Galvanic (Voltaic) and Electrolytic Cells: Study Guide (Chemistry 9.8)

Two kinds of electrochemical cell

In Topic 9.7 you saw that electrical energy can drive a reaction that would not happen by itself. An electrochemical cell is the device that links a redox reaction to an electric current. There are two kinds:

	Galvanic (voltaic) cell	Electrolytic cell
Redox reaction	thermodynamically favourable (ΔG° < 0)	thermodynamically unfavourable (ΔG° > 0)
Energy change	chemical energy → electrical energy	electrical energy → chemical energy
What is in the circuit	a voltmeter, bulb or other device that uses the current	a power supply (battery or DC source) that pushes the current
Everyday example	a battery powering a torch	recharging that battery; electroplating; extracting reactive metals from molten salts
Oxidation happens at	the anode	the anode
Reduction happens at	the cathode	the cathode

The last two rows are the same for both kinds. That is the most useful rule in this topic: anode = oxidation, reduction = cathode (“an ox, red cat”).

What each part of a galvanic cell does

The figure shows a galvanic cell built from lead and silver. One beaker holds a lead strip in lead(II) nitrate solution; the other holds a silver strip in silver nitrate solution.

• Electrodes. Solid conductors where electrons are lost or gained. Here both electrodes take part in the reaction (lead dissolves, silver is deposited). Some cells use inert electrodes, such as graphite or platinum, which only carry electrons.
• Half-cell solutions. They contain the ions that are oxidised or reduced (here Ag⁺) or that form (here Pb²⁺), and they let ions move.
• Wire and voltmeter. Electrons travel through the external wire. A voltmeter measures the potential difference between the electrodes; an ammeter or bulb shows that current is flowing.
• Salt bridge. A tube or paper strip soaked in a solution of an unreactive salt (KNO₃ here). It completes the circuit and keeps each beaker electrically neutral, while stopping the two solutions from mixing. If Ag⁺ ions reached the lead strip directly, they would react on the lead surface and the electrons would never pass through the wire.

Why ions move the way they do

As the cell runs, the anode beaker gains Pb²⁺ ions, so it would build up positive charge. The cathode beaker loses Ag⁺ ions, leaving extra NO₃⁻, so it would build up negative charge. Charge build-up would stop the current within moments. The salt bridge prevents this: NO₃⁻ ions move into the anode beaker and K⁺ ions move into the cathode beaker. The general rule is that anions move towards the anode and cations move towards the cathode.

The view at the particle level

• At the anode surface: Pb atoms each lose two electrons and become Pb²⁺ ions that move off into the solution. The electrons travel through the metal into the wire. Over time the lead strip gets thinner.
• At the cathode surface: Ag⁺ ions from the solution reach the silver surface, each gain one electron from the metal, and become Ag atoms stuck to the electrode. The silver strip gets heavier and the Ag⁺ concentration falls.

Electrolytic cells

In an electrolytic cell, a power supply replaces the voltmeter. It pulls electrons from one electrode and pushes them into the other, forcing an unfavourable reaction to happen.

The overall reaction, MgCl₂(l) → Mg(l) + Cl₂(g), is the reverse of the very favourable reaction between magnesium and chlorine, so it needs a continuous supply of electrical energy. Notice:

• Oxidation (of Cl⁻) is still at the anode; reduction (of Mg²⁺) is still at the cathode.
• Electrons still flow through the external circuit from the anode to the cathode; the power supply is what moves them.
• There is no salt bridge, because both half-reactions happen in the same molten salt. The ions of the melt carry the charge: Cl⁻ moves towards the anode and Mg²⁺ towards the cathode.

About electrode signs (not assessed). In a galvanic cell the anode is the negative electrode; in an electrolytic cell the anode is the one joined to the positive terminal of the supply. The signs swap between the two kinds of cell, and this course does not assess them. Use “oxidation at the anode” instead.

Worked example 1: reading a galvanic cell from observations

Question. The cell in Figure 1 runs for an hour. The silver electrode gains 0.863 g. (a) Identify the anode and cathode and write the half-reactions and the overall equation. (b) State the direction of electron flow and of K⁺ and NO₃⁻ movement. (c) Find the change in mass of the lead electrode.

(a) The silver electrode gains mass, so Ag⁺ ions are being reduced to Ag there. Reduction happens at the cathode, so silver is the cathode and lead is the anode.

