9.9 Cell Potential and Free Energy

Standard cell potential, $E^\circ_{\text{cell}}$, tells you whether a redox reaction in an electrochemical cell is thermodynamically favored. A positive $E^\circ_{\text{cell}}$ means the reaction is favored and $\Delta G^\circ$ is negative, while a negative $E^\circ_{\text{cell}}$ means the reaction is not favored and $\Delta G^\circ$ is positive. For AP Chemistry, connect the sign of cell potential to the sign of free energy before judging spontaneity.

Why This Matters for the AP Chemistry Exam

This topic ties electrochemistry back to everything you learned about thermodynamic favorability. On the AP Chemistry exam, you may need to calculate a standard cell potential from half-reactions, decide whether a cell is galvanic or electrolytic based on the sign of E°, and link that result to the sign of ΔG°. The exam usually gives you the reduction potentials you need, so the real work is selecting the correct half-reactions, identifying the anode and cathode, and following a clean computational path with correct units and significant figures.

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Key Takeaways

• E°cell = E°cathode - E°anode, where both values come from standard reduction potentials.
• Oxidation happens at the anode and reduction happens at the cathode (An Ox, Red Cat).
• A positive E°cell means the reaction is thermodynamically favored; a negative E°cell means it is not.
• ΔG° = -nFE°cell connects cell potential to free energy, where n is moles of electrons transferred and F is 96,485 C/mol e-.
• A positive E° gives a negative ΔG° (favored); a negative E° gives a positive ΔG° (not favored).
• The standard reduction potential of 2H+ + 2e- → H2 is defined as 0 V and serves as the reference for all other values.

What is Cell Potential?

Cell potential is the "pull" on electrons that drives a redox reaction in an electrochemical cell. The stronger that pull, the more thermodynamically favored the reaction. You measure cell potential in volts, and it is sometimes called the electromotive force.

One way to picture it: each chemical species has some tendency to either give up electrons or gain electrons. The species that wants electrons more strongly pulls them through the circuit. That tendency is what cell potential captures.

In a galvanic (voltaic) cell, oxidation and reduction happen on separate sides:

• The anode is where oxidation occurs.
• The cathode is where reduction occurs.
• Remember: An Ox, Red Cat.

Electrons travel through a wire from the anode to the cathode. The force on those electrons as they move is the cell potential.

Standard conditions mean 1 M concentrations, a temperature of 298 K, and a pressure of 1 atm. Under these conditions you calculate standard cell potential with:

E°cell = E°cathode - E°anode

In words, the standard cell potential is the reduction potential of the species at the cathode minus the reduction potential of the species at the anode.

Worked Example

Calculate the cell potential for the following cell given the electrochemical data below:

2AgBr + 2Hg → 2Ag + Hg2Br2

Hg2Br2 + 2e- → 2Hg + 2Br- (E = +0.140 V)

2AgBr + 2e- → 2Ag + 2Br- (E = +0.071 V)

Here AgBr is reduced to Ag and Br-, so AgBr is at the cathode. Hg is oxidized, so it is at the anode.

Plugging into the equation:

E°cell = E°cathode - E°anode = 0.071 V - 0.140 V = -0.069 V

Another approach is to flip the oxidation half-reaction and add the potentials. The actual oxidation half-reaction for Hg is 2Hg + 2Br- → Hg2Br2 + 2e-, which has a potential of -0.140 V after flipping the sign. Adding:

E°cell = +0.071 V + (-0.140 V) = -0.069 V

Same result either way.

Calculating Cell Potential Using Reduction Potentials

In that problem you were handed the specific data. For other questions you reference a table of standard reduction potentials, which is a long list of reduction reactions paired with their potentials.

To get the right value, you need the correct two half-reactions and you have to decide which one occurs at the anode and which at the cathode. Once you know that, apply E°cell = E°cathode - E°anode.

The AP Chemistry exam provides the data you need to solve for cell potential, so you do not have to memorize the table. Still, getting comfortable with how a reduction potential table is organized builds the conceptual understanding that helps on test day.

