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Chemistry · Ch 6 — Redox Reactions

Standard Electrode Potential

6.4.1

Standard Electrode Potential

The standard electrode potential, written E0E^0, is the electrode potential measured under one fixed, agreed set of conditions: every species taking part in the electrode reaction at unit concentration, and the temperature at 298 K. To make E0E^0 values comparable across different redox couples, one reference point is fixed by international convention: the standard hydrogen electrode is assigned E0=0.00E^0 = 0.00 V exactly, and every other couple's E0E^0 is measured relative to it. By convention, the electrode reaction used when tabulating a couple's E0E^0 is always written as a REDUCTION (oxidised form + n e⁻ → reduced form), so the value and sign of E0E^0 directly measure how strongly that couple's oxidised species tends to grab electrons and stay in its reduced form, compared with the H⁺/H2 reference couple. A large POSITIVE E0E^0 therefore marks a strong oxidising agent -- fluorine, at E0=+2.87E^0 = +2.87 V, has by far the greatest tendency of any species in Table 6.1 to gain an electron, making it the strongest oxidising agent listed. A large NEGATIVE E0E^0 marks a strong reducing agent, meaning that redox couple is a stronger reducing agent than H⁺/H2 itself -- the alkali metals sit at the most negative end of the table (lithium at E0=−3.05E^0 = -3.05 V, potassium at −2.93-2.93 V, sodium at −2.71-2.71 V), reflecting their group-wide tendency to give away their single outer electron easily and form cations, which is exactly what makes a metal a strong reducing agent. Given the E0E^0 values of two redox couples, it is possible to predict whether a reaction between them will happen spontaneously (Problem 6.9): write a reduction half reaction for one species using its tabulated E0E^0, write an oxidation half reaction for the other species by reve …

Table 6.1Standard electrode potentials of some redox couples (all written as reduction, at 298 K)

Table 6.1, oxidised form + n e⁻ → reduced form, with E0E^0 in volts, listed from most positive (strongest oxidising agent) to most negative (strongest reducing agent):

F2(g) + 2e⁻ → 2F⁻, E0E^0 = +2.87 V.

H2O2 + 2H⁺ + 2e⁻ → 2H2O, E0E^0 = +1.78 V.

MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O, E0E^0 = +1.51 V.

Cl2(g) + 2e⁻ → 2Cl⁻, E0E^0 = +1.36 V.

Cr2O7²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O, E0E^0 = +1.33 V.

O2(g) + 4H⁺ + 4e⁻ → 2H2O, E0E^0 = +1.23 V.

Br2 + 2e⁻ → 2Br⁻, E0E^0 = +1.09 V.

2Hg²⁺ + 2e⁻ → Hg2²⁺, E0E^0 = +0.92 V.

Fe³⁺ + e⁻ → Fe²⁺, E0E^0 = +0.77 V.

I2(s) + 2e⁻ → 2I⁻, E0E^0 = +0.54 V.

2H⁺ + 2e⁻ → H2(g), E0E^0 = 0.00 V (the reference).

Zn²⁺ + 2e⁻ → Zn(s), E0E^0 = -0.76 V.

Al³⁺ + 3e⁻ → Al(s), E0E^0 = -1.66 V.

Mg²⁺ + 2e⁻ → Mg(s), E0E^0 = -2.36 V.

Na⁺ + e⁻ → Na(s), E0E^0 = -2.71 V.

Ca²⁺ + 2e⁻ → Ca(s), E0E^0 = -2.87 V. …

Misc Problem 6.9Testing spontaneity using standard electrode potentials

Worked out. Worked example testing whether two proposed reactions are spontaneous, by writing a reduction half reaction for one species, an oxidation half reaction (E0 sign reversed from the table) for the other, and checking whether the two E0E^0 values sum to a positive number. (a) Fe³⁺(aq) and I⁻(aq): reduction 2Fe³⁺(aq) + 2e⁻ → 2Fe²⁺(aq), E0E^0 = +0.77 V; oxidation 2I⁻(aq) → I2(s) + 2e⁻, E0E^0 = -0.54 V (sign reversed from the table's +0.54 V for the reverse reduction). Sum = +0.77 + (-0.54) = +0.23 V, positive, so 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I2(s) is spontaneous. (b) Ag⁺(aq) and Cu(s): reduction 2Ag⁺(aq) + 2e⁻ → 2Ag(s), E0E^0 = +0.80 V; oxidation Cu(s) → Cu²⁺(aq) + 2e⁻, E0E^0 = -0.34 V. Sum = +0.80 + (-0.34) = +0.46 V, posit …