Electrochemistry connects chemical thermodynamics and chemical equilibrium to the flow of electrons in chemical reactions. This guide covers balancing redox equations, galvanic (voltaic) cells, standard cell potentials, the Nernst equation, and electrolysis — with detailed worked examples for the calculation types most commonly assigned at the university level.
1. Oxidation and Reduction: The Basics
- Oxidation: loss of electrons (LEO — Lose Electrons, Oxidation); oxidation number increases.
- Reduction: gain of electrons (GER — Gain Electrons, Reduction); oxidation number decreases.
- Oxidizing agent: the species that is reduced (it causes oxidation in another species by accepting electrons).
- Reducing agent: the species that is oxidized (it causes reduction in another species by donating electrons).
Worked Example: In Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), zinc is oxidized (0 → +2, loses electrons) and Cu²⁺ is reduced (+2 → 0, gains electrons). Zn is the reducing agent; Cu²⁺ is the oxidizing agent.
2. Assigning Oxidation Numbers
Key rules (in priority order): free elements = 0; monatomic ions = ion charge; oxygen = −2 (except peroxides, −1); hydrogen = +1 (except metal hydrides, −1); the sum of oxidation numbers in a neutral compound = 0, and in a polyatomic ion = the ion’s charge.
Worked Example: Find the oxidation number of Mn in MnO₄⁻.
Let x = oxidation number of Mn. 4(−2) + x = −1 → x − 8 = −1 → x = +7
3. Balancing Redox Equations (Half-Reaction Method)
Worked Example — Balance in acidic solution: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺
Step 1 — Split into half-reactions: Reduction: MnO₄⁻ → Mn²⁺ Oxidation: Fe²⁺ → Fe³⁺
Step 2 — Balance atoms other than O and H: (already balanced: 1 Mn, 1 Fe)
Step 3 — Balance O using H₂O, then H using H⁺: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Step 4 — Balance charge using electrons: Left side charge: (−1) + 8(+1) = +7. Right side charge: +2. Add 5 e⁻ to the left: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
For iron: Fe²⁺ → Fe³⁺ + e⁻
Step 5 — Equalize electrons and add half-reactions: Multiply the iron half-reaction by 5: 5Fe²⁺ → 5Fe³⁺ + 5e⁻
Add both half-reactions (electrons cancel): MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
Step 6 — Check: Atoms and charge both balance (left charge: −1+8+10=+17… let’s verify: −1 + 8(+1) + 5(+2) = −1+8+10 = +17; right charge: +2 + 5(+3) = +2+15 = +17 ✓)
For basic solution, balance as if acidic first, then add OH⁻ to both sides to neutralize every H⁺ (forming H₂O), and simplify.
4. Galvanic (Voltaic) Cells
A galvanic cell converts spontaneous redox reaction energy into electrical energy, using two separated half-cells connected by a wire (external circuit) and a salt bridge (to maintain charge neutrality).
- Anode: electrode where oxidation occurs (negative terminal in a galvanic cell).
- Cathode: electrode where reduction occurs (positive terminal in a galvanic cell).
- Electrons flow through the external wire from anode to cathode. Remember: “An Ox” (anode = oxidation) and “Red Cat” (reduction = cathode).
4.1 Cell Notation
Convention: Anode | Anode solution || Cathode solution | Cathode
Worked Example: Write the cell notation for Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).
Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s)
5. Standard Cell Potential (E°cell)
E°cell = E°cathode − E°anode (using standard reduction potentials from a table)
Worked Example: Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, find E°cell for the Zn-Cu galvanic cell.
Cu²⁺ is reduced (cathode), Zn is oxidized (anode): E°cell = E°cathode − E°anode = 0.34 − (−0.76) = +1.10 V
A positive E°cell confirms the reaction is spontaneous as written, consistent with it being a working galvanic cell.
5.1 Relating E°cell to ΔG° and K
Connecting back to chemical thermodynamics and chemical equilibrium:
ΔG° = −nFE°cell and ΔG° = −RT ln K, so E°cell = (RT/nF) ln K
where n = moles of electrons transferred, and F = Faraday’s constant = 96,485 C/mol.
Worked Example: Calculate ΔG° for the Zn-Cu cell above (n = 2).
ΔG° = −nFE°cell = −(2)(96,485)(1.10) = −212,267 J/mol ≈ −212.3 kJ/mol
The large negative ΔG° confirms strong spontaneity, corresponding to a very large equilibrium constant K.
