Electrochemistry Essentials: Standard Reduction Potentials and How to Use the Table

1) What a Standard Reduction Potential (E°) Represents

A standard reduction potential, written E°, is a measured voltage that tells how strongly a chemical species tends to be reduced (gain electrons) under a fixed reference set of conditions. It is always reported for a reduction half-reaction as written in a standard reduction potential table.

Two key interpretation points:

• E° is relative, not absolute. Values are measured relative to the standard hydrogen electrode (SHE), which is defined as E° = 0.00 V.
• More positive E° means the reduction is more favorable (stronger tendency to gain electrons) under standard conditions.

Standard conditions assumption (what “standard” means)

When you use E° values, you are assuming:

• Solutes: 1.0 M (activities approximated by concentrations in many problems)
• Gases: 1 bar
• Temperature: 25 °C (298 K)

These conditions matter because electrode potentials depend on concentration/pressure and temperature. E° is the “baseline” potential; real cell voltages often differ when conditions are not standard.

How to read a table entry

A typical entry looks like:

Continue in our app.

• Listen to the audio with the screen off.
• Earn a certificate upon completion.
• Over 5000 courses for you to explore!

Or continue reading below...

Download the app

Cu2+ + 2 e− → Cu(s)     E° = +0.34 V

Interpretation: under standard conditions, Cu2+ has a moderate tendency to be reduced to Cu(s). If paired with another half-reaction, the one with the higher E° (more positive) will tend to run as the reduction (cathode) in a galvanic cell.

2) Ranking Oxidizing and Reducing Strength from E° Values

Oxidizing agents (species that get reduced)

An oxidizing agent is the species that is reduced. In a reduction potential table, the oxidizing agent is typically on the left side of the reduction half-reaction (it accepts electrons).

Rule: The more positive the E° for a reduction half-reaction, the stronger the oxidizing agent (greater tendency to be reduced).

Example comparison:

Ag+ + e− → Ag(s)        E° = +0.80 V  (stronger oxidizing agent: Ag+)

Cu2+ + 2 e− → Cu(s)     E° = +0.34 V

Since +0.80 V is more positive than +0.34 V, Ag+ is a stronger oxidizing agent than Cu2+ under standard conditions.

Reducing agents (species that get oxidized)

A reducing agent is the species that is oxidized (it donates electrons). In a reduction potential table, the reducing agent is typically on the right side of the listed reduction (it is the reduced form that could be oxidized if the reaction is reversed).

Rule: The more negative the E° for the reduction as written, the stronger the reducing agent on the product side (because reversing a very negative reduction gives a very favorable oxidation).

Example comparison:

Zn2+ + 2 e− → Zn(s)     E° = −0.76 V  (stronger reducing agent: Zn(s))

Cu2+ + 2 e− → Cu(s)     E° = +0.34 V

Zn(s) is a stronger reducing agent than Cu(s) because the reduction potential for Zn2+/Zn is more negative, meaning zinc metal more readily gives up electrons (is oxidized) when paired appropriately.

Quick ranking drill (no calculation)

Given these reductions:

Cl2(g) + 2 e− → 2 Cl−(aq)     E° = +1.36 V

Fe2+ + 2 e− → Fe(s)           E° = −0.44 V

Al3+ + 3 e− → Al(s)           E° = −1.66 V

• Strongest oxidizing agent: Cl2(g) (most positive E°)
• Strongest reducing agent: Al(s) (most negative E° reduction corresponds to most favorable oxidation when reversed)

3) Selecting Half-Reactions for a Galvanic Cell (Using the Table Correctly)

Standard reduction potential tables are designed to be used in a consistent way. The most common mistakes are (a) flipping signs incorrectly, and (b) multiplying E° when balancing electrons.

Step-by-step workflow

1. Pick the two relevant half-reactions from the table (as reductions).
2. Decide which runs as reduction (cathode): choose the half-reaction with the more positive E° to remain as written.
3. Decide which runs as oxidation (anode): the other half-reaction must be reversed (because tables list reductions).
4. Reverse the anode half-reaction and change the sign of its E° (because reversing a half-reaction reverses the direction of electron flow).
5. Balance electrons by multiplying coefficients in the half-reactions as needed, but do not multiply E°.
6. Add the half-reactions to get the overall cell reaction.

Why you do not multiply E° when balancing

E° is an intensive property (like temperature), not an extensive one (like moles). Multiplying a half-reaction by 2 doubles electrons and species amounts, but it does not “double the voltage.” The potential reflects the driving force per unit charge, not the total amount of reaction.

Worked selection example (classic Zn/Cu cell)

Table entries:

Cu2+ + 2 e− → Cu(s)     E° = +0.34 V

Zn2+ + 2 e− → Zn(s)     E° = −0.76 V

Choose cathode (reduction): copper, because +0.34 V is more positive.

