Electrochemical vs Electrolytic Cells — The Difference That Confuses Everyone
Almost every student meets the same trap in the electrochemistry chapter: "the anode is negative" is memorised from the galvanic cell, and then the electrolysis question arrives where the anode is positive, and the whole answer collapses. The two cell types are not opposites in every respect — some things stay exactly the same. This article separates what changes from what does not, and works the standard numericals through step by step.
The one difference everything else follows from
Both devices connect chemistry to electricity, but they run in opposite directions:
- A galvanic (voltaic / electrochemical) cell runs a spontaneous redox reaction and produces electrical energy. A dry cell, a lead-acid battery and the Daniell cell are all galvanic.
- An electrolytic cell uses an external power supply to force a non-spontaneous reaction to happen. Electroplating, extraction of aluminium and electrolysis of water are all electrolytic.
Every other difference — the sign of Ecell, the sign of ΔG, the polarity of the electrodes — is a consequence of that single fact.
What stays the SAME in both (memorise this first)
Anions always move toward the anode. Cations always move toward the cathode.
Electrons always travel through the external wire, never through the solution.
These four statements are true in both cell types, with no exceptions. If you remember only them, you can rebuild the rest by reasoning. The Latin roots help: anode goes with anions, cathode goes with cations, and "AN OX / RED CAT" (anode = oxidation, reduction = cathode) is the standard memory hook.
Why the electrode polarity flips
In a galvanic cell nothing is pushing the electrons; the oxidation reaction itself pumps them out. Electrons pile up at the anode, so the anode is the negative terminal and the cathode is positive.
In an electrolytic cell the battery decides the polarity. The electrode wired to the positive terminal of the supply has electrons pulled out of it, so oxidation is forced there — making the positive electrode the anode. The electrode wired to the negative terminal is pushed full of electrons, so reduction is forced there, making the negative electrode the cathode.
So the label "anode" never changes meaning. Only the sign changes, because in one case the chemistry sets the polarity and in the other the power supply does.
Side-by-side comparison
| Feature | Galvanic / electrochemical cell | Electrolytic cell |
|---|---|---|
| Energy conversion | Chemical → electrical | Electrical → chemical |
| Reaction | Spontaneous | Non-spontaneous, forced |
| Ecell | Positive | Negative for the forced reaction; supply must exceed it |
| ΔG | Negative | Positive |
| Anode sign | Negative | Positive |
| Cathode sign | Positive | Negative |
| Electrolytes | Usually two half-cells joined by a salt bridge | Usually one electrolyte, both electrodes in it |
| External source | None | Battery or DC supply required |
| Typical use | Batteries, fuel cells, sensors | Electroplating, metal extraction, refining |
| Oxidation site | Anode | Anode |
| Reduction site | Cathode | Cathode |
Worked example 1 — EMF of the Daniell cell
Cell notation: Zn(s) | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu(s). By convention the anode is written on the left. Standard electrode potentials: E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V.
E°cell = E°cathode − E°anode
E°cell = (+0.34) − (−0.76) = 0.34 + 0.76 = +1.10 V
The value is positive, so the reaction Zn + Cu²⁺ → Zn²⁺ + Cu is spontaneous and this is a galvanic cell. Zinc, the more easily oxidised metal, is the anode.
Worked example 2 — turning EMF into Gibbs energy
For the Daniell cell, 2 electrons are transferred (Zn → Zn²⁺ + 2e⁻), so n = 2.
ΔG° = −(2)(96485 C mol⁻¹)(1.10 V)
2 × 96485 = 192970
192970 × 1.10 = 212267 J mol⁻¹
ΔG° = −212267 J mol⁻¹ ≈ −212.3 kJ mol⁻¹
Negative ΔG° confirms spontaneity. Reverse the same cell into an electrolytic arrangement — force Cu + Zn²⁺ → Cu²⁺ + Zn — and both signs flip: E° = −1.10 V and ΔG° = +212.3 kJ mol⁻¹, which is why an external supply of more than 1.10 V is needed.
Worked example 3 — how much metal an electrolytic cell deposits
Electrolytic cells are quantitative. Faraday's first law connects charge to amount:
A current of 2.00 A is passed through CuSO₄ solution for 30.0 minutes. How much copper is deposited at the cathode? (Cu²⁺ + 2e⁻ → Cu, so n = 2; M(Cu) = 63.55 g mol⁻¹.)
t = 30.0 × 60 = 1800 s
Q = 2.00 × 1800 = 3600 C
moles of electrons = 3600 / 96485 = 0.03731 mol
moles of Cu = 0.03731 / 2 = 0.018655 mol
mass = 0.018655 × 63.55 = 1.19 g
Sanity check: one full faraday (96485 C) would give half a mole of copper, about 31.8 g. Our charge is roughly 1/27 of a faraday, and 31.8/27 ≈ 1.18 g. The answer is the right size.
Common mistakes that cost marks
- "The anode is negative." True only for galvanic cells. In electrolysis the anode is positive. Say why in the answer — that alone often carries the mark.
- Swapping oxidation and reduction in electrolysis. Oxidation is at the anode in both cells. The site never moves.
- Subtracting the wrong way. E°cell = cathode − anode, and both values must be reduction potentials. Never reverse the sign of a potential and subtract as well — that is double counting.
- Multiplying E° when you balance electrons. Electrode potential is an intensive property. Doubling the equation does not double E°; it only changes n in ΔG° = −nFE°.
- Saying electrons flow through the solution. Inside the electrolyte, charge is carried by ions moving; electrons travel only through the wire.
- Forgetting the salt bridge's job. It completes the circuit and keeps both half-cells electrically neutral. Remove it and the current stops almost at once.
- Ignoring which ion is actually discharged. In aqueous electrolysis water itself can be oxidised or reduced, so the product is not always the obvious ion — state the competing electrode reactions before choosing.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 12 | Distinguish the two cells in five points; label a diagram in words; write electrode reactions |
| NEET / JEE | E°cell from a table of standard potentials; feasibility of a displacement reaction |
| IIT-JAM / CUET-PG | ΔG° from EMF, Nernst equation at non-standard concentrations |
| GATE / CSIR-NET | Faraday's laws with mixed electrolytes, overpotential, electrode process selectivity |
Test a cell's EMF at real concentrations. Standard potentials only apply at 1 M. The Nernst Equation calculator lets you put in E°, n and the actual concentrations and returns the working cell potential — the fastest way to see how a galvanic cell fades as the reactant runs out.
Open the Nernst Equation (Cell EMF) Calculator →Electrochemistry carries steady weight in the Class 12 syllabus and rewards regular numerical practice. ABC Chemistry runs Class 11–12 chemistry at the Gurugram centre and online across India — see abcchemistry.in.