Standard Cell Potential

E°cell = E°cathode − E°anode: subtract the anode's standard reduction potential from the cathode's to get the cell voltage.

E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}

E(cell) = E(cathode) - E(anode), both as standard reduction potentials

What it means

Every half-cell has a standard reduction potential measuring how strongly it pulls electrons. The cell voltage is simply the cathode's pull minus the anode's pull — both entered as reduction potentials, with the subtraction doing all the sign work. A positive E°cell means the reaction runs spontaneously as written and the cell can do electrical work; a negative one means you would have to drive it with an external voltage.

Variables

Symbol	Name	Unit	Meaning
$E°cell$	Standard cell potential	V	The voltage of the full electrochemical cell under standard conditions; positive means spontaneous.
$E°cathode$	Cathode reduction potential	V	The standard reduction potential of the half-cell where reduction happens.
$E°anode$	Anode reduction potential	V	The standard reduction potential of the half-cell where oxidation happens, still entered as a reduction potential.

Assumptions

Standard conditions: 1 M ion concentrations, 1 atm gas pressures, 25 °C.
Both half-cell values must be standard reduction potentials — never flip a sign yourself; the subtraction handles it.
The cathode is the electrode where reduction occurs; the anode is where oxidation occurs.
The cell reaction is written so that it runs spontaneously in the forward direction.

Worked example

Problem

A galvanic cell pairs a copper cathode (Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V) with a zinc anode (Zn → Zn²⁺ + 2e⁻; its reduction potential is −0.76 V). Calculate E°cell.

Given:

E°(Cu²⁺/Cu) = +0.34 V
E°(Zn²⁺/Zn) = −0.76 V
Cathode (reduction): Cu²⁺ + 2e⁻ → Cu
Anode (oxidation): Zn → Zn²⁺ + 2e⁻

Identify the cathode's reduction potential: E°cathode = +0.34 V.
Identify the anode's reduction potential: E°anode = −0.76 V (keep it as the reduction value).
Apply E°cell = E°cathode − E°anode = 0.34 V − (−0.76 V).
Subtracting a negative adds: 0.34 + 0.76 = 1.10 V.

E°cell = 1.10 V

The positive 1.10 V means zinc reduces copper(II) spontaneously — the classic Daniell cell delivers about 1.1 volts under standard conditions.