The big idea: We cannot measure the voltage of a single half-cell — only the difference between two. So chemists agree on one half-cell to act as the zero and measure everything else against it.
That reference is the standard hydrogen electrode (SHE), which is defined as exactly 0.00 V.
The voltage of any other half-cell, measured against the SHE under standard conditions, is its standard electrode potential, E°.
Each half-cell has a standard electrode potential E° measured against the SHE. The half-cell with the more positive E° is reduced (cathode); the more negative one is oxidised (anode). E°cell is the difference.
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Standard conditions and sign: Standard conditions for E° values: 298 K, solution concentrations of 1 mol dm⁻³, and gas pressures of 100 kPa. The little ° symbol on E° always means standard.
Every E° is quoted for the half-equation written as a reduction (electrons on the left):
- a positive E° means the half-cell is reduced more easily than H⁺, - a negative E° means it is reduced less easily than H⁺ (so it tends to be oxidised instead).
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An E° table lists half-equations as reductions, ordered by their E° value. The data booklet gives you these values — you just need to read them correctly.
The single rule: the more positive the E°, the stronger the species is as an oxidising agent (it pulls electrons toward itself and is reduced). The more negative the E°, the stronger the species is as a reducing agent (it readily gives electrons away and is oxidised).
| Reduction half-equation | E° / V |
|---|---|
| Mg²⁺(aq) + 2e⁻ ⇌ Mg(s) | −2.37 |
| Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) | −0.76 |
| Ni²⁺(aq) + 2e⁻ ⇌ Ni(s) | −0.26 |
| 2H⁺(aq) + 2e⁻ ⇌ H₂(g) (SHE) | 0.00 |
| Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) | +0.34 |
| Ag⁺(aq) + e⁻ ⇌ Ag(s) | +0.80 |
Top vs bottom of the table: Reading the table above (most negative at the top, most positive at the bottom):
- Most positive E° (Ag⁺, +0.80 V) — the strongest oxidising agent; Ag⁺ is reduced most readily. - Most negative E° (Mg²⁺/Mg, −2.37 V) — the metal Mg is the strongest reducing agent; it is oxidised most readily.
So a more reactive metal sits higher (more negative E°) and is the better reducing agent.
Worked example — comparing oxidising strength
Using the table, which is the stronger oxidising agent: Cu²⁺ or Zn²⁺?
Solution
- Read off the two E° values for the reduction half-equations:
- Rule: the more positive E° is the stronger oxidising agent (more readily reduced). +0.34 > −0.76.
Final answer
Cu²⁺ is the stronger oxidising agent (E° = +0.34 V is more positive than −0.76 V).
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When two half-cells are connected, the one with the more positive E° is reduced (it becomes the cathode); the one with the more negative E° is oxidised (the anode). The standard cell potential is their difference — and the data booklet gives you the equation.
- standard cell potential / EMF (V)
- standard electrode potential of the reduction half-cell (V)
- standard electrode potential of the oxidation half-cell (V)
Watch the sign — don't reverse the E°: Use the E° values straight from the table (the reduction values). Do not flip the sign of the anode's E° yourself — the minus sign in E°cell = E°(cathode) − E°(anode) already does that for you.
If you assign cathode and anode correctly (more positive = cathode), you will always get a positive E°cell.
Worked example — E°cell of a nickel–silver cell
A cell is built from a nickel half-cell, E°(Ni²⁺/Ni) = −0.26 V, and a silver half-cell, E°(Ag⁺/Ag) = +0.80 V. Calculate the standard cell potential.
Solution
- Decide the electrodes — the more positive E° is reduced (cathode):
- Formula first — write the given equation:
- Substitute the E° values:
- Work it out — subtracting a negative adds:
Final answer
E°cell = +1.06 V.
Positive E°cell = spontaneous: A positive E°cell means the redox reaction is spontaneous (feasible) as written — the cell will deliver a current. A negative value means the reaction will not happen spontaneously in that direction.
This connects to energy through the data-booklet equation ΔG° = −nFE°cell: a positive E°cell makes ΔG° negative, and a negative ΔG° is the thermodynamic condition for a spontaneous reaction.
- standard Gibbs energy change (J mol⁻¹)
- number of electrons transferred in the balanced equation
- Faraday's constant = 96 500 C mol⁻¹
- standard cell potential (V)
How this is tested: On Paper 2 (HL) this is a multi-step calculation: you are given two E° values and asked to (1) calculate E°cell, (2) state whether the reaction is feasible (positive = spontaneous), and (3) identify the oxidising and reducing agents and/or write the overall equation.
The markers want the right cathode/anode assignment, the correct substitution into E°cell = E°(cathode) − E°(anode), and a clear statement linking a positive E°cell to spontaneity.
Sure marks: Pick the more positive E° as the cathode every time. Substitute into the given equation with brackets around each value (so the double-negative is safe). Then: positive E°cell → spontaneous; the species reduced at the cathode is the oxidising agent, the species oxidised at the anode is the reducing agent.
IB-style question — copper–zinc cell (a)
A standard cell is set up from a copper half-cell, E°(Cu²⁺/Cu) = +0.34 V, and a zinc half-cell, E°(Zn²⁺/Zn) = −0.76 V. (a) Calculate the standard cell potential, E°cell, and state whether the cell reaction is spontaneous.
Solution
- The more positive E° is reduced (cathode); the more negative is oxidised (anode):
- Formula first:
- Substitute and solve:
- E°cell is positive, so the reaction is spontaneous (feasible).
Final answer
E°cell = +1.10 V; because it is positive, the cell reaction is spontaneous.
IB-style question — copper–zinc cell (b)
(b) Identify the oxidising agent and the reducing agent, and write the overall ionic equation for the spontaneous cell reaction.
Solution
- At the cathode Cu²⁺ is reduced, so Cu²⁺ is the oxidising agent:
- At the anode Zn is oxidised, so Zn is the reducing agent:
- Both transfer 2 electrons, so add the half-equations directly (electrons cancel):
Final answer
Oxidising agent = Cu²⁺; reducing agent = Zn. Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s).