What you'll be able to do: Use a table of standard reduction potentials to calculate a standard cell potential and predict whether a redox reaction is spontaneous.
Best after: Galvanic Cells and Cell Notation
A single number tells you whether a redox reaction will run on its own: the standard cell potential. Reading it off a table takes about thirty seconds once you know the two rules that govern the calculation.
These are recommended, not required. You can start this lesson at any time.
You cannot measure the potential of a single half-cell, only the difference between two. Chemists therefore fixed one half-cell as the zero point: the standard hydrogen electrode, 2 H⁺(1 M) + 2 e⁻ → H₂ at standard pressure 1 bar, assigned exactly 0 V.
Every tabulated value is that couple measured against hydrogen.
All values are written as reductions. A large positive value, such as +2.87 V for F₂, marks a powerful oxidizing agent that grabs electrons readily. A large negative value, such as -3.04 V for Li⁺, means the reduction is very unfavorable, which is another way of saying that lithium metal is an excellent reducing agent.
A table entry refers to the oxidized species on the reactant side of that reduction. A positive Ag⁺/Ag potential describes Ag⁺ as the electron acceptor; it does not mean Ag metal is a strong oxidant. The more negative Zn²⁺/Zn reduction potential means Zn metal can be a useful electron donor when paired with a suitable oxidant. Compare the actual reacting forms, not merely element names or absolute voltage magnitudes.
F₂ + 2 e⁻ → 2 F⁻, E° = +2.87 V
Given two half-cells, the one with the higher reduction potential wins the competition for electrons and becomes the cathode. The other one is forced to run backward as an oxidation and becomes the anode. This choice is exactly what makes E°cell positive.
E°cell = E°cathode - E°anode, where both numbers come straight from the table as reduction potentials. The subtraction already accounts for reversing the anode half-reaction, so flipping the sign yourself and then subtracting double-counts it. Adding a flipped anode value gives the same result, but mixing the two approaches is the most common source of sign errors.
For Al³⁺ + 3 e⁻ → Al at −1.66 V and Cu²⁺ + 2 e⁻ → Cu at +0.34 V, select Cu²⁺ reduction and reverse aluminum reduction. Multiply Al oxidation by two and Cu reduction by three to cancel six electrons: 2 Al + 3 Cu²⁺ → 2 Al³⁺ + 3 Cu. Then E°cell = +0.34 − (−1.66) = +2.00 V. If instead using oxidation potentials, add +1.66 V to +0.34 V. Either convention works; reversing a sign and then subtracting the reversed number mixes conventions.
E°cell = E°cathode - E°anode
If you multiply a half-reaction by 3 to balance electrons, the number of electrons changes but the potential does not. Potential is energy per unit charge, like temperature or density: doubling the amount of material does not change it. Only the total energy, which appears later as the free energy change, scales with the electron count.
Potential is joules per coulomb. Scaling the balanced reaction doubles both the free-energy change and charge transferred, so their ratio E° is unchanged. To combine redox reactions by thermodynamic addition, convert each potential to ΔG° = −nFE° first, add the energies with their equation multipliers, then divide by the target's −nF. Averaging or adding arbitrary cell potentials directly generally fails when electron counts differ.
Any oxidized species in the table can oxidize any reduced species that appears lower in the table. This is the reasoning behind the activity series: copper does not dissolve in HCl because Cu²⁺/Cu sits above H⁺/H₂, so hydrogen ions cannot oxidize copper, while zinc, which sits below, dissolves readily.
A positive E° supports forward change for standard activities, but actual direction depends on E = E° − RT ln Q/(nF). Surface oxide films, slow electrode kinetics, complexation, precipitation and dissolved oxygen can also affect observed metal corrosion. For example, a simple metal/H⁺ potential test isolates hydrogen-ion reduction; oxidizing acids or oxygen introduce different possible cathodic reactions. State those limits instead of turning one E° subtraction into a universal claim about every acid exposure.
Choose the half-reaction with larger standard reduction potential as the cathode and reverse the other for oxidation. Subtract the two listed reduction potentials; never multiply E° by stoichiometric coefficients. A positive E°cell predicts a favorable standard-state reaction, though kinetics can still affect what is observed.
The voltage of a half-reaction written as a reduction, measured against the standard hydrogen electrode at standard conditions.
The half-reaction with the more positive E° is the reduction.
E°cell = E°cathode - E°anode, using tabulated reduction values.
E°cell > 0 means the reaction is spontaneous as written under standard conditions.
Scaling a half-reaction does not change its potential.
E°cell = E°cathode - E°anode
2 H⁺(1 M) + 2 e⁻ → H₂ at standard pressure 1 bar, E° = 0 V
Given Cu²⁺ + 2 e⁻ → Cu, E° = +0.34 V and Zn²⁺ + 2 e⁻ → Zn, E° = -0.76 V, find E°cell for the zinc-copper cell and state whether it is spontaneous.
Try it first: Decide which half-cell is the cathode before touching the numbers.
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Test Ag(s) + H⁺ → Ag⁺ + H₂ after balancing, using E°(Ag⁺/Ag) = +0.80 V and E°(H⁺/H₂) = 0.00 V. Assess standard-state thermodynamic favorability of that specified reaction.
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For 2 Al(s) + 3 Cu²⁺ → 2 Al³⁺ + 3 Cu(s), with Al³⁺ + 3 e⁻ → Al at E° = -1.66 V and Cu²⁺ + 2 e⁻ → Cu at +0.34 V, find E°cell.
Try it first: Notice that the half-reactions had to be scaled by 2 and 3 to give 6 electrons.
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Can Fe³⁺ oxidize I⁻ under standard conditions? E°(Fe³⁺/Fe²⁺)=+0.77 V and E°(I₂/I⁻)=+0.54 V.
Try it first: Fe³⁺ is the stronger oxidizing agent by its higher reduction potential.
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| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Reversing the sign of the anode value and then subtracting it. | The subtraction already reverses that half-reaction, so the sign is applied twice. | Plug in tabulated reduction values exactly as printed. |
| Multiplying E° when a half-reaction is scaled. | Potential is intensive; only free energy scales with the amount of reaction. | Scale coefficients and electrons only, never the voltage. |
| Choosing the cathode as the half-reaction with the larger magnitude. | What matters is the more positive value, not the larger absolute value. | Compare positions on a number line, keeping signs. |
| Believing a negative E°cell means nothing at all happens. | The reverse reaction is spontaneous, and the forward direction can still be driven electrically. | Read a negative value as pointing to the reverse direction. |
| Treating standard potentials as valid at any concentration. | Modern standard values use unit activities and standard gas pressure 1 bar; 1 M is the ideal-solution shorthand. Other conditions need the Nernst equation. | Check the conditions before using E° directly. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
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Standard reduction potentials measure how strongly a species pulls electrons, all reported relative to the standard hydrogen electrode, which is defined as exactly 0 V. A more positive value means a stronger oxidizing agent, one more eager to be reduced. To build a spontaneous cell, the half-reaction with the higher (more positive) reduction potential runs as the reduction at the cathode, and the other is reversed to become the oxidation at the anode. The standard cell potential is E°cell = E°cathode - E°anode, using both values as tabulated for reduction and never flipping a sign for the reversed half-reaction. A positive E°cell means the reaction as written is spontaneous under standard conditions. Because potential is an intensity, not an amount, E° values are never multiplied when a half-reaction is scaled to balance electrons. The table also ranks reactivity: any species can oxidize anything that sits lower in reduction potential than itself.
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