Identify the half-cell reaction (oxidation):
H₂(g) → 2H⁺(aq) + 2e⁻
n = 2 electrons transferred
[H⁺] = 0·02 M (HCl is strong acid, fully dissociates)
P(H₂) = 2 atm
Nernst equation for oxidation potential:
E = E° − (0·059/n) × log([H⁺]²/P(H₂))
E = 0 − (0·059/2) × log((0·02)²/2)
E = −0·0295 × log(4×10⁻⁴/2)
E = −0·0295 × log(2×10⁻⁴)
Calculate log(2×10⁻⁴):
log(2×10⁻⁴) = log 2 + log 10⁻⁴ = 0·3010 − 4 = −3·699
E = −0·0295 × (−3·699) = +0·109 V
E = −(0·059/2) × log([H⁺]²/P_H₂)
= −0·0295 × log(2×10⁻⁴)
= −0·0295 × (−3·699) = +0·109 V
The Standard Hydrogen Electrode (SHE) is the universal reference electrode. Standard conditions: H₂ gas at 1 atm pressure, H⁺ concentration = 1 M (activity = 1), temperature = 298 K. E° = 0·000 V by convention (assigned). All other electrode potentials are measured relative to SHE. In this problem, the conditions are non-standard ([H⁺] = 0·02 M ≠ 1 M; P(H₂) = 2 atm ≠ 1 atm), so we must use the Nernst equation to calculate the actual EMF.
E = E° − (RT/nF) × ln Q
At 298 K: E = E° − (0·059/n) × log Q
For oxidation: H₂ → 2H⁺ + 2e⁻
Q = [H⁺]²/P(H₂)
E = 0 − (0·059/2) × log([H⁺]²/P(H₂))
The Nernst equation allows calculation of cell EMF under non-standard conditions. It shows that EMF depends on the concentrations/pressures of reactants and products. At equilibrium: E = 0, and Q = Keq → relationship between EMF and equilibrium constant: E° = (0·059/n) × log Keq. Higher E° → larger Keq → more spontaneous reaction. The equation is named after German physicist Walther Nernst (Nobel Prize 1920).
📌 Standard Reduction Potential (SRP or E°red): measured with reduction half-reaction. IUPAC standard. Higher E°red = stronger oxidising agent.
📌 Standard Oxidation Potential (SOP or E°ox): E°ox = −E°red
📌 Cell EMF: E°cell = E°cathode − E°anode (both in reduction potentials)
📌 OR: E°cell = E°ox(anode) + E°red(cathode)
📌 Spontaneous cell: E°cell > 0 → ΔG° < 0 → Keq > 1
📌 ΔG° = −nFE°cell (n = moles of electrons, F = 96487 C/mol)
The electrochemical series arranges elements in order of their standard reduction potentials. At the top (most negative E°red): Li, K, Na, Mg, Al — strongest reducing agents, weakest oxidising agents. At the bottom (most positive E°red): F₂, Au, Pt — strongest oxidising agents, weakest reducing agents. Applications: (1) Predict if a reaction is spontaneous (higher E°red species oxidises lower E°red species). (2) Determine which metal displaces another (reactive metal displaces less reactive). (3) Calculate cell EMF. (4) Identify cathode and anode.
📌 Galvanic cell: Spontaneous reaction → produces electrical energy. E°cell > 0. Example: Daniel cell (Zn|ZnSO₄||CuSO₄|Cu).
📌 Electrolytic cell: Non-spontaneous → requires external electrical energy. E°cell < 0. Example: electrolysis of water, copper refining.
📌 In both: oxidation at anode, reduction at cathode
📌 Galvanic: anode is negative (−), cathode is positive (+)
📌 Electrolytic: anode is positive (+), cathode is negative (−)
📌 Salt bridge: maintains electrical neutrality, allows ion flow, prevents mixing of solutions
First law: Mass deposited (m) is proportional to charge passed (Q): m = ZQ = Z×I×t, where Z = electrochemical equivalent (g/C). Second law: When the same charge is passed through different electrolytes, masses deposited are proportional to their equivalent weights. Equivalent weight = Molar mass/n-factor. Faraday's constant F = 96487 ≈ 96500 C/mol. One Faraday deposits one gram-equivalent of substance. For Cu (n=2): 96500 C deposits 63/2 = 31·75 g Cu. For Ag (n=1): 96500 C deposits 108 g Ag.
Resistance (R) = ρ × (l/A), where ρ = resistivity. Conductance G = 1/R = κ × (A/l), where κ = conductivity (specific conductance, S/cm). Molar conductance Λm = κ × 1000/M (where M = molarity). Kohlrausch's law: at infinite dilution, molar conductance = sum of individual ion conductances. For strong electrolytes: Λm = Λ°m − b√C (decreases linearly with √C). For weak electrolytes: Λm increases sharply near zero concentration (degree of dissociation increases). Degree of dissociation α = Λm/Λ°m for weak electrolytes.
ΔG = −nFE (for any conditions). ΔG° = −nFE° (for standard conditions). At equilibrium: ΔG = 0 → E = 0 (cell stops working). Relationship with Keq: ΔG° = −RT ln Keq = −nFE°. Therefore: E° = (RT/nF) ln Keq = (0·059/n) log Keq at 298 K. For temperature dependence: (∂E/∂T)P = ΔS/nF — if entropy increases (ΔS > 0), EMF increases with temperature. For the hydrogen electrode: ΔG° = 0 (by convention), E° = 0 V.