Electrons in Motion
Electrochemistry is about one fundamental process: the transfer of electrons between substances. This guide builds from OIL RIG through oxidation numbers, half-reaction balancing, galvanic cells, standard potentials, and electrolysis — with the reasoning behind every step.
Electrochemistry is built on one concept: electrons move from one species to another. The species that loses electrons is oxidized. The species that gains electrons is reduced. These two processes always happen together — electrons must go somewhere.
Agents: The reducing agent donates electrons → it is itself oxidized. The oxidizing agent accepts electrons → it is itself reduced. This reversal trips up many students — the oxidizing agent gets reduced.
Electrons cannot float free in solution. For one species to lose electrons, those electrons must immediately be captured by another species. You can’t have a “half-redox” — both half-reactions are always happening simultaneously.
| Process | What happens | Oxidation state | Species is the… |
|---|---|---|---|
| Oxidation | Loses e&sup-; | Increases | Reducing agent |
| Reduction | Gains e&sup-; | Decreases | Oxidizing agent |
Mg is oxidized → Mg is the reducing agent. Cu²+ is reduced → Cu²+ (CuSO⊂4;) is the oxidizing agent.
b) True or false: “A reducing agent gains electrons.” Explain.
c) Describe in one sentence why oxidation and reduction must always occur together.
a) Al: 0 → +3 (oxidized, reducing agent). Cl: 0 → −1 (reduced, oxidizing agent). AlCl⊂3; half-reactions: Al → Al³+ + 3e&sup-; (oxidation); Cl⊂2; + 2e&sup-; → 2Cl&sup-; (reduction).
b) False. A reducing agent donates electrons (loses them), so it is itself oxidized. It reduces another species by giving it electrons.
c) Electrons cannot exist freely in solution; every electron released by an oxidized species must immediately be captured by another species that is reduced.
Oxidation numbers (oxidation states) are a bookkeeping tool that lets us track which atoms gained or lost electrons in a reaction. They are assigned, not measured — they represent a hypothetical charge if the compound were purely ionic.
The Rules (in priority order)
- Rule 1: Pure element → oxidation number = 0. (Na, O⊂2;, P⊂4;, Fe are all 0.)
- Rule 2: Monatomic ion → oxidation number = ionic charge. (Cl&sup-; = −1, Ca²+ = +2.)
- Rule 3: Fluorine is always −1.
- Rule 4: Oxygen is −2 in most compounds. Exceptions: peroxides (e.g. H⊂2;O⊂2;): −1; OF⊂2;: +2.
- Rule 5: Hydrogen is +1 bonded to nonmetals; −1 in metal hydrides (NaH, CaH⊂2;).
- Rule 6: Sum of all oxidation numbers = 0 (neutral compound) or = net charge (ion).
2Cr = 12
Cr = +6
Products: Fe(s) = 0
Change: +3 → 0 (decrease) → Fe is REDUCED
CO⊂2;: C + 2(−2) = 0 → C = +4
Change: +2 → +4 (increase) → C is OXIDIZED
Yes, it is redox. Fe is reduced (Fe⊂2;O⊂3; is the oxidizing agent). CO is oxidized (CO is the reducing agent).
b) In the peroxide H⊂2;O⊂2;, what is the oxidation state of O? Why is it different from the usual −2?
c) Identify the OS change in: Cl⊂2; + 2KBr → 2KCl + Br⊂2;. What is oxidized and what is reduced?
a) (i) NO⊂2;: N + 2(−2) = 0 → N = +4. (ii) NO⊂3;&sup-;: N + 3(−2) = −1 → N = +5. (iii) N⊂2;H⊂4;: 2N + 4(+1) = 0 → 2N = −4 → N = −2.
b) O in H⊂2;O⊂2; = −1. In H⊂2;O⊂2;: 2(+1) + 2O = 0 → O = −1. The two oxygen atoms are bonded to each other (O–O bond) in addition to H, so they don’t draw a full 2 extra electrons as they would in a normal compound. The O–O bond shares electrons symmetrically between two identical atoms.
c) Cl⊂2;: Cl = 0 → KCl: Cl = −1 (decreases: Cl⊂2; is reduced, oxidizing agent). KBr: Br = −1 → Br⊂2;: Br = 0 (increases: Br&sup-; is oxidized, reducing agent).
Redox equations must balance for both atoms and charge. The half-reaction method splits the reaction into two separate equations and balances them independently before combining.
Steps for Acidic Solution
- Step 1: Write the unbalanced net ionic equation. Remove spectator ions.
- Step 2: Split into oxidation half-reaction and reduction half-reaction.
- Step 3: Balance atoms other than O and H in each half-reaction.
- Step 4: Add H⊂2;O to balance oxygen atoms.
- Step 5: Add H&sup+; to balance hydrogen atoms.
- Step 6: Add e&sup-; to balance charge on each side of each half-reaction.
