What is a galvanic cell?
A galvanic cell (also called a voltaic cell) makes electricity from a redox reaction that happens by itself. We keep the two half-reactions in two separate beakers. This forces the electrons to go through a wire, where we can use them.
- Anode: oxidation happens here (atoms lose electrons). In a galvanic cell it is the negative (−) side.
- Cathode: reduction happens here (ions gain electrons). It is the positive (+) side.
- Salt bridge: a tube of a jelly with KCl or KNO₃. Its ions move into the beakers to keep each one electrically neutral. Without it, the current stops in a moment.
For the Daniell cell: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Written short: Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu. The anode goes on the left, a single line means a boundary between phases, and the double line is the salt bridge.
Electrode potential
When a metal rod dips in a solution of its own ions, a small voltage builds up between the metal and the solution. This is the electrode potential. It shows how strongly that electrode wants to take electrons.
By agreement we always write it as a reduction potential (Mⁿ⁺ + ne⁻ → M). If all ions are at 1 M, gases at 1 bar and the temperature is 298 K, it is the standard electrode potential E°.
- A big positive E° (like Ag⁺/Ag, +0.80 V) means the ion is easy to reduce. It is a good oxidising agent.
- A big negative E° (like Mg²⁺/Mg, −2.37 V) means the metal is easy to oxidise. It is a good reducing agent.
A list of E° values from low to high is the electrochemical series.
Standard hydrogen electrode (SHE)
We cannot measure one electrode alone. A voltmeter always needs two ends. So scientists picked one reference electrode and gave it the value zero.
The standard hydrogen electrode is a platinum foil coated with platinum black. It sits in 1 M H⁺ solution, and pure hydrogen gas at 1 bar bubbles over it at 298 K. Its reaction is 2H⁺(aq) + 2e⁻ → H₂(g), and E° = 0.00 V by definition.
To find E° of copper, we join Pt | H₂(1 bar) | H⁺(1 M) || Cu²⁺(1 M) | Cu. The meter reads 0.34 V, and copper is the cathode. So E°(Cu²⁺/Cu) = +0.34 V. With zinc, zinc becomes the anode and the reading is 0.76 V, so E°(Zn²⁺/Zn) = −0.76 V.
EMF of a cell
The emf (electromotive force) is the voltage of the cell when no current is drawn. Use the two reduction potentials:
E°cell = E°cathode − E°anode = E°right − E°left
Daniell cell: E°cell = 0.34 − (−0.76) = 1.10 V. If you get a negative answer, the reaction does not go as written. The real cell runs the other way.
Do not multiply E° when you multiply an equation by 2. E° is an intensive property. It does not depend on how much stuff there is.
Nernst equation
E° is for 1 M solutions only. Real solutions have other concentrations. The Nernst equation corrects for this.
For one electrode Mⁿ⁺ + ne⁻ → M: E = E° − (RT/nF) ln(1/[Mⁿ⁺]). At 298 K this becomes E = E° − (0.059/n) log(1/[Mⁿ⁺]).
For a whole cell aA + bB → cC + dD: Ecell = E°cell − (0.059/n) log Q, where Q = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ. Leave out pure solids and liquids from Q.
Here n = number of electrons moved in the balanced equation, F = 96 500 C mol⁻¹ (Faraday constant), R = 8.314 J K⁻¹ mol⁻¹. The number 0.059 = 2.303 RT/F at 298 K.
Equilibrium constant from E°
At equilibrium the cell is "flat": Ecell = 0 and Q = K. So E°cell = (0.059/n) log K, or log K = nE°/0.059.
Concentration cell
Two same electrodes with different concentrations still make a voltage. E° = 0, so E = (0.059/n) log(C₂/C₁), with C₂ the stronger solution on the cathode side.
Cell emf and Gibbs energy
The electrical work a cell can do is charge × voltage. For n moles of electrons, the charge is nF. The most useful work equals the fall in Gibbs energy:
ΔG = −nFEcell and ΔG° = −nFE°cell
- E positive → ΔG negative → the reaction is spontaneous.
- Also ΔG° = −2.303 RT log K, which matches the K formula above.
Here ΔG does depend on n. If you double the equation, ΔG doubles, but E stays the same.
Try it: a lemon cell
Push a clean zinc (galvanised) nail and a copper coin into a lemon, 2 cm apart. Touch both to the ends of a cheap LED or a multimeter. Predict first: which metal is the anode? (Hint: look at the E° ladder in step 3.) Then check the meter sign. Try in the 3D too: pick Mg | Cu, then move the slider and predict whether E goes up or down before you let go.
