Chemistry · Ch 6 — Redox Reactions
Standard Electrode Potential
Standard Electrode Potential
The standard electrode potential, written , 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 values comparable across different redox couples, one reference point is fixed by international convention: the standard hydrogen electrode is assigned V exactly, and every other couple's is measured relative to it. By convention, the electrode reaction used when tabulating a couple's is always written as a REDUCTION (oxidised form + n e⁻ → reduced form), so the value and sign of 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 therefore marks a strong oxidising agent -- fluorine, at 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 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 V, potassium at V, sodium at 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 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 , write an oxidation half reaction for the other species by reve …
Table 6.1, oxidised form + n e⁻ → reduced form, with in volts, listed from most positive (strongest oxidising agent) to most negative (strongest reducing agent):
F2(g) + 2e⁻ → 2F⁻, = +2.87 V.
H2O2 + 2H⁺ + 2e⁻ → 2H2O, = +1.78 V.
MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O, = +1.51 V.
Cl2(g) + 2e⁻ → 2Cl⁻, = +1.36 V.
Cr2O7²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O, = +1.33 V.
O2(g) + 4H⁺ + 4e⁻ → 2H2O, = +1.23 V.
Br2 + 2e⁻ → 2Br⁻, = +1.09 V.
2Hg²⁺ + 2e⁻ → Hg2²⁺, = +0.92 V.
Fe³⁺ + e⁻ → Fe²⁺, = +0.77 V.
I2(s) + 2e⁻ → 2I⁻, = +0.54 V.
2H⁺ + 2e⁻ → H2(g), = 0.00 V (the reference).
Zn²⁺ + 2e⁻ → Zn(s), = -0.76 V.
Al³⁺ + 3e⁻ → Al(s), = -1.66 V.
Mg²⁺ + 2e⁻ → Mg(s), = -2.36 V.
Na⁺ + e⁻ → Na(s), = -2.71 V.
Ca²⁺ + 2e⁻ → Ca(s), = -2.87 V. …
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 values sum to a positive number. (a) Fe³⁺(aq) and I⁻(aq): reduction 2Fe³⁺(aq) + 2e⁻ → 2Fe²⁺(aq), = +0.77 V; oxidation 2I⁻(aq) → I2(s) + 2e⁻, = -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), = +0.80 V; oxidation Cu(s) → Cu²⁺(aq) + 2e⁻, = -0.34 V. Sum = +0.80 + (-0.34) = +0.46 V, posit …