- Step 7: Multiply each half-reaction by integers so electrons cancel exactly.
- Step 8: Add the two half-reactions. Cancel electrons and any species appearing on both sides.
- Step 9: Check: all atoms balanced, net charge balanced.
Oxidation: Fe²+ → Fe³+
b) How many electrons are transferred in total per formula unit of Cr⊂2;O⊂7;²− that reacts?
a) Cr: +6 → +3 (reduced −3 per Cr, −6 total for Cr⊂2;). C: +3 → +4 (oxidized +1 per C, +2 per C⊂2;O⊂4;²−).
Reduction: Cr⊂2;O⊂7;²− + 14H&sup+; + 6e&sup-; → 2Cr³+ + 7H⊂2;O
Oxidation: C⊂2;O⊂4;²− → 2CO⊂2; + 2e&sup-;
Multiply oxidation × 3: 3C⊂2;O⊂4;²− → 6CO⊂2; + 6e&sup-;
Cr⊂2;O⊂7;²− + 14H&sup+; + 3C⊂2;O⊂4;²− → 2Cr³+ + 7H⊂2;O + 6CO⊂2;
b) 6 electrons transferred per Cr⊂2;O⊂7;²− (3 per Cr, ×2 Cr).
A galvanic cell separates the two half-reactions into different compartments. Electrons flow through an external wire from the anode to the cathode, producing an electric current. The spontaneous redox reaction drives the current.
ANode = OXidation. REDuction at CAThode.
In a galvanic cell: anode is the negative electrode (electrons leave here). Cathode is the positive electrode (electrons arrive here).
Electrons flow: anode → wire → cathode. Conventional current is opposite.
Standard Reduction Potentials and E°cell
E°cell = +0.80 − (−0.25) = +0.80 + 0.25 = +1.05 V
E°cell = +1.05 V > 0 → the reaction is spontaneous ✓. This is a galvanic cell.
b) A cell has E°cell = −0.45 V. Is it spontaneous? What type of cell is it?
c) Why does a salt bridge contain ions rather than being a plain wire?
a) Fe has lower E° → Fe is the anode. Cu is the cathode. E°cell = +0.34 − (−0.44) = +0.78 V. Cell notation: Fe | Fe²+(aq) || Cu²+(aq) | Cu
b) E°cell < 0 → not spontaneous. It would require external electrical energy to run → it is an electrolytic cell.
c) As the cell operates, the anode half-cell builds up positive ions (e.g. Zn²+) and the cathode half-cell is depleted of positive ions. Without a salt bridge, charge imbalance would stop the current. The salt bridge allows ions to migrate between compartments to maintain electrical neutrality without mixing the solutions.
Electrolysis uses an external electrical power source to force a non-spontaneous redox reaction to occur. The process is the reverse of a galvanic cell — we push electrons in the unfavored direction.
The external battery drives electrons away from the anode (connected to the positive terminal of the battery). This makes the anode electron-deficient and positive, which attracts anions from solution. At the cathode (connected to the negative terminal), electrons are pushed in, making it negative — it attracts cations which then gain those electrons and are reduced. The half-reaction identities (anode=oxidation, cathode=reduction) remain the same; only the battery polarity has changed.
Faraday’s Law of Electrolysis
The amount of substance deposited or released at an electrode is directly proportional to the amount of charge passed through the cell.
Q = I × t = 3.00 A × 2700 s = 8100 C
n(e&sup-;) = 9000 / 96500 = 0.09326 mol
V(H⊂2;) = 0.04663 × 22.4 = 1.044 L
V(O⊂2;) = 0.02331 × 22.4 = 0.522 L
H⊂2;:O⊂2; = 1.044:0.522 = 2:1 ✓ (as expected from the overall reaction 2H⊂2;O → 2H⊂2; + O⊂2;)
b) How long (in minutes) must a 1.50 A current flow through AlCl⊂3; to deposit 1.00 g of Al? (M(Al) = 26.98 g/mol; Al³+ + 3e&sup-; → Al)
c) In electrolysis of CuCl⊂2;, what happens at the anode? Write the half-reaction.
a) Q = 2.50 × (20.0×60) = 3000 C. n(e&sup-;) = 3000/96500 = 0.03109 mol. n(Cu) = 0.03109/2 = 0.01554 mol.
m(Cu) = 0.01554 × 63.55 = 0.988 g ≈ 0.99 g
b) n(Al) = 1.00/26.98 = 0.03706 mol. n(e&sup-;) = 0.03706 × 3 = 0.11118 mol. Q = 0.11118 × 96500 = 10729 C.
t = Q/I = 10729/1.50 = 7152 s = 119 min
c) At the anode, Cl&sup-; ions are oxidized: 2Cl&sup-; → Cl⊂2; + 2e&sup-;. Cl⊂2; gas is produced at the anode.