Galvanic vs Electrolytic Cells: The Nernst Equation Worked
Galvanic (voltaic) cells generate electricity from spontaneous redox; electrolytic cells use electricity to drive non-spontaneous reactions. Here is exactly how each works with Nernst equation and three worked cell calculations.
Learning Objectives
- ✓Distinguish galvanic from electrolytic cells by sign convention, polarity, and direction of current flow.
- ✓Compute cell potential from standard reduction potentials and Nernst-equation corrections.
- ✓Apply Faraday\u2019s laws of electrolysis to predict mass deposited.
1. Direct Answer: Spontaneous vs Driven
A GALVANIC CELL (or voltaic cell) generates electric current from a SPONTANEOUS redox reaction — ΔG < 0 and E°_cell > 0. Anode is NEGATIVE (oxidation), cathode is POSITIVE (reduction); electrons flow from anode to cathode through the external circuit. Standard examples: Daniell cell (Zn|Zn²⁺||Cu²⁺|Cu), most batteries, fuel cells. An ELECTROLYTIC CELL drives a NON-SPONTANEOUS reaction by applying external voltage — ΔG > 0 and E°_cell < 0 in the spontaneous direction. Anode is POSITIVE (oxidation forced), cathode is NEGATIVE (reduction forced); electrons still go from external source through cathode. Examples: electroplating, electrolysis of water, refining of copper, charging a battery. The MNEMONIC: in a galvanic cell, the anode is the unhappy electrode (negative) because oxidation is reluctant; in an electrolytic cell, you FORCE oxidation, so the anode is happy and you bias it positive.
Key Points
- •Galvanic: spontaneous, generates current, E°_cell > 0, anode negative.
- •Electrolytic: driven by external voltage, anode positive, E°_cell < 0 in spontaneous direction.
- •In BOTH: oxidation at anode, reduction at cathode.
2. Standard Reduction Potentials
Standard reduction potentials (E°, in volts) are tabulated for half-reactions written as reductions, all at 25°C, 1 atm, 1 M solutions. Hydrogen reduction (2H⁺ + 2e⁻ → H₂) is defined as 0.00 V. Above hydrogen in the table (more positive E°) are easily-reduced species like F₂ (+2.87 V) — strong oxidizing agents. Below hydrogen (more negative E°) are easily-oxidized species like Li (E°_red = −3.05 V for Li⁺/Li, so Li metal is a strong reducing agent). For any galvanic cell, E°_cell = E°_cathode − E°_anode (using BOTH as reduction values; do NOT flip the anode sign). If E°_cell > 0, the cell as written is spontaneous and galvanic. If E°_cell < 0, the cell is non-spontaneous and would need driving by an external source.
Key Points
- •All E° values tabulated as reductions; 25°C, 1 atm, 1 M.
- •E°_cell = E°_cathode − E°_anode (no flipping signs).
- •Positive E°_cell → spontaneous galvanic; negative → must be driven electrolytically.
3. The Nernst Equation
Real cells rarely operate at standard 1 M concentrations. The NERNST EQUATION corrects E for actual concentrations: E = E° − (RT/nF) × ln Q, where R = 8.314 J/mol·K, T is absolute temperature, n is electrons transferred per cell reaction, F = 96,485 C/mol, and Q is the reaction quotient. At 25°C this simplifies to E = E° − (0.0257/n) × ln Q, or equivalently E = E° − (0.0592/n) × log₁₀ Q. As products accumulate Q grows and E drops — the driving force decreases as the cell discharges. At equilibrium Q = K and E = 0; no further work can be extracted. The Nernst equation is also the basis for pH meters (concentration cells with hydrogen reduction at known and unknown pH) and ion-selective electrodes.
Key Points
- •Nernst (25°C): E = E° − (0.0592/n) × log₁₀ Q.
- •As products accumulate, Q grows, E drops, cell discharges.
