how far a reaction will go at equilibrium directly from cell potential data.

**E°cell and K Relationship** — The simplified base-10 log form is used almost exclusively for AP Chemistry calculations at 25°C, the standard temperature for most exam problems.

*Notation:* $E^\circ_{\text{cell}} = \frac{0.0592 \ \text{V}}{n} \log K \ (\text{at } 25^\circ \text{C})$

The same spontaneity rules apply here: $E^\circ_{\text{cell}} > 0$ means $\log K > 0$, so $K > 1$, meaning products are favored at equilibrium. $E^\circ_{\text{cell}} < 0$ means $K < 1$, so reactants are favored.

**Worked example:** Calculate $K$ for the reaction $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$ at 25°C, given $E^\circ_{\text{cell}} = 1.10$ V.

1. Rearrange the simplified equation to solve for $\log K$:
2. $$\log K = \frac{n E^\circ_{\text{cell}}}{0.0592 \ \text{V}}$$
3. Substitute $n=2$ and $E^\circ_{\text{cell}}=1.10$ V:
4. $$\log K = \frac{(2)(1.10)}{0.0592} \approx 37.16$$
5. Solve for $K$ by taking the antilog:
6. $$K = 10^{37.16} \approx 1.4 \times 10^{37}$$
7. Interpret: $K \gg 1$, so the reaction goes almost to completion, consistent with the negative $\Delta G^\circ$ calculated earlier.

> **warning**
>
> This relationship only applies to standard state conditions. Non-standard conditions require the Nernst equation, covered in a separate subtopic.

## Summarizing Spontaneity Rules

All three thermodynamic quantities ($E^\circ_{\text{cell}}$, $\Delta G^\circ$, $K$) give consistent information about the spontaneity of a redox reaction under standard conditions. The table below summarizes all sign relationships:

| $E^\circ_{\text{cell}}$ | $\Delta G^\circ$ | $K$ | Spontaneity (standard conditions) |
| --- | --- | --- | --- |
| $> 0$ | $< 0$ | $> 1$ | Spontaneous |
| $= 0$ | $= 0$ | $= 1$ | At equilibrium |
| $< 0$ | $> 0$ | $< 1$ | Non-spontaneous |

**Check your understanding**

Test your understanding of the relationships:

1. A redox reaction has $K = 2.5 \times 10^{-3}$. What is true of $E^\circ_{\text{cell}}$ and $\Delta G^\circ$?

   - $E^\circ$ positive, $\Delta G^\circ$ positive
   - $E^\circ$ negative, $\Delta G^\circ$ positive
   - $E^\circ$ positive, $\Delta G^\circ$ negative
   - $E^\circ$ negative, $\Delta G^\circ$ negative

   *Why:* Correct! $K < 1$ means the reaction favors reactants, so it is non-spontaneous under standard conditions, requiring $E^\circ < 0$ and $\Delta G^\circ > 0$.

2. What is $\Delta G^\circ$ for a reaction with $E^\circ_{\text{cell}} = 0$ V?

   - $\Delta G^\circ > 0$
   - $\Delta G^\circ < 0$
   - $\Delta G^\circ = 0$
   - Cannot be determined without $n$

   *Why:* Correct! $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, so any value of $n$ and $F$ multiplied by $E^\circ_{\text{cell}} = 0$ gives $\Delta G^\circ = 0$.

## Common pitfalls

- **Wrong:** Forgetting the negative sign in $\Delta G^\circ = -nFE^\circ_{\text{cell}}$, leading to reversed spontaneity
  - Why it fails: The negative sign is required to match the sign conventions for $\Delta G$ (negative = spontaneous) and $E^\circ_{\text{cell}}$ (positive = spontaneous). Without it, all signs are flipped.
  - Correct: Always remember the negative sign: positive $E^\circ_{\text{cell}}$ → negative $\Delta G^\circ$ → spontaneous reaction.
- **Wrong:** Using $n$ equal to the electrons from only one half-reaction instead of the total balanced reaction
  - Why it fails: $n$ is the total moles of electrons transferred in the full balanced reaction, which must cancel out between oxidation and reduction half-reactions.
  - Correct: Balance the full redox reaction before identifying $n$ to get the correct total number of electrons transferred.
- **Wrong:** Using the $0.0592$ V simplified equation for temperatures other than 25°C
  - Why it fails: The simplified log form is derived assuming 298 K (25°C), so it is not valid at other temperatures.
  - Correct: Use the full form $E^\circ_{\text{cell}} = \frac{RT}{nF} \ln K$ if the reaction temperature is not 25°C.
- **Wrong:** Assuming a non-spontaneous reaction ($E^\circ_{\text{cell}}$ negative) can never occur
  - Why it fails: $E^\circ_{\text{cell}}$ only describes spontaneity under standard conditions. Non-standard conditions can make the reaction spontaneous.
  - Correct: Remember that $E^\circ_{\text{cell}}$ only tells you about spontaneity under standard state conditions, not all possible conditions.

## Cheatsheet

| Relationship | Formula (25°C) | Spontaneity Rule |
| --- | --- | --- |
| $\Delta G^\circ$ from $E^\circ_{\text{cell}}$ | $\Delta G^\circ = -nFE^\circ_{\text{cell}}$ | $E^\circ > 0 \to \Delta G^\circ < 0 \to$ Spontaneous |
| $K$ from $E^\circ_{\text{cell}}$ | $\log K = \frac{n E^\circ_{\text{cell}}}{0.0592 \text{ V}}$ | $E^\circ > 0 \to K > 1 \to$ Favors products |
| All Sign Rules | N/A | $E^\circ > 0 = \Delta G^\circ < 0 = K > 1 \to$ Spontaneous |
|  |  | $E^\circ < 0 = \Delta G^\circ > 0 = K < 1 \to$ Non-spontaneous |

## What's next

Now that you understand how cell potential relates to Gibbs free energy and equilibrium, you can extend this knowledge to non-standard conditions, where cell potential changes as reaction concentrations shift. This relationship is described by the Nernst equation, which is critical for solving problems involving concentration cells and batteries that are not at standard state. You can also apply these concepts to electrolysis, where non-spontaneous redox reactions are driven by an external voltage, a common topic in AP Chemistry FRQs. These concepts build directly on the relationships you learned here, connecting all of thermodynamics and electrochemistry into a unified framework.

- [Cell potential under nonstandard conditions](https://www.owlsprep.com/study/ap-chemistry-u9-cell-potential-under-nonstandard-conditions/)
- [Electrolysis and Faraday’s Law](https://www.owlsprep.com/study/ap-chemistry-u9-electrolysis-and-faraday-s-law/)

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