Standard reduction potentials E° at 298 K, measured against the standard hydrogen electrode. Pick any two half-cells to build a cell and read E°cell, ΔG° and K live, or tap rows in the full series below.
Cell builder
Choose two half-cells
The couple with the higher E° is reduced (cathode); the other is oxidised (anode). Tap the scale or the series to pick.
Tap any two rows to send them to the cell builder. The bar shows how far each E° sits from the SHE (0 V).
Reduction half-reactionE° relative to SHEE° / VWhat it tells you
Values follow the NCERT Class 12 table where a couple is listed there (e.g. Li⁺/Li −3.05 V, Mg²⁺/Mg −2.36 V, Br₂/Br⁻ +1.09 V, Au³⁺/Au +1.40 V, Co³⁺/Co²⁺ +1.81 V) and NCERT’s +1.74 V for Ce⁴⁺/Ce³⁺; the rest are standard data-book values rounded to 0.01 V. Data books differ slightly for a few couples (modern tables give Au³⁺/Au ≈ +1.50 V and Co³⁺/Co²⁺ ≈ +1.92 V). Use the values given in your exam paper.
Combine potentials · JEE Advanced
Add half-reactions through ΔG°, not E°
E° is intensive, so potentials of consecutive steps cannot simply be added. Add n·E° (which is −ΔG°/F) and divide by the total electrons. Tap two species to get the E° of that couple.
E° for the couple you picked
Disproportionation check
Quick drill
Ten-second questions from the table
New questions are generated from the series each time: strongest oxidant or reductant, will it react, and E°cell values.
Score 0 / 0
Tip: decide from the table first, then check with the builder.
Exam notes
What examiners ask about the series
How to read the table
More positive E° → the species on the left (oxidised form) is a stronger oxidising agent; F₂ is the strongest in the table. More negative E° → the species on the right (reduced form) is a stronger reducing agent; Li is the strongest. In a cell the couple with the higher E° is the cathode, and E°cell = E°cathode − E°anode (both as reduction potentials). E°cell > 0 means ΔG° < 0, so the cell reaction is spontaneous as written.
Why lithium is the strongest reducing agent in water
Lithium has the highest ionisation enthalpy of the alkali metals, yet the most negative E°. The very small Li⁺ ion has a very large hydration enthalpy, which more than makes up for the energy needed to remove the electron, so Li → Li⁺(aq) is the most favourable change in aqueous solution.
Which metals give H₂ with dilute acids?
Metals with a negative E° (above hydrogen: Mg, Zn, Fe, Ni, Sn …) reduce H⁺ to H₂ in dilute HCl or H₂SO₄. Metals with a positive E° (Cu, Ag, Au) do not. Copper still dissolves in nitric acid because the oxidant there is NO₃⁻ (NO₃⁻/NO, +0.97 V), not H⁺, so the gas is NO or NO₂, not H₂.
Displacement: which metal replaces which?
A metal with a more negative E° reduces the ions of a metal with a more positive E°: Zn displaces Cu from CuSO₄ (E°cell = +1.10 V), Cu displaces Ag from AgNO₃ (+0.46 V), but Cu does not displace Zn. The same rule explains why Fe³⁺ oxidises I⁻ to I₂ (+0.23 V) but not Br⁻ (−0.32 V).
Electrolysis of aqueous NaCl: why H₂ and Cl₂?
At the cathode, water (2H₂O + 2e⁻ → H₂ + 2OH⁻, −0.83 V) is reduced in preference to Na⁺ (−2.71 V). At the anode, O₂ from water is thermodynamically favoured (+1.23 V vs +1.36 V for Cl₂), but oxygen has a large overpotential, so Cl₂ is liberated from concentrated brine.
Corrosion and sacrificial protection
Rusting is an electrochemical cell: Fe is oxidised (Fe²⁺/Fe, −0.44 V) while O₂ is reduced (O₂ + 4H⁺ + 4e⁻ → 2H₂O, +1.23 V), giving E°cell = +1.67 V. Zinc (−0.76 V) is more easily oxidised than iron, so a zinc coating (galvanising) or a Mg/Zn block corrodes first and protects the iron.
E° does not change when you multiply a half-equation
2Ag⁺ + 2e⁻ → 2Ag still has E° = +0.80 V. E° is an intensive property; ΔG° = −nFE° is extensive and does scale with n. That is why potentials of successive steps are combined through n·E° (see “Combine potentials”), while E°cell for a full cell is a simple difference.
