Electrolytic Cells: Driving a Reaction with an Applied Voltage

A reaction with a negative E°_cell does not happen by itself, but it can be driven with a power supply. The supply raises the Fermi level of one electrode and lowers that of the other, so that electrons can flow downhill at both electrodes while the supply lifts them back uphill through the wire. Here you recharge a Daniell cell, Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s), and then split water, 2H₂O(l) → 2H₂(g) + O₂(g). Work through the three stages in order.

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Predict, then see: recharging a Daniell cell,

Or explore freely:

Supply voltage 1.10 V

The blue lines are the Fermi levels of the two electrodes; the red dashed lines are the levels of the Zn²⁺/Zn and Cu²⁺/Cu couples. Moving dots show electrons crossing each interface.
Under each electrode is the half-reaction taking place there. At 1.10 V the supply just balances the cell; above 1.10 V zinc is deposited from Zn²⁺(aq) at the cathode and copper dissolves as Cu²⁺(aq) at the anode.
Without the supply, joined by a wire, the cell would discharge: electrons flow downhill from the zinc electrode (higher Fermi level) to the copper electrode.

Remember

Fermi level: E_F = −4.44 eV − eE, with E in V vs SHE. A more negative potential raises E_F.
The level of a redox couple is −4.44 eV − eE_eq. Electrons flow downhill in energy: from an electrode into a lower empty level (reduction), or from a higher filled level into the electrode (oxidation).
In an electrolytic cell the cathode is connected to the negative terminal and the anode to the positive terminal. The supply lifts electrons from the anode's Fermi level to the cathode's: E_F(cathode) − E_F(anode) = eV.
The minimum voltage is |E°_cell| (1.10 V to recharge a Daniell cell, 1.23 V to split water). Below it, a battery discharges; water simply does not react.
The dashed levels belong to the couples at each electrode surface: Zn²⁺ ions in solution and zinc atoms in the electrode (Zn²⁺(aq)/Zn(s)), and likewise for copper. The zinc metal is in the electrode, not in the solution.

The model shares the voltage above the minimum equally between the two electrodes, as equal overpotentials. In a real cell the share depends on the kinetics of each electrode (Lectures 5 and 6). Solutions are at 1 M and 298 K.