Spontaneity and Cell Potential

Core Concept

A redox reaction is spontaneous when its cell potential is positive: E°cell > 0 ⇒ spontaneous. This is the electrochemical restatement of ΔG° < 0, since ΔG° = −nFE°cell. Whenever you pair a stronger oxidizing agent (higher E°) with a stronger reducing agent (lower E°), the spontaneous direction is the one that yields a positive E°cell. The sign of E°cell therefore predicts, in one step, which way a redox reaction runs and whether it can power a battery.

Key Ideas

• Positive voltage = spontaneous. E°cell > 0 ⇔ ΔG° < 0 ⇔ K > 1.
• Oxidizer reduced, reducer oxidized. In the spontaneous direction, the species with the higher E° is reduced and the lower E° is oxidized.
• Same sign as ΔG° flipped. Because ΔG° = −nFE°cell, a positive E°cell is a negative ΔG°.
• Bigger E°cell ⇒ farther from equilibrium. Larger voltage means larger K and a more product-favored reaction.
• Practical prediction tool. "Will Zn dissolve in acid? Will Cu plate onto Fe?" are answered by computing E°cell.

Equations and Variables

E°cell = E°cathode − E°anode, and ΔG° = −nFE°cell.

How It Works

1. Write the two half-reactions as reductions and note their E° values.
2. Assign the more positive E° to the cathode (reduction) and the other, reversed, to the anode (oxidation).
3. Compute E°cell = E°cathode − E°anode.
4. If E°cell > 0, the reaction as written is spontaneous; if E°cell < 0, the reverse is spontaneous; if E°cell = 0, the system is at equilibrium.

Worked Example

Will solid zinc spontaneously reduce Cu²⁺ to Cu in aqueous solution?

Candidate reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

• Cathode (reduction): Cu²⁺ + 2 e⁻ → Cu, E° = +0.34 V
• Anode (oxidation): Zn → Zn²⁺ + 2 e⁻ (reverse of −0.76 V)
• E°cell = +0.34 − (−0.76) = +1.10 V

E°cell > 0, so yes — zinc spontaneously reduces Cu²⁺. (This is why a zinc strip dipped in blue CuSO₄ solution becomes coated with reddish copper while the solution fades.)

For contrast, consider Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s). Here E°cell = −1.10 V, so that direction is nonspontaneous — copper will not reduce Zn²⁺.

How it works

1. Write the two half-reactions as reductions and note their E° values.
2. Assign the more positive E° to the cathode (reduction) and the other, reversed, to the anode (oxidation).
3. Compute E°cell = E°cathode − E°anode.
4. If E°cell > 0, the reaction as written is spontaneous; if E°cell < 0, the reverse is spontaneous; if E°cell = 0, the system is at equilibrium.

Common confusions

• "E°cell negative means it goes forward slowly." — Wrong. Negative E°cell means the reaction as written is nonspontaneous; its reverse is spontaneous.
• "Spontaneity depends on current or resistance." — Wrong. The thermodynamic direction depends only on E°cell; current/resistance affect how fast, not which way.
• "E°cell = 0 means nothing happens." — Wrong. It means the system is at equilibrium (ΔG° = 0, K = 1).
• "A large positive E°cell guarantees a fast reaction." — Wrong. Voltage says nothing about kinetics.
• "Sign convention flips for electrolytic cells." — The spontaneity test is the same; electrolytic cells simply run a nonspontaneous reaction with external power.

Quick review

• E°cell > 0 ⇒ spontaneous (ΔG° < 0, K > 1).
• Spontaneous direction reduces the higher-E° species.
• E°cell = E°cathode − E°anode.
• Sign predicts direction; magnitude predicts how product-favored.
• ΔG° = −nFE°cell.