Cell potential and free energy — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: The quantitative and qualitative relationships between standard cell potential ($E_{cell}^{\circ }$), Gibbs free energy change ($\Delta G$), and the equilibrium constant ($K$), including core formulas $\Delta G=-nF{E}_{cell}$ and interconversion of all three properties under standard and non-standard conditions.

You should already know: How to calculate standard cell potential from standard reduction potentials. The relationship between Gibbs free energy and reaction spontaneity. How to balance redox half-reactions and calculate reaction quotient Q.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Cell potential and free energy?

This topic connects two foundational AP Chemistry concepts: electrochemistry (cell potential, the measurable voltage an electrochemical cell produces) and thermodynamics (Gibbs free energy, the measure of usable energy that determines reaction spontaneity). Per the AP Chemistry CED, this topic makes up 7-9% of the total exam weight, as part of Unit 9: Applications of Thermodynamics. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically tests qualitative sign and magnitude relationships between $E^{\circ }$, $\Delta G^{\circ }$, and $K$, while FRQ requires numerical calculation and conceptual connection to spontaneity or reaction favorability. The core insight of this topic is that cell potential is a directly measurable quantity that lets us calculate thermodynamic properties of redox reactions that would otherwise be difficult to measure directly. It also unifies three separate frameworks for predicting reaction behavior: cell potential, Gibbs free energy, and equilibrium.

2. The Core Relationship Between Gibbs Free Energy and Cell Potential

The first key relationship links the spontaneity criteria from electrochemistry and thermodynamics. The derivation comes from the maximum electrical work a cell can do: $w_{max}=-\text{(total charge transferred)}\times E_{cell}$. Total charge transferred is the product of $n$ (moles of electrons transferred per mole of reaction) and $F$ (Faraday's constant, ~$96500\text{C/mol}{e}^{-}$, provided on the AP formula sheet). From thermodynamics, the maximum non-expansion work done by a system equals the change in Gibbs free energy: $\Delta G=w_{max}$. Substituting gives the core formula: $$\Delta G=-nF{E}_{cell}$$ For standard state conditions (1 M solutes, 1 atm gases, 298 K), this simplifies to the standard form: $$\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$$ The negative sign is critical to matching spontaneity rules: if a reaction is spontaneous (galvanic cell), $E_{cell}$ is positive, so $\Delta G$ becomes negative, which aligns with the thermodynamic rule for spontaneity. For non-spontaneous reactions (electrolytic cells requiring external voltage), $E_{cell}$ is negative, so $\Delta G$ is positive, which also matches. This unification means any spontaneous redox reaction must have a positive cell potential and negative Gibbs free energy change.

Worked Example

A galvanic cell runs on the balanced reaction $Mg(s)+P{b}^{2+}(aq)\to M{g}^{2+}(aq)+Pb(s)$, with $E_{cell}^{\circ }=2.23V$. Calculate $\Delta G^{\circ }$ for the reaction and confirm spontaneity.

1. Find $n$ by writing half-reactions: Oxidation: $Mg(s)\to M{g}^{2+}+2e^{-}$; Reduction: $P{b}^{2+}+2e^{-}\to Pb(s)$. So $n=2\text{mol}{e}^{-}$ per mole of reaction.
2. Use the standard formula $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$. Recall 1 V = 1 J/C, so units cancel correctly to give energy in joules.
3. Substitute values: $\Delta G^{\circ }=-(2)(96500C/mol)(2.23V)=-430190J/mol=-430kJ/mol$.
4. $\Delta G^{\circ }$ is negative, so the reaction is spontaneous under standard conditions, which matches the positive $E_{cell}^{\circ }$.

Exam tip: Always calculate $n$ from the fully balanced total reaction, not just a single half-reaction. If you multiply half-reactions to balance electrons, $n$ scales with the total reaction, so never leave $n$ as the value from the unmultiplied half-reaction.

3. Interconverting $E_{cell}^{\circ }$, $\Delta G^{\circ }$, and $K$

We already know from thermodynamics that standard Gibbs free energy is related to the equilibrium constant $K$ by: $$\Delta G^{\circ }=-RT\ln K$$ Since we also have $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$, we can set these equal and cancel the negative signs to get a direct relationship between standard cell potential and equilibrium constant: $$E_{cell}^{\circ }=\frac{RT}{nF}\ln K$$ At 298 K (standard temperature assumed on the AP exam unless stated otherwise), this simplifies to two common forms: $E^{\circ }=\frac{0.0257V}{n}\ln K$ (natural log) or $E^{\circ }=\frac{0.0592V}{n}\log K$ (base-10 log). This relationship lets us immediately predict the extent of a redox reaction from its standard cell potential:

• If $E^{\circ }>0$, $K>1$, products are favored at equilibrium
• If $E^{\circ }<0$, $K<1$, reactants are favored at equilibrium
• If $E^{\circ }=0$, $K=1$, products and reactants are equally favored

Worked Example

For the reaction $S{n}^{4+}(aq)+2F{e}^{2+}(aq)\to S{n}^{2+}(aq)+2F{e}^{3+}(aq)$, $E_{cell}^{\circ }=-0.62V$ at 298 K. Calculate the equilibrium constant $K$.

