The Electrochemical Series — Predicting Whether a Reaction Happens
The table of standard electrode potentials is the most productive single page in inorganic chemistry. With it you can answer, in about twenty seconds and without any experiment, whether copper dissolves in hydrochloric acid, whether Fe³⁺ oxidises iodide, whether Cu⁺ survives in water, and how far a reaction goes at equilibrium. This article shows the reasoning, then does five of those calculations in full.
The convention, stated precisely
Every value in the series is a reduction potential, measured against the standard hydrogen electrode (SHE, defined as exactly 0.000 V) at 298 K with all species at unit activity and any gas at 1 bar.
= E°(species being reduced) − E°(species being oxidised)
ΔG° = −nFE°cell, F = 96485 C mol⁻¹
E°cell > 0 ⟺ ΔG° < 0 ⟺ K > 1 ⟺ spontaneous under standard conditions
Read the table as a ladder of oxidising power. A couple high in the table has a strong appetite for electrons, so its oxidised form is a strong oxidising agent. A couple low in the table gives electrons away readily, so its reduced form is a strong reducing agent. Any species will oxidise the reduced form of any couple below it.
| Half-reaction (reduction) | E° / V | Half-reaction (reduction) | E° / V |
|---|---|---|---|
| F₂ + 2e⁻ → 2F⁻ | +2.87 | Cu²⁺ + 2e⁻ → Cu | +0.34 |
| MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O | +1.51 | 2H⁺ + 2e⁻ → H₂ | 0.00 |
| Cl₂ + 2e⁻ → 2Cl⁻ | +1.36 | Ni²⁺ + 2e⁻ → Ni | −0.25 |
| Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O | +1.33 | Fe²⁺ + 2e⁻ → Fe | −0.44 |
| O₂ + 4H⁺ + 4e⁻ → 2H₂O | +1.23 | Zn²⁺ + 2e⁻ → Zn | −0.76 |
| Br₂ + 2e⁻ → 2Br⁻ | +1.09 | Al³⁺ + 3e⁻ → Al | −1.66 |
| NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O | +0.96 | Mg²⁺ + 2e⁻ → Mg | −2.37 |
| Ag⁺ + e⁻ → Ag | +0.80 | Na⁺ + e⁻ → Na | −2.71 |
| Fe³⁺ + e⁻ → Fe²⁺ | +0.77 | K⁺ + e⁻ → K | −2.93 |
| I₂ + 2e⁻ → 2I⁻ | +0.54 | Li⁺ + e⁻ → Li | −3.04 |
Worked example 1 — the Daniell cell, all the way to K
Q. For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), find E°cell, ΔG° and K at 298 K.
Step 1 — identify the electrodes. Cu²⁺ is reduced (cathode), Zn is oxidised
(anode).
E°cell = +0.34 − (−0.76) = +1.10 V → spontaneous.
Step 2 — Gibbs energy. Two electrons are transferred, so n = 2.
ΔG° = −nFE° = −2 × 96485 × 1.10 = −212 267 J mol⁻¹ =
−212.3 kJ mol⁻¹.
Step 3 — equilibrium constant. At 298 K,
log K = nE°/0.0592:
log K = (2 × 1.10)/0.0592 = 2.20/0.0592 = 37.16
K = 1037.16 ≈ 1.4 × 10³⁷
A cell voltage of just over one volt corresponds to an equilibrium constant of 10³⁷ — the reaction is complete for all practical purposes. This is why small potential differences matter so much: every 0.0592 V per electron multiplies K by ten.
Worked example 2 — will copper dissolve in acid?
Q. Cu + 2H⁺ → Cu²⁺ + H₂. Predict, then repeat for dilute nitric acid.
In HCl. H⁺ is reduced, Cu oxidised:
E°cell = 0.00 − 0.34 = −0.34 V → ΔG° = +65.6 kJ mol⁻¹ →
not spontaneous. Copper does not dissolve in hydrochloric acid, and no amount
of concentration will fix a deficit that large.
In HNO₃. Here the oxidising agent is nitrate, not the proton:
E°cell = 0.96 − 0.34 = +0.62 V → spontaneous.
