Standard hydrogen electrode and standard electrode potentials
Electrode potentials are measured against the standard hydrogen electrode (E° = 0 V). The more negative the E°, the stronger the metal is as a reducing agent.
Why a reference electrode is needed
The potential of a single half-cell cannot be measured on its own; only the difference between two electrodes can. So every half-cell is paired with a reference, the standard hydrogen electrode (SHE). By convention, its potential is zero at all temperatures. Following the IUPAC convention, electrode potentials are listed as reduction potentials.
How to draw the standard hydrogen electrode
- Draw a beaker of acid containing H⁺ ions at 1 M (for example, HCl).
- Draw a glass tube dipping into the acid, with a platinum foil coated with platinum black at its lower end.
- Show pure H₂ gas at 1 bar entering through a side inlet and bubbling over the platinum.
- Show a wire leading from the platinum to the outside.
- Label: H₂ at 1 bar, Pt black electrode, 1 M H⁺, and write H⁺(aq) + e⁻ → ½H₂(g) beside it.
Measuring E° of a half-cell
- 1Join the half-cell to the SHE through a salt bridge, with the SHE written on the left. Measure the EMF.
- 2Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M) || Cu²⁺(aq, 1 M) | Cu(s) gives +0.34 V. Copper is the cathode, so E°(Cu²⁺/Cu) = +0.34 V.
- 3With zinc in place of copper, the value is −0.76 V. The negative sign means the reaction really runs the other way: zinc reduces H⁺. So E°(Zn²⁺/Zn) = −0.76 V.
- 4For couples with no metal, such as Fe³⁺/Fe²⁺, an inert electrode of platinum or gold carries the electrons.
Some standard reduction potentials at 298 K
| Half-reaction | E° (V) |
|---|---|
| F₂(g) + 2e⁻ → 2F⁻(aq) | +2.87 |
| MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) | +1.51 |
| Cl₂(g) + 2e⁻ → 2Cl⁻(aq) | +1.36 |
| O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) | +1.23 |
| Br₂(l) + 2e⁻ → 2Br⁻(aq) | +1.09 |
| Ag⁺(aq) + e⁻ → Ag(s) | +0.80 |
| Fe³⁺(aq) + e⁻ → Fe²⁺(aq) | +0.77 |
| I₂(s) + 2e⁻ → 2I⁻(aq) | +0.54 |
| Cu²⁺(aq) + 2e⁻ → Cu(s) | +0.34 |
| 2H⁺(aq) + 2e⁻ → H₂(g) | 0.00 |
| Ni²⁺(aq) + 2e⁻ → Ni(s) | −0.25 |
| Fe²⁺(aq) + 2e⁻ → Fe(s) | −0.44 |
| Zn²⁺(aq) + 2e⁻ → Zn(s) | −0.76 |
| 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) | −0.83 |
| Mg²⁺(aq) + 2e⁻ → Mg(s) | −2.36 |
| Li⁺(aq) + e⁻ → Li(s) | −3.05 |
Reading the series
- Going up the table, the oxidised form is a stronger oxidising agent: F₂ is the strongest.
- Going down the table, the reduced form is a stronger reducing agent: Li is the strongest.
- Metals with a negative E° (Zn, Fe, Mg) can displace hydrogen from dilute acids. Copper and silver, with positive E°, cannot.
- A metal displaces from solution the ions of a metal with a higher E°: zinc displaces copper from CuSO₄ solution.
- In any pair, the half-cell with the higher E° is the cathode, and E°cell = E°cathode − E°anode.
Worked example: will it react?
Using E° values, decide (a) whether CuSO₄ solution can be stored in a zinc vessel, and (b) whether the cell Ni(s) | Ni²⁺(1 M) || Ag⁺(1 M) | Ag(s) works, and find its E°cell. E°(Zn²⁺/Zn) = −0.76 V; E°(Cu²⁺/Cu) = +0.34 V; E°(Ni²⁺/Ni) = −0.25 V; E°(Ag⁺/Ag) = +0.80 V.
- (a) Possible reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). E°cell = 0.34 − (−0.76) = +1.10 V.
- A positive E°cell means the reaction is spontaneous, so the zinc vessel would dissolve. CuSO₄ cannot be stored in it.
- (b) E°cell = E°cathode − E°anode = 0.80 − (−0.25) = +1.05 V.
- The value is positive, so the cell works, with nickel as the anode and silver as the cathode.
(a) No: zinc reacts with Cu²⁺ (E°cell = +1.10 V); (b) yes, E°cell = +1.05 V
Lemon and tamarind pickles are kept in glass or glazed ceramic jars, not in iron tins: iron (E° = −0.44 V) reacts with the acid in them and spoils the taste.
2 marks: “What is the standard hydrogen electrode? Write its notation.” A platinum foil coated with platinum black, dipped in 1 M H⁺ solution, with pure H₂ gas at 1 bar bubbled over it. Its potential is taken as zero at all temperatures. Pt(s) | H₂(g, 1 bar) | H⁺(aq, 1 M).
3 marks: “Given E°(Cl₂/Cl⁻) = +1.36 V, E°(Br₂/Br⁻) = +1.09 V and E°(I₂/I⁻) = +0.54 V, arrange the halogens in order of oxidising power, and say whether Br₂ can oxidise I⁻ ions.” Oxidising power: Cl₂ > Br₂ > I₂, since a higher E° means a stronger oxidising agent. For Br₂ + 2I⁻ → 2Br⁻ + I₂, E°cell = 1.09 − 0.54 = +0.55 V, which is positive, so yes.
1 marks: “Can copper displace hydrogen from dilute HCl? Give a reason.” No. E°(Cu²⁺/Cu) = +0.34 V is higher than E°(H⁺/H₂) = 0 V, so copper is a weaker reducing agent than hydrogen.
Higher E°: Hungry for electrons (oxidising agent). Lower E°: Loses electrons (reducing agent). The higher E° sits at the cathode.