9.5 Oxidation, Reduction & Electrochemistry

Oxidation States

Quick Answer: The oxidation state (oxidation number) is the charge an atom would have if all bonds were ionic. Oxidation increases the oxidation state (loss of e−); reduction decreases it (gain of e−). The mnemonic LEO the lion says GER — Lose Electrons = Oxidation, Gain Electrons = Reduction — captures the direction. Or use OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Rules (apply in order)

1. Free element in any form (O2, Na, Fe, S8): 0.
2. Monatomic ion: oxidation state equals the ion charge (Na+ = +1, Cl− = −1).
3. H is +1 with nonmetals, −1 with metals (metal hydrides: NaH → H is −1).
4. O is −2 in most compounds, −1 in peroxides (H2O2), 0 in O2, +2 with F (OF2).
5. Sum of oxidation states = 0 for a neutral compound, = the ion's charge for a polyatomic ion.

Worked example: Mn in KMnO4. K is +1 (rule 2), O is −2 (rule 4), and the compound is neutral. Let x = Mn's state: +1 + x + 4(−2) = 0 → x = +7. Mn in MnO4− is also +7 because the sum must equal −1: x + 4(−2) = −1 → x = +7.

Balancing Redox by Half-Reactions

In Acidic Solution

Balance Cr2O7²− + Fe2+ → Cr3+ + Fe3+.

1. Split:
   • Oxidation: Fe2+ → Fe3+
   • Reduction: Cr2O7²− → Cr3+
2. Balance atoms other than O/H: Cr2O7²− → 2 Cr3+.
3. Balance O with H2O: Cr2O7²− → 2 Cr3+ + 7 H2O.
4. Balance H with H+: 14 H+ + Cr2O7²− → 2 Cr3+ + 7 H2O.
5. Balance charge with e−: left charge = 14 + (−2) = +12; right = +6. Add 6 e− to the left: 6 e− + 14 H+ + Cr2O7²− → 2 Cr3+ + 7 H2O.
6. Oxidation half: Fe2+ → Fe3+ + e− (charge +2 → +3, lose 1 e−).
7. Equalize electrons (×6 on oxidation): 6 Fe2+ → 6 Fe3+ + 6 e−.
8. Add and simplify: 6 Fe2+ + 14 H+ + Cr2O7²− → 2 Cr3+ + 6 Fe3+ + 7 H2O.

Verify atoms (Fe 6, Cr 2, O 7, H 14) and charge (left +24, right +24) — balanced.

In Basic Solution

After balancing in acid, add OH− to both sides equal to the number of H+, combine H+ + OH− → H2O, and cancel excess water.

Galvanic (Voltaic) Cells

A galvanic cell harnesses a spontaneous redox reaction to produce electricity. Two half-cells are connected by a wire (electron path) and a salt bridge (ion path).

• Anode: oxidation occurs here; electrons leave; marked negative (−).
• Cathode: reduction occurs here; electrons arrive; marked positive (+).
• Electron flow: anode → cathode (through the external wire).
• Ion flow: anions migrate toward the anode; cations toward the cathode (through the salt bridge).

Classic Zn/Cu Cell

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

• Anode (oxidation): Zn(s) → Zn2+(aq) + 2 e−
• Cathode (reduction): Cu2+(aq) + 2 e− → Cu(s)
• Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Standard reduction potentials: E°(Cu2+/Cu) = +0.34 V; E°(Zn2+/Zn) = −0.76 V.

E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V

Positive E°cell → spontaneous as written. This is the Daniell cell, the prototype voltaic cell.

Electrolytic Cells

An electrolytic cell uses an external power source to drive a nonspontaneous reaction. The signs flip vs a galvanic cell because the external supply forces the reaction: the anode is positive (pulls electrons out) and the cathode is negative (pushes electrons in). But the chemistry at each electrode stays the same: oxidation at the anode, reduction at the cathode.

Applications: electroplating (Cu on a cheaper metal), refining impure copper, splitting water into H2 and O2 (electrolysis), and the Downs cell for producing sodium metal from molten NaCl.

Electrolysis Calculation (Faraday's Law)

Moles of substance = (current × time) / (n × F), where F = 96,485 C/mol e− and n = electrons per ion.

Example: How many grams of Cu deposit at the cathode when 3.00 A flows for 30.0 min through CuSO4(aq)?

1. Charge Q = 3.00 C/s × 30.0 × 60 s = 5400 C.
2. Moles of e− = 5400 / 96,485 = 0.05599 mol e−.
3. Cu2+ + 2 e− → Cu, so moles Cu = 0.05599 / 2 = 0.02800 mol.
4. Mass = 0.02800 × 63.55 = 1.78 g Cu.

Standard Reduction Potentials and Spontaneity

All standard potentials are written as reductions. A species higher in the table is a stronger oxidizing agent (more easily reduced). To get E°cell:

E°cell = E°(cathode, reduction) − E°(anode, reduction)

Equivalently, flip the anode reaction (sign of E° flips) and add. Spontaneous when E°cell > 0, which also means ΔG° = −nFE° < 0.

The Nernst Equation

Real cells run at nonstandard concentrations. At 25 °C with base-10 form:

E = E° − (0.0592 / n) log Q

where Q is the reaction quotient. When E = 0 the cell is dead and Q = K (the equilibrium constant). For the Zn/Cu cell at nonstandard [Zn2+]/[Cu2+], plugging into this equation gives the corrected voltage — a common PA-CAT-style calculation.

Worked Nernst Example

For the Zn/Cu cell with [Zn2+] = 0.10 M and [Cu2+] = 1.0 × 10−4 M, n = 2:

E = 1.10 − (0.0592/2) log([Zn2+]/[Cu2+])
  = 1.10 − 0.0296 × log(0.10 / 1.0 × 10−4)
  = 1.10 − 0.0296 × log(1000)
  = 1.10 − 0.0296 × 3
  = 1.10 − 0.0888
  = 1.0112 V ≈ 1.01 V

Lower product / higher reactant concentration raises Q's effect: here [Cu2+] is small, which makes the cell voltage drop slightly below standard.