Standard Cell Potential of the Daniell Cell

Medium

Problem

A Daniell cell is constructed as Zn(s)|Zn2+(1 M)||Cu2+(1 M)|Cu(s). Given E(Zn2+/Zn) = -0.76 V and E(Cu2+/Cu) = +0.34 V, calculate the standard cell potential, identify the anode and cathode, and state the direction of electron flow.

Given

E(Zn2+/Zn) = -0.76 V
E(Cu2+/Cu) = +0.34 V
Standard conditions: 1 M ions, 25 C, solids pure.

Find

E cell under standard conditions, which electrode is the anode and which is the cathode, and the direction of electron flow.

Principle

In a galvanic cell, the half-reaction with the higher (more positive) reduction potential runs as the reduction at the cathode, while the other runs in reverse as the oxidation at the anode. The standard cell potential is the cathode potential minus the anode potential, both taken as tabulated reductions. A positive E cell means the overall reaction is spontaneous.

Formula

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

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

Solution

Compare the reduction potentials: +0.34 V (copper) is higher than -0.76 V (zinc), so Cu2+ is reduced at the cathode and Zn is oxidized at the anode.
Write the half-reactions: cathode: Cu2+ + 2e- -> Cu; anode: Zn -> Zn2+ + 2e-.
Apply E cell = E cathode - E anode = (+0.34 V) - (-0.76 V) = 0.34 + 0.76 = 1.10 V.
Electrons are produced at the zinc anode and consumed at the copper cathode, so they flow from Zn to Cu through the external wire.

E cell = 1.10 V; zinc is the anode (oxidized), copper is the cathode (reduced); electrons flow from Zn to Cu through the wire.

Unit check

Both half-cell potentials are in volts, so their difference is in volts. Subtracting a negative potential adds its magnitude, giving 1.10 V — a positive value with the correct unit for an electromotive force.

What this means physically

The positive 1.10 V means the cell reaction is spontaneous — this is why the Daniell cell was a practical battery. Zinc, the more easily oxidized metal, dissolves at the anode while copper plates out at the cathode, and the 1.10 V is the maximum voltage the cell can deliver before concentrations drift from standard values.

Common mistake

Students often add the two reduction potentials instead of subtracting, getting 0.34 + (-0.76) = -0.42 V, or flip the sign of the anode potential twice. The formula E(cathode) - E(anode) uses both potentials exactly as tabulated — as reductions.