6. The Nernst Equation: Non-Standard Conditions
Real cells rarely operate at standard 1 M concentrations. The Nernst equation adjusts cell potential for actual conditions:
Ecell = E°cell − (RT/nF) ln Q, or at 25°C using base-10 log:
Ecell = E°cell − (0.0592/n) log Q
Worked Example: For the Zn-Cu cell (E°cell = 1.10 V, n = 2), find Ecell when [Zn²⁺] = 0.50 M and [Cu²⁺] = 0.020 M.
Q = [Zn²⁺]/[Cu²⁺] = 0.50/0.020 = 25.0
Ecell = 1.10 − (0.0592/2) log(25.0) = 1.10 − (0.0296)(1.398) = 1.10 − 0.0414 = 1.059 V
Notice the cell potential decreases slightly as product concentration ([Zn²⁺]) increases relative to reactant concentration — consistent with Le Chatelier reasoning from chemical equilibrium, where the reaction becomes less “forward-favored” as it proceeds.
6.1 Concentration Cells
A concentration cell has identical electrodes but different ion concentrations in each half-cell, so E°cell = 0, and the entire potential arises from the concentration difference (Q ≠ 1).
Worked Example: A Cu concentration cell has [Cu²⁺] = 1.0 M in the cathode compartment and 0.010 M in the anode compartment (n = 2).
Ecell = 0 − (0.0592/2) log(0.010/1.0) = −(0.0296)(−2) = +0.0592 V
7. Electrolytic Cells and Electrolysis
Unlike galvanic cells (spontaneous, ΔG° < 0), electrolytic cells use an external electrical power source to force a non-spontaneous reaction to occur (ΔG° > 0). The definitions of anode (oxidation) and cathode (reduction) remain the same, but the polarity is reversed compared to a galvanic cell (anode is now positive, cathode is now negative).
7.1 Faraday’s Laws and Quantitative Electrolysis
moles of electrons = It / F, where I = current (amps), t = time (seconds), F = 96,485 C/mol
Worked Example: How many grams of copper are deposited by passing 2.00 A of current through a Cu²⁺ solution for 1.00 hour?
t = 3600 s Charge Q = It = 2.00 × 3600 = 7200 C Moles of electrons = 7200/96,485 = 0.0746 mol e⁻
Since Cu²⁺ + 2e⁻ → Cu (n = 2): Moles Cu = 0.0746/2 = 0.0373 mol Mass Cu = 0.0373 mol × 63.55 g/mol = 2.37 g
8. Common Assignment Pitfalls
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- Forgetting that E° values are intensive properties — do NOT multiply half-reaction potentials by stoichiometric coefficients when combining half-reactions (unlike ΔH or ΔG, which are extensive).
- Mixing up anode/cathode polarity between galvanic cells (anode negative) and electrolytic cells (anode positive).
- Forgetting to balance electrons before adding half-reactions (Section 3, Step 5).
- Using the wrong sign convention in the Nernst equation — always write Q with products over reactants, matching the overall balanced cell reaction.
- Forgetting that a spontaneous reaction has E°cell > 0, which corresponds to ΔG° < 0 and K > 1 — three equivalent ways of expressing the same thermodynamic favorability.
9. Full Worked Problem
Question: A galvanic cell is built from Ag⁺/Ag (E° = +0.80 V) and Ni²⁺/Ni (E° = −0.25 V) half-cells. Determine the cell reaction, E°cell, and whether the reaction is spontaneous.
Solution: Ag⁺ has the higher (more positive) reduction potential, so it is reduced at the cathode; Ni is oxidized at the anode.
Cathode (×2 to balance electrons): 2Ag⁺ + 2e⁻ → 2Ag Anode: Ni → Ni²⁺ + 2e⁻
Overall: 2Ag⁺(aq) + Ni(s) → 2Ag(s) + Ni²⁺(aq)
E°cell = E°cathode − E°anode = 0.80 − (−0.25) = +1.05 V
Since E°cell > 0, the reaction is spontaneous as written.
Electrochemistry beautifully ties together chemical thermodynamics (via ΔG°), chemical equilibrium (via K), and redox principles that also appear in organic reaction mechanisms involving oxidation states of carbon. Practice both half-reaction balancing and Nernst equation calculations, since assignments typically test both skills together in multi-part problems.
