Oxidation/Reduction Vocabulary, Calculations, Ecell

Oxidation

The loss of electrons by a species; its oxidation number increases

Reduction

The gain of electrons by a species; oxidation number decreases

Oxidizing Agent

The species that is reduced (it accepts electrons and causes the other species to be oxidized)

Reducing Agent

The species that is oxidized (it donates electrons and causes th other species to be reduced)

Consider the two standard half-reactions:

(1) Ag⁺(aq) + e- → Ag(s), E° = +0.80 V

(2) Cu²⁺ (aq) + 2e- → Cu(s), E° = +0.34 V

Under standard conditions, which reaction occurs at the cathode (reduction), which species is oxidized, which are the oxidizing and reducing agents, and what is E°cell?

Cathode (reduced): Ag⁺/Ag.

Anode (oxidized): Cu/Cu²⁺.

Oxidizing agent: Ag⁺.

Reducing agent: Cu(s). E=0.80 - 0.34 = +0.46 V

Consider:

(1) Ag⁺ + e- → Ag, +0.80 V

(2) Zn²⁺ + 2e- → Zn, -0.76 V

Identify cathode/anode, oxidized/reduced species, OA/RA, and E°cell

Cathode: Ag⁺/Ag

Anode: Zn/Zn²⁺

OA: Ag⁺

RA: Zn(s)

E°cell = 0.80 - (-0.76) = +1.56 V

Consider:

(1) Cu²⁺ + 2e- → Cu, +0.34 V

(2) Zn²⁺ + 2e- → Zn, -0.76 V

Determine roles and E°cell

Cathode: Cu²⁺/Cu.

Anode: Zn/Zn²⁺.

OA: Cu²⁺.

RA: Zn(s).

E°cell = 0.34 - (-0.76) = +1.10 V.

Consider:

(1) Fe³⁺ + e- → Fe²⁺, +0.77 V

(2) Cu²⁺ + 2e- → Cu, +0.34 V

Determine roles and E°cell

Cathode: Fe³⁺/Fe²⁺

Anode: Cu/Cu²⁺

OA: Fe³⁺

RA: Cu(s)

E°cell = 0.77 - 0.34 = +0.43 V.

Consider:

(1) Fe³⁺ + e- → Fe²⁺, +0.77 V

(2) Fe²⁺ + 2e- → Fe(s), -0.44 V

Determine roles and E°cell

Cathode: Fe³⁺/Fe²⁺.

Anode: Fe/Fe²⁺.

OA: Fe³⁺. RA: Fe(s).

E°cell = 0.77 — (-0.44) = +1.21 V.

Consider:

(1) Br₂(l) + 2e- → 2Br⁻, +1.09 V

(2) l₂(s) +2e- → 2I⁻, +0.54 V

Determine roles and E°cell

Cathode: Br₂/Br⁻.

Anode: I⁻/l₂.

OA: Br₂.

RA: I⁻

Е°cell = 1.09 - 0.54 = +0.55 V.

Consider:

(1) Cl₂(g) + 2e- → 2Cl⁻, +1.36 V

(2) Br₂(l) + 2e- → 2Br⁻, +1.09 V

Determine roles and E°cell

Cathode: Cl₂/CI⁻

Anode: Br⁻/Br₂

OA: Cl₂

RA: Br⁻

E°cell = 1.36 — 1.09 = +0.27 V.

Consider:

(1) 2H⁺ + 2e- → H₂(g), 0.00 V

(2) Zn²⁺ + 2e- → Zn, -0.76 V

Determine roles and E°cell

Cathode: H⁺/H₂

Anode: Zn/Zn²⁺

OA: H⁺

RA: Zn(s)

E°cell = 0.00 - (-0.76) = +0.76 V

Consider:

(1) Ag⁺ + e- → Ag, +0.80 V

(2) Fe²⁺ + 2e- → Fe, -0.44 V

Determine roles and E°cell

Cathode: Ag⁺/Ag

Anode: Fe/Fe²⁺

OA: Ag⁺

RA: Fe(s)

E°ell = 0.80 — (-0.44) = +1.24 V

Consider:

(1) Cu²⁺ + 2e- → Cu, +0.34 V

(2) Pb²⁺ + 2e- → Pb, -0.13 V

Determine roles and E°cell

Cathode: Cu²⁺/Cu

Anode: Pb/Pb²⁺

OA: Cu²⁺

RA: Pb(s)

E°cell = 0.34 - (-0.13) = +0.47 V

Consider:

(1) Sn²⁺ + 2e- → Sn, -0.14 V

(2) Pb²⁺ + 2e- → Pb, -0.13 V

Determine roles and E°cell

Cathode: Pb²⁺/Pb

Anode: Sn/Sn²⁺

OA: Pb²⁺

RA: Sn(s)

E°cell = -0.13 — (-0.14) = +0.01 V

Consider:

(1) Fe³⁺ + e- → Fe²⁺, +0.77 V

(2) Sn²⁺ + 2e- → Sn, -0.14 V

Determine roles and E°cell

Cathode: Fe³⁺/Fe²⁺

Anode: Sn/Sn²⁺

OA: Fe³⁺

RA: Sn(s)

E°cell = 0.77- (-0.14) = +0.91 V

Consider:

(1) Ce⁴⁺ + e- → Ce³⁺, +1.61 V

(2) Fe²⁺ + 2e- → Fe, -0.44 V

Determine roles and E°cell

Cathode: Ce²⁺/Ce³⁺

Anode: Fe/Fe²⁺

OA: Ce⁴⁺

RA: Fe(s)

E°cell = 1.61 - (-0.44) = +2.05 V

Consider:

(1) MnO₄ + 8H⁺ + 5e- → Mn²⁺ + 4H₂O, +1.51 V

(2) l₂ + 2e- → 21, +0.54 V

Determine roles and E°cell

Cathode: MnO4/Mn²*.

Anode: I⁻/l₂

OA: MnO₄

RA: I⁻

E°cell = 1.51 - 0.54 = +0.97 V.

Consider:

(1) Cr₂O₇²⁻ + 14H⁺ + 6e- → 2Cr³⁺ + 7H₂O, +1.33 V

(2) Fe²⁺ + 2e- → Fe, -0.44 V

Determine roles and E°cell

Cathode: Cr₂O₇²⁻/Cr³⁺

Anode: Fe/Fe²⁺

OA: Cr₂O₇²⁻

RA: Fe(s)

E°cell = 1.33 - (-0.44) = +1.77 V

Consider:

(1) O₂(g) + 4H⁺ + 4e- → 2H₂O(I), +1.23 V

(2) Cu²⁺ + 2e- → Cu, +0.34 V

Determine roles and E°cell

Cathode: O₂/H₂O

Anode: Cu/Cu²⁺

OA: O₂

RA: Cu(s)

E°cell 1.23 — 0.34 = +0.89 V