• Anode (oxidation): Pb(s) → Pb²⁺(aq) + 2e⁻
• Cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s)
• Multiply the cathode equation by 2 so the electrons cancel: Pb(s) + 2Ag⁺(aq) → Pb²⁺(aq) + 2Ag(s)

(b) Electrons flow through the wire from the lead electrode to the silver electrode. In the salt bridge, NO₃⁻ moves towards the lead beaker and K⁺ moves towards the silver beaker.

(c)

1. Moles of Ag deposited: 0.863 g ÷ 107.87 g mol⁻¹ = 0.008000 mol.
2. Each Ag needs 1 electron, so 0.008000 mol of electrons passed.
3. Each Pb gives 2 electrons, so moles of Pb oxidised = 0.008000 ÷ 2 = 0.004000 mol.
4. Mass of Pb lost: 0.004000 mol × 207.2 g mol⁻¹ = 0.829 g.

Answer. The lead electrode loses 0.829 g.

Check. Two Ag are deposited for every Pb that dissolves, but a Pb atom is almost twice as heavy as an Ag atom, so the two mass changes should be similar in size. They are.

Worked example 2: an electrolytic cell for magnesium

Question. The cell in Figure 2 produces 4.86 g of magnesium. (a) Explain why this cell needs a power supply. (b) Write the half-reactions and say where gas forms. (c) What volume would the chlorine occupy at 298 K and 1.00 atm?

(a) Magnesium and chlorine react together very favourably to form MgCl₂. The cell runs the reverse of that reaction, which is therefore thermodynamically unfavourable (ΔG° > 0). It happens only while the power supply provides electrical energy, so this is an electrolytic cell.

(b) Anode (oxidation): 2Cl⁻ → Cl₂(g) + 2e⁻. Cathode (reduction): Mg²⁺ + 2e⁻ → Mg. Gas (chlorine) forms at the anode.

(c)

1. Moles of Mg: 4.86 g ÷ 24.31 g mol⁻¹ = 0.1999 mol.
2. Both half-reactions involve 2 electrons, so 1 mol of Cl₂ forms for every 1 mol of Mg: n(Cl₂) = 0.1999 mol.
3. V = nRT / P = 0.1999 mol × 0.08206 L atm mol⁻¹ K⁻¹ × 298 K ÷ 1.00 atm = 4.89 L.

Interpretation. The electron count links the two electrodes: whatever passes into the cathode must have left the anode.

Changing the set-up: what happens?

Exam questions often ask how a change to the apparatus changes the results. Reason from the job each part does.

Change	Effect	Reason
Remove the salt bridge	Current stops almost at once	Charge builds up in each beaker; there is no path for ions to balance it
Fill the bridge with distilled water	Little or no current	Pure water has almost no ions to carry charge
Use a bridge salt whose ions react with a half-cell ion (e.g. KCl next to Ag⁺)	Performance falls	A precipitate (AgCl) forms, removing Ag⁺ and blocking the bridge
Use a larger anode	Cell runs for longer; voltage unchanged	More metal is available to oxidise, but each atom behaves the same way
Replace the voltmeter with a power supply pushing the other way	The cell becomes electrolytic	Electrical energy now drives the reverse, unfavourable reaction (like recharging a battery)
Let the two solutions mix	Little or no current in the wire	Electrons transfer directly between particles in solution, and the energy is lost as heat

Common misconceptions

• “Electrons flow through the salt bridge.” No. Electrons only move through the wire and electrodes. Ions carry charge through the salt bridge and the solutions.
• “The salt bridge supplies the reacting ions.” Its ions are chosen not to react. They only balance charge.
• “The anode is always negative.” That is true only for galvanic cells, and signs are not assessed. Anode = oxidation is always true.
• “In an electrolytic cell, electrons flow from cathode to anode.” In both types they flow through the external circuit from the anode to the cathode.
• “The cathode always gains mass.” Only if a metal is deposited there. A gas or a dissolved ion can also be the product.
• “Bigger electrodes give a bigger voltage.” Electrode size changes how much material can react, not the voltage.
• “Current flowing means the reaction is favourable.” In an electrolytic cell, the current flows because a power supply forces an unfavourable reaction.

Where this leads

Next, Topic 9.9 puts numbers on cells: you will calculate cell potentials from half-reaction data and link the sign of E°cell to ΔG°. Later topics use the same cells to explain non-standard conditions and to relate current to the amount of product. Try the practice questions now, then use the revision notes and the checklist to consolidate.