Anode (oxidation): reverse zinc:

Zn(s) → Zn2+ + 2 e−     E°ox = +0.76 V

Electrons already match (2 e−), so add:

Zn(s) + Cu2+ → Zn2+ + Cu(s)

4) Calculating E°cell and Using the Sign to Predict Spontaneity

For a galvanic cell under standard conditions:

E°cell = E°cathode − E°anode (where both E° values are taken as reduction potentials from the table)

Equivalent approach:

E°cell = E°red(cathode) + E°ox(anode)

Step-by-step calculation (Zn/Cu)

Using reduction potentials:

• Cathode reduction: Cu2+/Cu, E° = +0.34 V
• Anode reduction potential (as listed): Zn2+/Zn, E° = −0.76 V

Compute:

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

Interpreting the sign

• E°cell > 0: reaction is spontaneous in the galvanic direction under standard conditions.
• E°cell < 0: the written direction is nonspontaneous under standard conditions (the reverse direction would be spontaneous).
• E°cell = 0: equilibrium under standard conditions (no net driving force).

Spontaneity check without arithmetic (quick logic)

If you choose the more positive E° as the cathode reduction and the less positive as the anode (reversed), E°cell will come out positive. If you accidentally assign them the other way, you will get a negative value—use that as a diagnostic.

5) “More Positive” vs “More Negative” in Real Terms (Worked Examples + Ranking Exercises)

Real meaning of “more positive E°”

When a reduction half-reaction has a more positive E° than another, it means (under standard conditions) that species has a stronger tendency to take electrons from something else. In a galvanic pairing, it will more strongly “pull” electrons through the external circuit, acting as the cathode reaction.

Real meaning of “more negative E°”

A more negative reduction potential means the reduction is relatively unfavorable under standard conditions. The reverse process (oxidation of the reduced form) is relatively favorable, so the reduced form is a stronger electron donor (stronger reducing agent).

Worked example A: Will Fe(s) reduce Cu2+(aq)?

Use table entries:

Cu2+ + 2 e− → Cu(s)     E° = +0.34 V

Fe2+ + 2 e− → Fe(s)     E° = −0.44 V

If Fe(s) reduces Cu2+, then:

• Cathode (reduction): Cu2+ → Cu (more positive E°)
• Anode (oxidation): Fe → Fe2+ (reverse of the listed reduction)

Compute:

E°cell = (+0.34) − (−0.44) = +0.78 V

Since E°cell is positive, Fe(s) will spontaneously reduce Cu2+(aq) under standard conditions (iron dissolves to Fe2+, copper plates out).

Worked example B: Can Ag+(aq) oxidize Cl−(aq) to Cl2(g)?

Relevant reductions:

Ag+ + e− → Ag(s)                 E° = +0.80 V

Cl2(g) + 2 e− → 2 Cl−(aq)        E° = +1.36 V

To oxidize Cl− to Cl2, you would need the anode to be 2 Cl− → Cl2 + 2 e− (reverse of the chlorine reduction). Then:

• Cathode: Ag+ → Ag, E° = +0.80 V
• Anode reduction potential (as listed): Cl2/Cl−, E° = +1.36 V

Compute:

E°cell = (+0.80) − (+1.36) = −0.56 V

Negative means nonspontaneous as written: Ag+ is not a strong enough oxidizing agent to pull electrons from Cl− to make Cl2 under standard conditions. In fact, Cl2 is the stronger oxidizing agent (more positive E°).

Mini-exercise set 1: Identify strongest oxidizing agent

Rank the oxidizing agents (strongest to weakest) using E° values:

Br2 + 2 e− → 2 Br−      E° = +1.07 V

Ag+ + e− → Ag           E° = +0.80 V

Fe3+ + e− → Fe2+        E° = +0.77 V

Answer: Br2 > Ag+ > Fe3+ (most positive E° corresponds to strongest oxidizing agent).

Mini-exercise set 2: Identify strongest reducing agent

Rank the reducing agents (strongest to weakest). Remember: look at the products of the reductions and use “more negative E° means stronger reducing agent.”

Mg2+ + 2 e− → Mg        E° = −2.37 V

Zn2+ + 2 e− → Zn        E° = −0.76 V

Cu2+ + 2 e− → Cu        E° = +0.34 V

Answer: Mg(s) > Zn(s) > Cu(s).

Mini-exercise set 3: Choose anode/cathode quickly

Given:

Ni2+ + 2 e− → Ni(s)     E° = −0.25 V

Ag+ + e− → Ag(s)        E° = +0.80 V

• Cathode (reduction): Ag+ → Ag (more positive)
• Anode (oxidation): Ni → Ni2+ + 2 e− (reverse nickel reduction)
• E°cell: (+0.80) − (−0.25) = +1.05 V

Common table-usage pitfalls (fast fixes)

• Pitfall: Multiplying E° when multiplying a half-reaction. Fix: Never scale E° with coefficients.
• Pitfall: Forgetting to flip the sign when reversing a half-reaction. Fix: Reversal changes the sign of E°.
• Pitfall: Mixing up “strong oxidizing agent” with “strong reducing agent.” Fix: Oxidizing agents correspond to more positive E° (left side species); reducing agents correspond to more negative E° (right side species).
• Pitfall: Using E° to predict behavior under nonstandard conditions without checking concentrations/pressures. Fix: Treat E° as the standard baseline; real conditions can shift direction and voltage.