Key formulas and definitions
- E°cell = E°cathode − E°anode
- Ecell = E°cell − (0.059/n) log Q (at 298 K)
- E(Mⁿ⁺/M) = E° − (0.059/n) log (1/[Mⁿ⁺])
- E(H⁺/H₂) = −0.059 pH (at 1 bar H₂)
- log K = nE°cell / 0.059
- ΔG = −nFEcell ; ΔG° = −nFE°cell = −2.303 RT log K
- F = 96 500 C mol⁻¹
Worked examples
1. Find E°cell for Zn | Zn²⁺ || Cu²⁺ | Cu. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.
Step 1: Cathode = Cu (higher E°), anode = Zn. Step 2: E°cell = E°cathode − E°anode = 0.34 − (−0.76). Step 3: E°cell = 1.10 V.
2. A cell uses Cu and Ag. E°(Cu²⁺/Cu) = 0.34 V, E°(Ag⁺/Ag) = 0.80 V. Write the cell and find E°cell.
Step 1: Ag has the higher E°, so Ag⁺ is reduced (cathode). Cu is the anode. Step 2: Cell: Cu | Cu²⁺ || Ag⁺ | Ag. Reaction: Cu + 2Ag⁺ → Cu²⁺ + 2Ag. Step 3: E°cell = 0.80 − 0.34 = 0.46 V. (We do not double 0.80 even though 2Ag⁺ appear.)
3. Find the emf of Zn | Zn²⁺(0.1 M) || Cu²⁺(0.001 M) | Cu at 298 K.
Step 1: Zn + Cu²⁺ → Zn²⁺ + Cu, n = 2, E° = 1.10 V. Step 2: Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.001 = 100, so log Q = 2. Step 3: E = 1.10 − (0.059/2) × 2 = 1.10 − 0.059. Step 4: E = 1.041 V.
4. Find the potential of a hydrogen electrode in a solution of pH 3 (H₂ at 1 bar, 298 K).
Step 1: 2H⁺ + 2e⁻ → H₂. E = 0 − (0.059/2) log (1/[H⁺]²). Step 2: This simplifies to E = −0.059 × pH. Step 3: E = −0.059 × 3 = −0.177 V.
5. Find ΔG° for the Daniell cell (E° = 1.10 V).
Step 1: n = 2, F = 96 500 C mol⁻¹. Step 2: ΔG° = −nFE° = −2 × 96 500 × 1.10 J mol⁻¹. Step 3: ΔG° = −212 300 J mol⁻¹ = −212.3 kJ mol⁻¹. Negative, so the reaction is spontaneous.
6. Find log K for Zn + Cu²⁺ ⇌ Zn²⁺ + Cu at 298 K (E° = 1.10 V).
Step 1: log K = nE°/0.059. Step 2: log K = 2 × 1.10 / 0.059 = 2.20 / 0.059. Step 3: log K ≈ 37.3, so K ≈ 2 × 10³⁷. The reaction goes almost fully to the right.
7. Find Ecell for Mg | Mg²⁺(0.001 M) || Cu²⁺(0.0001 M) | Cu. E°(Mg²⁺/Mg) = −2.37 V, E°(Cu²⁺/Cu) = 0.34 V.
Step 1: E° = 0.34 − (−2.37) = 2.71 V, n = 2. Step 2: Q = [Mg²⁺]/[Cu²⁺] = 0.001/0.0001 = 10, log Q = 1. Step 3: E = 2.71 − (0.059/2) × 1 = 2.71 − 0.0295. Step 4: E ≈ 2.68 V.
8. A concentration cell: Cu | Cu²⁺(0.01 M) || Cu²⁺(1 M) | Cu. Find E at 298 K.
Step 1: Same electrodes, so E° = 0, n = 2. Step 2: E = (0.059/2) log (1/0.01) = 0.0295 × 2. Step 3: E = 0.059 V. The stronger solution is the cathode.
Common mistakes
- Multiplying E° by the coefficient when balancing (E° of Ag⁺/Ag stays 0.80 V even for 2Ag⁺). E° does not depend on amount; ΔG does.
- Using oxidation potentials in E°cathode − E°anode. Always use both as reduction potentials.
- Putting solids like Zn(s) or Cu(s) into Q in the Nernst equation. Pure solids and liquids are left out.
- Forgetting n, or using the wrong n. Count electrons in the balanced full equation (Cu + 2Ag⁺ → … has n = 2).