A cell has E^_cathode = +0.34 V and E^_anode = -0.13 V. Calculate E^_cell and state whether it is galvanic. [2 points]

A cell reaction transfers n = 1 mol of electrons with E^ = +0.50 V. Calculate G^. [2 points]

United StatesChemistrySyllabus dot point

How is the standard cell potential calculated, and how does it relate to the free energy change of a cell reaction?

Topic 9.9 Cell Potential and Free Energy: calculate the standard cell potential from standard reduction potentials, and relate it to the free energy change with delta G standard equals minus n F E standard.

A focused answer to AP Chemistry Topic 9.9, covering the standard cell potential from standard reduction potentials, the sign of the cell potential and spontaneity, and the relationship delta G standard equals minus n F E standard, with full worked examples.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The standard cell potential measures the driving force of a cell's redox reaction and is found from standard reduction potentials: E standard cell equals E standard cathode minus E standard anode, where the cathode is the half-reaction that is reduced and the anode the one that is oxidized. A positive cell potential means the cell reaction is spontaneous (favorable), which is the case for a galvanic cell; a negative cell potential means it is non-spontaneous and would require an electrolytic cell. The cell potential is linked to the free energy change by delta G standard equals minus n F E standard, where n is the moles of electrons transferred and F is Faraday's constant (96500 coulombs per mole). Because of the minus sign, a positive cell potential gives a negative free energy change, confirming spontaneity. Standard reduction potentials are intensive, so they are not multiplied when half-reactions are scaled.

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What this topic is asking
Calculating the standard cell potential
Cell potential and spontaneity
Relating cell potential to free energy
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What this topic is asking

The College Board (Topic 9.9) wants you to calculate the standard cell potential from standard reduction potentials, and to relate it to the free energy change with $\Delta G^\circ = -nFE^\circ$ . This ties the electrochemistry of Topic 9.8 to the thermodynamics of the whole unit: a positive cell potential means a spontaneous reaction.

Calculating the standard cell potential

To use the formula, identify which species is reduced (the cathode, with the more positive reduction potential in a spontaneous cell) and which is oxidized (the anode), then subtract the anode's reduction potential from the cathode's. Standard reduction potentials are intensive: they do not change when you scale a half-reaction, so never multiply a potential by a coefficient.

Cell potential and spontaneity

So a galvanic cell always has a positive cell potential, and an electrolytic cell is run on a reaction with a negative cell potential. This parallels the sign of $\Delta G$ : positive potential and negative free energy both signal a spontaneous reaction, as the next relationship makes precise.

Relating cell potential to free energy

The cell potential and the free energy change are connected by

\Delta G^\circ = -nFE^\circ

where $n$ is the number of moles of electrons transferred in the balanced cell reaction and $F = 96500\ \text{C mol}^{-1}$ is Faraday's constant. The minus sign makes a positive $E^\circ$ give a negative $\Delta G^\circ$ , so a spontaneous cell (positive potential) has a favorable free energy change. This equation also links electrochemistry to equilibrium: combined with $\Delta G^\circ = -RT\ln K$ , it relates $E^\circ$ to $K$ .

Cell potential and free energy

A cell uses $\text{Ag}^+ + e^- \rightarrow \text{Ag}$ ( $E^\circ = +0.80$ V) at the cathode and $\text{Ni}^{2+} + 2e^- \rightarrow \text{Ni}$ ( $E^\circ = -0.25$ V) with nickel oxidized. Find $E^\circ_\text{cell}$ and $\Delta G^\circ$ ( $n = 2$ ).

step 1 Identify cathode and anode

Silver is reduced (cathode, $E^\circ = +0.80$ ); nickel is oxidized (anode, $E^\circ = -0.25$ ).

step 2 Calculate the cell potential

$E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} = 0.80 - (-0.25) = +1.05$ V. (Note the silver potential is not doubled despite scaling its half-reaction.)

step 3 Confirm spontaneity

$E^\circ_\text{cell} > 0$ , so the cell reaction is spontaneous (galvanic).

step 4 Calculate the free energy

$\Delta G^\circ = -nFE^\circ = -(2)(96500)(1.05) = -2.03 \times 10^{5}\ \text{J mol}^{-1} = -203\ \text{kJ mol}^{-1}$ .

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Q1. A cell has $E^\circ_\text{cathode} = +0.34$ V and $E^\circ_\text{anode} = -0.13$ V. Calculate $E^\circ_\text{cell}$ and state whether it is galvanic. [2 points]

Cue. $E^\circ_\text{cell} = 0.34 - (-0.13) = +0.47$ V; positive, so it is a galvanic (spontaneous) cell.

Q2. A cell reaction transfers $n = 1$ mol of electrons with $E^\circ = +0.50$ V. Calculate $\Delta G^\circ$ . [2 points]

Cue. $\Delta G^\circ = -(1)(96500)(0.50) = -4.8 \times 10^{4}\ \text{J mol}^{-1} = -48\ \text{kJ mol}^{-1}$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A cell uses

\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}

(

E^\circ = +0.34

V) and

\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}

(

E^\circ = -0.76

V). Zinc is oxidized. (

F = 96500\ \text{C mol}^{-1}

.) (a) Calculate the standard cell potential. (b) State whether the cell reaction is spontaneous, and justify. (c) Calculate

\Delta G^\circ

for the cell reaction (

n = 2

). (d) Justify the relationship between the sign of

E^\circ

and the sign of

\Delta G^\circ

Show worked answer →

A 4-point quantitative FRQ on cell potential and free energy.

(a) Cell potential (1 point): $E^\circ_\text{cell} = E^\circ_\text{cathode} - E^\circ_\text{anode} = (+0.34) - (-0.76) = +1.10$ V (copper is reduced at the cathode, zinc oxidized at the anode).
(b) Spontaneity (1 point): $E^\circ_\text{cell} > 0$ , so the cell reaction is spontaneous (thermodynamically favorable).
(c) Free energy (1 point): $\Delta G^\circ = -nFE^\circ = -(2)(96500)(1.10) = -2.12 \times 10^{5}\ \text{J mol}^{-1} = -212\ \text{kJ mol}^{-1}$ .
(d) Justify (1 point): from $\Delta G^\circ = -nFE^\circ$ , a positive $E^\circ$ gives a negative $\Delta G^\circ$ , so a positive cell potential corresponds to a spontaneous reaction; the minus sign and the positive $n$ and $F$ make the signs opposite.

Markers reward the cell potential, the spontaneous conclusion, the free energy with the unit, and the sign relationship.

AP 2021 (style)1 marksSection I (multiple choice). A galvanic cell has a standard cell potential of

+0.90

V. The standard free energy change for the cell reaction is (A) positive (B) negative (C) zero (D) cannot be determined. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

From $\Delta G^\circ = -nFE^\circ$ , a positive $E^\circ$ gives a negative $\Delta G^\circ$ (since $n$ and $F$ are positive), so the reaction is spontaneous. The trap is (A): the minus sign flips the sign of $E^\circ$ to give $\Delta G^\circ$ .

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)