- •At equilibrium Q = K and E = 0 — battery is dead.
4. Worked Example 1: Daniell Cell
Zn(s) | Zn²⁺(0.10 M) || Cu²⁺(1.00 M) | Cu(s). Standard reduction potentials: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V; Zn²⁺ + 2e⁻ → Zn, E° = −0.76 V. E°_cell = +0.34 − (−0.76) = +1.10 V (spontaneous, galvanic). n = 2. Q = [Zn²⁺] / [Cu²⁺] = 0.10 / 1.00 = 0.10. Nernst correction: E = 1.10 − (0.0592/2) × log(0.10) = 1.10 − 0.0296 × (−1) = 1.10 + 0.0296 = 1.13 V. Lower [Zn²⁺] than [Cu²⁺] makes the cell run slightly hotter than standard, because the unfavorable buildup of Zn²⁺ hasn\u2019t accumulated yet. As the cell discharges, [Zn²⁺] grows and [Cu²⁺] drops, Q rises, and E falls toward 0.
Key Points
- •Daniell standard E°_cell = +1.10 V.
- •Nernst correction is small for modest concentration deviations.
- •Cell voltage drops as products accumulate during discharge.
5. Worked Example 2: Concentration Cell
A concentration cell has IDENTICAL half-reactions on both sides differing only in concentration. Example: Cu | Cu²⁺(0.01 M) || Cu²⁺(1.0 M) | Cu. E° = 0 because both half-cells use the same metal. n = 2. Q = [Cu²⁺]_anode / [Cu²⁺]_cathode = 0.01 / 1.0 = 0.01. Nernst: E = 0 − (0.0592/2) × log(0.01) = 0 − 0.0296 × (−2) = +0.0592 V. The cell drives current spontaneously even though E° = 0, because the system favors equalizing concentrations — dilute side oxidizes (loses Cu, raises [Cu²⁺]), concentrated side reduces (deposits Cu, lowers [Cu²⁺]). pH meters work on the same principle with hydrogen half-cells at different [H⁺].
Key Points
- •Concentration cells have E° = 0 but generate current from concentration gradient.
- •Direction: dilute side oxidizes, concentrated side reduces.
- •pH meters use this principle with hydrogen reduction at known/unknown [H⁺].
6. Worked Example 3: Electroplating Mass Calculation
Electroplating silver onto a spoon: Ag⁺ + e⁻ → Ag, n = 1. Pass current of 2.0 A for 30 minutes. Total charge Q = I × t = 2.0 × 1800 = 3600 C. Moles of electrons = Q / F = 3600 / 96,485 = 0.0373 mol. Since n = 1 for Ag, moles of Ag deposited = 0.0373 mol. Mass = 0.0373 × 107.87 g/mol = 4.03 g of silver deposited. FARADAY\u2019S FIRST LAW: mass deposited is proportional to charge passed. FARADAY\u2019S SECOND LAW: for a given charge, mass deposited is proportional to equivalent weight (molar mass / n). Industrial electrolysis (aluminum production via Hall-Héroult, chlorine via chlor-alkali) uses these laws to design current and time requirements.
Key Points
- •Q (charge in coulombs) = I × t.
- •Moles of electrons = Q / F (F = 96,485 C/mol).
- •Moles of product = moles of electrons / n.
7. Cell Diagrams and Sign Conventions
The shorthand for a galvanic cell is anode | anode solution || cathode solution | cathode, where single bar means phase boundary and double bar is the salt bridge. The Daniell cell: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Reading left to right gives oxidation, then reduction — the anode-then-cathode order. SIGN CONVENTIONS in a galvanic cell: anode is NEGATIVE (the source of electrons that flow OUT through the wire), cathode is POSITIVE (where electrons flow IN). The conventional CURRENT flows the OPPOSITE direction (from + to −) in the external circuit because the convention was set before electrons were discovered. Always check the sign of E°_cell against the diagram direction — if E°_cell as written is negative, you wrote the diagram backwards and would need to flip it (or apply external voltage to drive it as written).
