Electrochemical cells

Every electrochemical cell has two electrodes sitting in electrolyte(s), joined outside by a metal wire and inside by an ionic path (a salt bridge, or a shared electrolyte). The overall reaction is split into two half-reactions, one happening at each electrode, and this split is what makes the electron flow usable:

• Oxidation — loss of electrons — always happens at the anode.
• Reduction — gain of electrons — always happens at the cathode.

This anode/cathode rule never changes. What changes between the two cell families is where the driving energy comes from, and that difference flips the sign of each electrode.

Galvanic (voltaic) cell. The redox reaction is spontaneous — $\Delta G < 0$ — so it happens on its own and pushes electrons out through the wire, converting chemical energy into electrical energy. Because the anode gives up electrons freely, it becomes electron-rich and is the negative terminal; the cathode, which consumes electrons, is the positive terminal. This is the arrangement inside every torch cell, car battery, and button cell.

Electrolytic cell. The reaction is non-spontaneous — it will not happen by itself — so an external battery is wired in to force electrons through in a chosen direction, converting electrical energy into chemical energy. The battery's positive terminal pulls electrons out of the electrode it is connected to, so that electrode loses electrons (oxidation) and is the anode; correspondingly it is wired positive. The electrode connected to the battery's negative terminal receives the pushed electrons (reduction) and is the cathode, wired negative. This is how molten salts are electrolysed and how metals are electroplated.

Cell notation. A galvanic cell is written in a fixed shorthand, illustrated by the Daniell cell:

where: the single vertical line $\mid$ marks a phase boundary (solid electrode touching its ionic solution), the double line $\parallel$ marks the salt bridge separating the two half-cells, and by convention the anode (oxidation) is always written on the left, the cathode (reduction) always on the right — regardless of how the apparatus actually sits on the bench.

Mechanism of current production. In the Daniell cell, zinc metal oxidises, $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$, releasing electrons into the zinc electrode. Those electrons travel through the external wire to the copper electrode, where they are consumed by $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$. Meanwhile, inside the cell, the salt bridge — an inert electrolyte such as $\text{KNO}_3$ in agar gel — keeps both half-cells electrically neutral: as $\text{Zn}^{2+}$ builds up in the anode compartment, anions from the bridge drift in to balance it; as $\text{Cu}^{2+}$ is used up in the cathode compartment, cations drift in to replace the lost positive charge. The salt bridge carries no electrons itself — only ions — and without it the half-cells would charge up and the current would stop almost immediately.

Understanding why the sign of each electrode flips between the two cell types matters more than memorising it, because a rule you understand cannot be misremembered under exam pressure.

Step 1. Start from the definition each family shares: oxidation happens at the anode, reduction at the cathode, in every electrochemical cell without exception. This part of the rule is fixed by the chemistry, not by the wiring.

Step 2. In a galvanic cell, nothing forces the electrons — the reaction itself is spontaneous and wants to happen. The anode metal (zinc, in the Daniell cell) genuinely gives away electrons because doing so is energetically favourable, so the anode accumulates negative charge relative to the cathode: the anode is the negative terminal, the cathode positive. The cell's own chemistry sets the polarity.

Step 3. In an electrolytic cell, the reaction does not want to happen — it is non-spontaneous, so nothing internal will make an electrode give up its electrons. Instead, an external battery is connected, and the battery's polarity dictates everything: its positive terminal actively withdraws electrons from the electrode it touches, forcing that electrode to oxidise and become the anode, while its negative terminal supplies electrons to the other electrode, forcing it to reduce and become the cathode. The wiring sets the polarity, and consequently it is now the anode that is positive and the cathode negative — the exact reverse of the galvanic case.

Step 4. So the anode/cathode identity (defined by which reaction happens there) never changes, but the positive/negative labelling flips because the source of the driving force flips — from the reaction itself in a galvanic cell, to an external battery in an electrolytic cell. Recognising which situation you are in — is a reaction driving the current, or is a battery forcing it — immediately tells you every polarity in the diagram.

Problem: A student sets up two experiments. Experiment A is the Daniell cell, $\text{Zn} \mid \text{Zn}^{2+} \parallel \text{Cu}^{2+} \mid \text{Cu}$. Experiment B is the electrolysis of molten $\text{NaCl}$ using inert electrodes wired to a battery. For each, identify the half-reactions, name the anode and cathode, state their polarity, and classify the cell type.

Step 1. Experiment A — read the notation. By convention the left electrode is the anode: $\text{Zn(s)} \rightarrow \text{Zn}^{2+}(aq) + 2e^-$ (oxidation). The right electrode is the cathode: $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu(s)}$ (reduction). Adding them, the electrons cancel and the overall reaction is $\text{Zn(s)} + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu(s)}$, which is spontaneous — this is a galvanic cell. Electrons flow Zn to Cu in the external wire, so the zinc anode is negative and the copper cathode is positive.

Step 2. Experiment B — no reaction happens without the battery, so molten NaCl electrolysis is an electrolytic cell. $\text{Na}^+$ ions migrate to the electrode wired to the battery's negative terminal and are reduced: $\text{Na}^+ + e^- \rightarrow \text{Na}$ — this electrode is the cathode, and it is negative. $\text{Cl}^-$ ions migrate to the electrode wired to the battery's positive terminal and are oxidised: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$ — this electrode is the anode, and it is positive. Overall: $2\text{NaCl(l)} \xrightarrow{\text{electrolysis}} 2\text{Na(l)} + \text{Cl}_2(g)$.

Step 3. Compare the two. Both obey oxidation-at-anode, reduction-at-cathode. But the polarity is exactly reversed: in A the anode is negative and the cathode positive (the reaction drives the electrons); in B the anode is positive and the cathode negative (the battery drives the electrons). Answer: Experiment A is a galvanic cell with Zn as the negative anode and Cu as the positive cathode; Experiment B is an electrolytic cell with the Cl₂-forming electrode as the positive anode and the Na-forming electrode as the negative cathode.

• Assuming the anode is always negative — true only for galvanic cells; in an electrolytic cell the anode is positive because it is wired to the battery's positive terminal.
• Writing the cell notation with oxidation on the wrong side — the anode (oxidation) must be on the left and the cathode (reduction) on the right, no matter how the actual apparatus is laid out on the bench.
• Thinking the salt bridge carries electrons — it only carries ions, maintaining electrical neutrality in each half-cell; the electrons travel exclusively through the external wire.
• Forgetting that "oxidation at anode, reduction at cathode" holds in every electrochemical cell — only the sign convention of the electrodes changes between galvanic and electrolytic types.

Ask the tutor: if you would like more practice writing cell notations for other metal-ion pairs, or a step-by-step walkthrough of electroplating as an electrolytic cell.