Electrode: a metal in its ion solution
Dip a zinc rod in zinc sulphate solution. Some Zn atoms leave the rod as Zn²⁺ and leave their electrons behind. Some Zn²⁺ take electrons and come back. A tiny charge difference forms between the rod and the solution.
This metal–ion pair is called an electrode or half-cell, and is written as the redox couple Zn²⁺/Zn (oxidised form / reduced form).
The charge difference is its electrode potential. When the ion solution is 1 M, gases are at 1 bar and the temperature is 298 K, it is the standard electrode potential E°.
Try it: step 1 of the 3D shows one half-cell by itself.
Daniell cell: redox split into two beakers
In a beaker, Zn + Cu²⁺ → Zn²⁺ + Cu gives only heat. If we keep Zn in ZnSO₄ and Cu in CuSO₄, and join them with a wire, the electrons must travel through the wire. This is a Daniell cell.
- Zinc electrode: Zn → Zn²⁺ + 2e⁻ (oxidation). This is the anode, the negative end.
- Copper electrode: Cu²⁺ + 2e⁻ → Cu (reduction). This is the cathode, the positive end.
- Electrons flow in the wire from anode to cathode. The voltmeter shows about 1.10 V.
Cell notation: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). Anode on the left, | is a boundary, || is the salt bridge.
Try it: at step 2 of the 3D, follow the yellow electrons along the wire.
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As Zn²⁺ enters the left beaker, it becomes positive. As Cu²⁺ leaves the right beaker, it becomes negative. This charge would soon stop the flow.
A salt bridge (a U-tube of KCl or KNO₃ in jelly) lets ions move: negative ions go to the anode side and positive ions go to the cathode side. It keeps both solutions neutral and completes the circuit, without letting the two solutions mix.
Try it: at step 3 the bridge is out and the meter shows 0 V.
Standard hydrogen electrode (SHE)
We cannot measure one electrode alone; we always need a pair. So we pick one electrode as the zero: the standard hydrogen electrode. Platinum is dipped in 1 M H⁺ and pure H₂ gas at 1 bar is bubbled over it. Its potential is taken as 0.00 V at every temperature.
Join any electrode to the SHE. The meter reading, with sign, is that electrode’s standard potential. With copper it reads 0.34 V and copper is the cathode, so E°(Cu²⁺/Cu) = +0.34 V. With zinc it reads 0.76 V but zinc is the anode, so E°(Zn²⁺/Zn) = −0.76 V.
Electrochemical series and E°cell
Listing reduction potentials in order gives the electrochemical series:
Li⁺/Li −3.04 · Mg²⁺/Mg −2.37 · Zn²⁺/Zn −0.76 · Fe²⁺/Fe −0.44 · H⁺/H₂ 0.00 · Cu²⁺/Cu +0.34 · Ag⁺/Ag +0.80 · F₂/F⁻ +2.87 V
- More negative E° → stronger reducing agent (gives electrons easily). Li is the strongest reducing agent in this list.
- More positive E° → stronger oxidising agent (takes electrons easily). F₂ is the strongest oxidising agent.
- A metal with negative E° can push H₂ out of dilute acid; a metal with positive E° cannot.
- The electrode with lower E° becomes the anode.
E°cell = E°(cathode) − E°(anode). A positive E°cell means the reaction runs by itself.
Try it: at step 5 choose Mg on the left and Ag on the right. Predict E°cell first.
Key formulas and definitions
- E°cell = E°(cathode) − E°(anode)
- Anode: oxidation (negative terminal); cathode: reduction (positive terminal)
- E°(SHE) = 0.00 V by choice
- Cell notation: anode | anode ion || cathode ion | cathode
- E°cell > 0 → reaction is spontaneous
Worked examples
1. Find E°cell for the Daniell cell. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.
Zn is lower → anode. E°cell = 0.34 − (−0.76) = +1.10 V.
2. Find E°cell for a Mg–Cu cell. E°(Mg²⁺/Mg) = −2.37 V.
Mg is anode. E°cell = 0.34 − (−2.37) = +2.71 V.
3. Find E°cell for a Zn–Ag cell. E°(Ag⁺/Ag) = +0.80 V.
Zn is anode. E°cell = 0.80 − (−0.76) = +1.56 V. Reaction: Zn + 2Ag⁺ → Zn²⁺ + 2Ag.
4. Find E°cell for a Fe–Cu cell. E°(Fe²⁺/Fe) = −0.44 V.
Fe is anode. E°cell = 0.34 − (−0.44) = +0.78 V.
5. Zinc joined to the SHE gives 0.76 V and zinc is the anode. What is E°(Zn²⁺/Zn)?
E°cell = E°(SHE) − E°(Zn) → 0.76 = 0 − E°(Zn) → E°(Zn²⁺/Zn) = −0.76 V.
6. Can Cu reduce Ag⁺? Can Ag reduce Cu²⁺?
Cu + 2Ag⁺ → Cu²⁺ + 2Ag: E° = 0.80 − 0.34 = +0.46 V > 0, yes. Ag + Cu²⁺: E° = 0.34 − 0.80 = −0.46 V < 0, no.
7. Write the cell notation and the cell reaction for a cell made from Fe²⁺/Fe and Ag⁺/Ag.
Fe is lower → anode. Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s). Reaction: Fe + 2Ag⁺ → Fe²⁺ + 2Ag. E°cell = 0.80 − (−0.44) = +1.24 V.
Common mistakes
- Multiplying E° by the number of electrons when balancing. E° does not change when you multiply the equation.
- Calling the anode positive. In a cell that makes electricity, the anode is negative.
- Subtracting in the wrong order. Always cathode minus anode.
- Thinking the salt bridge carries electrons. It carries ions; electrons go only through the wire.