Key Points
- •Cell diagram: anode | anode solution || cathode solution | cathode.
- •Galvanic: anode negative, cathode positive; electrons flow anode→cathode externally.
- •Conventional current flows opposite to electrons (cathode→anode externally).
8. Running Cell Calculations in ChemistryIQ
Provide the half-reactions and concentrations and ChemistryIQ computes E°_cell from standard reduction potentials, applies the Nernst equation for the given concentrations, identifies whether the cell is galvanic or electrolytic, and produces the cell diagram and direction of electron flow. For electrolysis problems, ChemistryIQ converts between charge, current, time, and mass deposited using Faraday\u2019s laws. This content is for educational purposes only.
Key Points
- •E°_cell + Nernst correction with cell-direction check.
- •Diagram and electron-flow direction automatically determined.
- •Faraday\u2019s laws applied for electroplating and electrolysis problems.
High-Yield Facts
- ★Galvanic: spontaneous, E°_cell > 0, anode negative; Electrolytic: driven, E°_cell < 0, anode positive.
- ★E°_cell = E°_cathode − E°_anode (no flipping signs).
- ★Nernst (25°C): E = E° − (0.0592/n) × log Q.
- ★At equilibrium Q = K and E = 0.
- ★Faraday\u2019s laws: moles electrons = Q/F = (I × t) / 96,485.
Practice Questions
1. Zn²⁺/Zn has E° = −0.76 V; Ag⁺/Ag has E° = +0.80 V. What is E°_cell for Zn | Zn²⁺ || Ag⁺ | Ag?
2. For a hydrogen concentration cell with [H⁺]_anode = 1 × 10⁻⁷ M (pH 7) and [H⁺]_cathode = 1.0 M (pH 0), find E at 25°C.
3. How much copper deposits on a cathode in 1 hour with a current of 1.5 A in CuSO₄ solution?
FAQs
Common questions about this topic
Without a salt bridge, the anode solution would become positively charged (as it loses electrons but gains metal cations from oxidation) and the cathode solution negatively charged (as metal cations are reduced). This charge imbalance would stop the reaction immediately. The salt bridge supplies inert ions (typically KCl or KNO₃) that flow into each half-cell to maintain electroneutrality, allowing the reaction to continue.
A discharging battery approaches equilibrium between its reactants and products. As reactants deplete and products accumulate, Q approaches K and the Nernst equation drives E toward 0. When E = 0 no more electrical work can be extracted regardless of remaining mass. For most consumer batteries the practical end of life arrives before the chemical equilibrium because internal resistance grows and voltage drops below the device\u2019s working threshold.
Aluminum has a very negative E° of reduction (Al³⁺ + 3e⁻ → Al, E° = −1.66 V), meaning Al³⁺ is HARD to reduce — no common chemical reducer is strong enough at industrial scale without producing byproducts. The Hall-Héroult process uses electrolysis of molten aluminum oxide in cryolite to drive the reduction. The high cost of electricity is why aluminum was a luxury metal in the 19th century (the cap of the Washington Monument was aluminum, then more valuable than silver) and remained so until cheap electricity made it ubiquitous.
It means the REDUCTION half-reaction as written is non-spontaneous relative to the standard hydrogen electrode at standard conditions. The reverse — OXIDATION — would be spontaneous. So a metal with a very negative E°_red is a good REDUCING agent (it gives up electrons easily). Li is the strongest common reducing agent at E°_red = −3.05 V.
Yes. Provide the half-reactions (handwritten or typed) and concentrations and ChemistryIQ computes E°_cell from standard reduction potentials, applies the Nernst equation, identifies galvanic vs electrolytic, and produces the cell diagram and electron-flow direction. For electrolysis problems it handles Faraday\u2019s laws conversions automatically. This content is for educational purposes only.