A few things to know about these tables:

• The reduction of H+ into H2 is a defined value of 0 V. Every other reduction potential is measured relative to this one. You can read each value as answering: is this reaction more or less favored than 2H+ + 2e- → H2?
• All values in the table are written as reductions, so if you reverse a reaction to make it an oxidation, you flip the sign of the potential.
• A negative reduction potential means that species (like Na or Li) is hard to reduce but is an excellent reducing agent. When you flip these to oxidations, the negative potential becomes positive, showing that oxidation is favored for them.

Standard Cell Potentials and Thermodynamic Favorability

The main conclusion you want from a cell potential calculation is whether the redox reaction is thermodynamically favored. If E°cell is positive, the reaction is favored: the electromotive force is strong enough to pull electrons off the reducing agent and onto the oxidizing agent. If E°cell is negative, the reaction is not favored and needs an externally applied potential to proceed.

This directly predicts the sign of ΔG°:

• If E°cell > 0, the reaction is thermodynamically favored and ΔG° is negative.
• If E°cell < 0, the reaction is not favored and ΔG° is positive.

This is also the difference between a galvanic cell (positive E°, favored reaction) and an electrolytic cell (negative E°, unfavored reaction that requires energy input).

Calculating ΔG° Using E°cell

You can also calculate ΔG° directly from cell potential:

ΔG° = -nFE°cell

In this equation:

• ΔG° is the standard Gibbs free energy change.
• E°cell is the standard cell potential.
• n is the number of moles of electrons transferred in the reaction.
• F is Faraday's constant, 96,485 coulombs per mole of electrons. A coulomb is a unit of electric charge.

Worked example:

Suppose we have a galvanic cell for which E°cell = 1.02 V. In the reaction, 1 mol of electrons transfers. At 298 K, what is ΔG° for this cell? What is the equilibrium constant, K?

First, find ΔG° with ΔG° = -nFE°cell:

ΔG° = -(1)(96,485)(1.02) = -98,414.7 J = -98.414 kJ

Next, use the relationship between ΔG° and K:

K = e^(-ΔG°/RT)

K = e^(98,414.7 / ((8.314)(298))) = 1.78 × 10^17

The large positive K confirms what the negative ΔG° and positive E°cell already told you: products are strongly favored at equilibrium.

How to Use This on the AP Chemistry Exam

Problem Solving

• Identify the reduction and oxidation half-reactions first, then assign cathode and anode.
• Apply E°cell = E°cathode - E°anode using the reduction potentials as given. Do not flip the sign of the cathode value.
• When you reverse a half-reaction to write it as an oxidation, flip the sign of its potential. Do not flip the sign for stoichiometric scaling.
• For ΔG° = -nFE°cell, count n carefully as moles of electrons transferred, and watch your units. Charge in coulombs times volts gives joules, so convert to kJ when needed.

Free Response

• State a clear claim about whether the cell is favored and back it up with the sign of E° or ΔG°.
• Connect E°, ΔG°, and K together: a positive E° means a negative ΔG° and K greater than 1.
• When asked to justify, reason from the chemistry (which species is more easily reduced) rather than just plugging numbers.

Common Trap

• E°cell is an intensity property and does not change when you multiply the balanced equation by a coefficient. The value of n in ΔG° = -nFE°cell does change, which changes ΔG°.

Common Misconceptions

• E°cell scales with stoichiometry. It does not. If you double all the coefficients in the overall reaction, E°cell stays the same, but n doubles, so ΔG° doubles.
• You always flip the sign of the cathode potential. Use the cathode reduction potential as written in E°cell = E°cathode - E°anode. The subtraction already handles the anode reversal.
• A negative reduction potential means a species is easy to reduce. It is the opposite. A very negative reduction potential means the species resists reduction and is a strong reducing agent in its reduced form.
• Nonspontaneous means nothing happens. A negative E° reaction can still be driven forward with an externally applied potential, which is exactly what an electrolytic cell does.
• A favored reaction (positive E°) must be fast. Thermodynamic favorability says nothing about rate. A reaction can be favored yet slow if it is under kinetic control.

Related AP Chemistry Guides

• 9.1 Introduction to Entropy
• 9.3 Gibbs Free Energy and Thermodynamic Favorability
• 9.2 Absolute Entropy and Entropy Change
• 9.8 Galvanic (Voltaic) and Electrolytic Cells
• 9.4 Thermodynamic and Kinetic Control
• Unit 9 Review: Thermodynamics and Electrochemistry