Standard Reduction Potentials

Core Concept

A standard reduction potential (E°) is the voltage of a half-reaction, written as a reduction, measured against the standard hydrogen electrode (SHE), which is assigned exactly 0.00 V. Stronger oxidizers have more positive E°; stronger reducers have more negative E°. For any cell, the half-reaction with the more positive E° proceeds as reduction (cathode) and the less positive (more negative) proceeds as oxidation (anode). The cell potential is E°cell = E°cathode − E°anode, using the tabulated (reduction) values for both — and, critically, these potentials are intensive properties, so they are never multiplied by the coefficients used to balance the electrons.

Key Ideas

• SHE is the zero. 2 H⁺(aq, 1 M) + 2 e⁻ ⇌ H₂(g, 1 atm) is defined as E° = 0.00 V.
• Tables list reductions. Every E° is written with electrons on the reactant side.
• Positive = good oxidizer, negative = good reducer. F₂ (+2.87 V) is the strongest common oxidizer; Li⁺ (−3.04 V) the weakest (Li is the strongest reducer).
• Cathode picks the higher E°. The more positive reduction potential is the cathode; flip the other sign when you reverse it for oxidation.
• Never scale E° by coefficients. Doubling a half-reaction doubles electrons and ΔG but leaves E° unchanged.

Equations and Variables

E°cell = E°cathode − E°anode

How It Works

1. Look up the two reduction half-reactions and their E° values.
2. Identify the larger (more positive) E° — that half-reaction is the cathode (reduction, as written).
3. The other is the anode; write it in reverse (oxidation).
4. Compute E°cell = E°(cathode) − E°(anode), keeping both values as their tabulated reduction potentials.
5. Balance electrons with coefficients only after the potentials are set — E° values stay as they are.

Worked Example

Find E°cell for Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).

Tabulated reductions:

• Cu²⁺ + 2 e⁻ → Cu, E° = +0.34 V
• Zn²⁺ + 2 e⁻ → Zn, E° = −0.76 V

Cu²⁺ has the more positive E°, so it is reduced at the cathode. Zn is oxidized at the anode (reverse of the tabulated reaction).

E°cell = E°cathode − E°anode = +0.34 V − (−0.76 V) = +1.10 V

Note that we did not multiply either potential by anything. The same answer results for 2 Al + 3 Cu²⁺ → 2 Al³⁺ + 3 Cu: E°cell = +0.34 − (−1.66) = +2.00 V, even though the balanced equation has coefficients 2 and 3.

How it works

1. Look up the two reduction half-reactions and their E° values.
2. Identify the larger (more positive) E° — that half-reaction is the cathode (reduction, as written).
3. The other is the anode; write it in reverse (oxidation).
4. Compute E°cell = E°(cathode) − E°(anode), keeping both values as their tabulated reduction potentials.
5. Balance electrons with coefficients only after the potentials are set — E° values stay as they are.

Common confusions

• "Multiply E° by the balancing coefficient." — Wrong. E° is intensive; only n (moles of electrons) scales ΔG and the electron count, never the voltage.
• "E°cell = E°anode − E°cathode." — Wrong. It is cathode minus anode.
• "The anode half-reaction is used as written in the table." — Wrong. The anode is the reverse of the tabulated reduction, so its tabulated E° is subtracted.
• "SHE is 0 V because it has no potential." — Wrong. It is 0 V by convention (a chosen reference), not because it has no tendency.
• "A large negative E° means a strong oxidizer." — Wrong. Large negative E° means a strong reducing agent (it readily loses electrons).

Quick review

• E° is measured against SHE (0.00 V).
• Reductions are tabulated; more positive E° = stronger oxidizer.
• E°cell = E°cathode − E°anode.
• Never scale E° by coefficients.
• Positive E°cell ⇒ spontaneous.