Copper dissolves in nitric acid because nitrate is the oxidant, which is also why the gas evolved is a nitrogen oxide rather than hydrogen. The series tells you not only whether the reaction goes, but what the products must be.
Worked example 3 — the halogen–iron test
Q. Does Fe³⁺ oxidise I⁻? Does it oxidise Br⁻?
With iodide: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
E°cell = 0.77 − 0.54 = +0.23 V → yes, iodine is liberated.
With bromide: 2Fe³⁺ + 2Br⁻ → 2Fe²⁺ + Br₂
E°cell = 0.77 − 1.09 = −0.32 V → no reaction.
Two questions, one subtraction each. Note that E° was not doubled even though the balanced equation carries two electrons — see the mistakes section.
Worked example 4 — disproportionation of Cu⁺
Q. Given Cu⁺ + e⁻ → Cu (E° = +0.52 V) and Cu²⁺ + e⁻ → Cu⁺ (E° = +0.16 V), is 2Cu⁺ → Cu²⁺ + Cu spontaneous in water?
Working. Cu⁺ is reduced in one half-reaction and oxidised in the other:
E°cell = E°(Cu⁺/Cu) − E°(Cu²⁺/Cu⁺) = 0.52 − 0.16 = +0.36 V
ΔG° = −1 × 96485 × 0.36 = −34.7 kJ mol⁻¹ → spontaneous.
That is precisely why simple Cu(I) salts are unstable in aqueous solution while Cu(II) salts are the norm; Cu(I) survives only when locked into an insoluble or strongly complexed form such as CuCl or [Cu(NH₃)₂]⁺. The general test: a species disproportionates when the potential for its reduction exceeds the potential for its oxidation.
Worked example 5 — leaving standard conditions
Q. What is the potential of a silver electrode in 1.0 × 10⁻⁵ M Ag⁺?
For Ag⁺ + e⁻ → Ag, n = 1 and Q = 1/[Ag⁺]:
E = 0.80 − 0.0592 log(1/10⁻⁵) = 0.80 − 0.0592 × 5 = 0.80 − 0.296 =
+0.504 V
Diluting by five orders of magnitude costs about 0.30 V. For a one-electron couple, each factor of ten in concentration is worth 59.2 mV — worth memorising, because it lets you judge at a glance whether a near-zero E° can be reversed by concentration. A cell at −0.34 V cannot be rescued this way; a cell at −0.03 V easily can.
Common mistakes
- Multiplying E° by a stoichiometric coefficient. Potential is an intensive property. Doubling the equation doubles ΔG° and n, leaving E° unchanged.
- Adding half-cell potentials. Use E°cathode − E°anode. If you prefer to flip the anode reaction and add, remember to change its sign first — but never do both.
- Combining two half-reactions of the same element by averaging E°. Combine them through ΔG° = −nFE° (which is additive) and convert back at the end.
- Confusing thermodynamics with kinetics. Aluminium has E° = −1.66 V yet cooking pans survive: a passivating oxide layer blocks the reaction. A positive E°cell says the reaction can happen, never how fast.
- Using standard values for non-standard conditions. Any pH-dependent couple — MnO₄⁻/Mn²⁺, Cr₂O₇²⁻/Cr³⁺, NO₃⁻/NO — changes markedly when [H⁺] is not 1 M.
- Forgetting the solvent. Water itself is oxidised above about +1.23 V and reduced below about 0 V, so couples outside that window are limited by the solvent's own stability.
Summary
| You want to know | Use | Criterion |
|---|---|---|
| Does it go? | E°cell = E°cat − E°an | Positive → yes |
| How much energy? | ΔG° = −nFE° | Negative → spontaneous |
| How far? | log K = nE°/0.0592 | K > 1 when E° > 0 |
| At real concentrations? | E = E° − (0.0592/n) log Q | 59.2 mV per decade, per electron |
| Is the species stable? | Compare its reduction and oxidation potentials | Reduction > oxidation → disproportionates |
Check your cell potentials instantly. The Nernst Equation calculator handles E°, n, the reaction quotient and the 0.0592/n term at 298 K, so you can test a prediction in seconds instead of re-deriving it.
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