1. Balance half-reactions to get $n$: Oxidation: $2F{e}^{2+}\to 2F{e}^{3+}+2e^{-}$; Reduction: $S{n}^{4+}+2e^{-}\to S{n}^{2+}$. So $n=2$.
2. Use the base-10 log simplification for 298 K: $E^{\circ }=\frac{0.0592V}{n}\log K$. Rearrange to solve for $\log K$: $\log K=\frac{n{E}^{\circ }}{0.0592V}$.
3. Substitute values: $\log K=\frac{(2)(-0.62)}{0.0592}\approx -20.95$.
4. Calculate $K$: $K=10^{-20.95}\approx 1.1\times 10^{-21}$. This matches the negative $E^{\circ }$, as $K<1$ confirms reactants are favored.

Exam tip: The $0.0257V$ and $0.0592V$ simplifications only work at 298 K. If the exam gives a different temperature, use the full $E^{\circ }=\frac{RT}{nF}\ln K$ formula with the given temperature.

4. Non-Standard Conditions and the Nernst Equation

The relationships above apply only to standard state conditions. For real-world cells with non-standard concentrations (like fresh or discharged batteries), we use the Nernst equation, derived by combining $\Delta G=\Delta G^{\circ }+RT\ln Q$ (Q = reaction quotient) with $\Delta G=-nF{E}_{cell}$. Rearranging gives the general Nernst equation: $$E_{cell}=E_{cell}^{\circ }-\frac{RT}{nF}\ln Q$$ At 298 K, this simplifies to: $$E_{cell}=E_{cell}^{\circ }-\frac{0.0592V}{n}\log Q$$ The Nernst equation explains why batteries lose voltage over time: as the reaction proceeds, reactant concentration decreases and product concentration increases, so Q increases. This makes the subtracted term larger, so $E_{cell}$ decreases until $E_{cell}=0$ at equilibrium, when the battery can no longer do work. When $E_{cell}=0$, $Q=K$, which brings us back to the $E^{\circ }-K$ relationship from the previous section.

Worked Example

A Mg/Pb cell with $E^{\circ }=2.23V$ (reaction: $Mg(s)+P{b}^{2+}(aq)\to M{g}^{2+}(aq)+Pb(s)$) has non-standard concentrations $[P{b}^{2+}]=0.050M$, $[M{g}^{2+}]=1.50M$ at 298 K. Calculate $E_{cell}$.

1. $n=2$, and solids are excluded from Q, so $Q=\frac{[M{g}^{2+}]}{[P{b}^{2+}]}=\frac{1.50}{0.050}=30$.
2. Use the 298 K Nernst equation: $E_{cell}=E^{\circ }-\frac{0.0592}{n}\log Q$.
3. Calculate $\log 30\approx 1.477$, so $\frac{0.0592}{2}(1.477)\approx 0.044V$.
4. Substitute: $E_{cell}=2.23-0.044=2.19V$. This matches our expectation: higher product concentration than standard reduces cell potential.

Exam tip: Solids and pure liquids are never included in Q, just like in equilibrium calculations. Accidentally adding them will change your Q value and lead to an incorrect $E_{cell}$.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using $E_{cell}$ (non-standard) instead of $E_{cell}^{\circ }$ when calculating $K$. Why: Students confuse the properties of equilibrium, which is only related to standard state Gibbs free energy and standard cell potential. Correct move: Always explicitly confirm you are using the standard cell potential $E^{\circ }$ when solving for $K$, and reserve $E_{cell}$ for non-standard concentration problems.
• Wrong move: Mixing up natural log and base-10 log constants, using 0.0257 V with log or 0.0592 V with ln. Why: Two common simplified forms are taught, and students rarely check which constant pairs with which log. Correct move: Write the pairing explicitly before plugging in values: $0.0257V$ goes with $\ln K$, $0.0592V$ goes with $\log K$.
• Wrong move: Getting the sign of $\Delta G$ wrong for a positive $E_{cell}$. Why: Students forget the negative sign in $\Delta G=-nF{E}_{cell}$, ending up with a positive $\Delta G$ for a spontaneous reaction. Correct move: After any calculation, check sign consistency: positive $E$ = negative $\Delta G$ = $K>1$. If your signs don't match this rule, correct the negative sign error.
• Wrong move: Using $\Delta G$ in kJ/mol with $R=8.314J/(mol\cdot K)$ without unit conversion. Why: Most calculations report $\Delta G$ in kJ, but R is almost always given in J, leading to a 3-order-of-magnitude error. Correct move: Convert $\Delta G$ to J before calculating $\ln K$, or convert R to $0.008314kJ/(mol\cdot K)$ to match kJ units for $\Delta G$.
• Wrong move: Using the 0.0592 V Nernst simplification at a temperature other than 298 K. Why: Students memorize the simplified form and forget it is temperature-dependent. Correct move: If the problem gives a temperature other than 25°C (298 K), use the full general Nernst equation with $R$, $T$, and $F$.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

For a spontaneous redox reaction under standard conditions, which set of properties is always correct? A) $E_{cell}^{\circ }>0$, $\Delta G^{\circ }>0$, $K>1$ B) $E_{cell}^{\circ }>0$, $\Delta G^{\circ }<0$, $K>1$ C) $E_{cell}^{\circ }>0$, $\Delta G^{\circ }<0$, $K<1$ D) $E_{cell}^{\circ }<0$, $\Delta G^{\circ }>0$, $K<1$

Worked Solution: We use the consistent sign rules for spontaneity under standard conditions. A spontaneous reaction has a positive $E_{cell}^{\circ }$ by electrochemistry rules. Using $\Delta G^{\circ }=-nF{E}^{\circ }$, a positive $E^{\circ }$ gives a negative $\Delta G^{\circ }$. Using $E^{\circ }=\frac{RT}{nF}\ln K$, a positive $E^{\circ }$ gives a positive $\ln K$, so $K>1$. All three conditions match option B. Correct answer: B.

Question 2 (Free Response)

Given these two standard reduction potentials at 298 K: $C{r}^{3+}(aq)+3e^{-}\to Cr(s)$ $E^{\circ }=-0.74V$ $F{e}^{2+}(aq)+2e^{-}\to Fe(s)$ $E^{\circ }=-0.44V$ (a) Write the balanced overall reaction for a galvanic cell constructed from these half-reactions, and calculate $E_{cell}^{\circ }$. (b) Calculate $\Delta G^{\circ }$ for the reaction, in kJ/mol. (c) Calculate the equilibrium constant $K$ for the reaction at 298 K.

Worked Solution: (a) For a galvanic cell, $E_{cell}^{\circ }$ must be positive, so we reverse the half-reaction with lower $E^{\circ }$ to make it oxidation. Oxidation: $2Cr(s)\to 2C{r}^{3+}(aq)+6e^{-}$. Reduction: $3F{e}^{2+}(aq)+6e^{-}\to 3Fe(s)$. Balanced overall: $2Cr(s)+3F{e}^{2+}(aq)\to 2C{r}^{3+}(aq)+3Fe(s)$. $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=-0.44V-(-0.74V)=0.30V$. (b) $n=6$ mol e⁻ per mole of reaction. $\Delta G^{\circ }=-nF{E}^{\circ }=-(6)(96500C/mol)(0.30V)=-173700J/mol=-174kJ/mol$. (c) Use $E^{\circ }=\frac{0.0592V}{n}\log K$, so $\log K=\frac{n{E}^{\circ }}{0.0592V}=\frac{(6)(0.30)}{0.0592}\approx 30.4$. $K=10^{30.4}\approx 2.5\times 10^{30}$.

Question 3 (Application / Real-World Style)

A lead-acid car battery has the overall reaction: $Pb(s)+Pb{O}_2(s)+2H_{2}S{O}_4(aq)\to 2PbS{O}_4(s)+2H_{2}O(l)$. The standard cell potential $E^{\circ }=2.04V$ at 298 K, and $n=2$ mol e⁻ per mole of reaction. A fully charged car battery has an $H_{2}S{O}_4$ concentration of 5.0 M, giving a reaction quotient $Q=0.0080$. Calculate the actual cell potential of the fully charged battery at 298 K, and explain why it differs from the standard potential.

Worked Solution: Use the Nernst equation for 298 K: $E_{cell}=E^{\circ }-\frac{0.0592V}{n}\log Q$. Substitute values: $\log Q=\log (0.0080)\approx -2.10$. $E_{cell}=2.04-\frac{0.0592}{2}(-2.10)=2.04+0.062=2.10V$. Interpretation: The fully charged battery has a higher concentration of $H_{2}S{O}_4$ (reactant) than the standard 1 M, so Q is less than 1, leading to a higher cell potential than the standard value.

7. Quick Reference Cheatsheet

8. What's Next

This chapter unifies thermodynamics, equilibrium, and electrochemistry, which is a common cross-unit topic tested heavily on the AP exam. Next, you will apply these relationships to concentration cells, electrolysis, and Faraday's laws of electrolytic production, where you will calculate the mass of product formed and the energy required for non-spontaneous redox reactions. Without mastering the sign rules and interconversion of $E^{\circ }$, $\Delta G^{\circ }$, and $K$, you will not be able to solve electrolysis problems or predict the behavior of real-world electrochemical systems. This topic also builds on the fundamental free energy and equilibrium relationships you learned earlier in Unit 9, and it will be referenced in any future problem involving redox reactions or cell design.

• Free energy and thermodynamic favorability
• Electrolysis and Faraday's laws
• Galvanic and electrolytic cell design
• Free energy and equilibrium