Electrochemical cells

Electrolysis

Electrolysis uses electricity to break down an ionic compound that is molten or dissolved in water (the electrolyte), so its ions can move.

• Positive ions go to the cathode (negative electrode) and gain electrons: reduction.
• Negative ions move to the anode (positive electrode) and lose electrons: oxidation.

Molten lead bromide gives lead and bromine. In aqueous solutions water also supplies H^+ and OH^-:

• At the cathode, hydrogen forms unless the metal is less reactive than hydrogen (such as copper or silver).
• At the anode, oxygen forms unless halide ions are present, in which case the halogen forms.

Electrolysis of brine (sodium chloride solution) makes hydrogen, chlorine and sodium hydroxide. Aluminium is extracted by electrolysing aluminium oxide dissolved in molten cryolite, which lowers the operating temperature.

Cells and batteries

A chemical cell makes electricity from a reaction. Two different metals in an electrolyte produce a voltage, which is bigger when their reactivities differ more. A battery is two or more cells connected together. Non-rechargeable cells stop when a reactant is used up.

A hydrogen fuel cell combines hydrogen and oxygen to make electricity, with only water as the product.

Half-cells and the standard hydrogen electrode

A half-cell contains an element in two oxidation states, such as Zn(s) in Zn^{2+}(aq). Two half-cells joined by a wire and a salt bridge (which lets ions flow and completes the circuit) form a cell. Each half-cell's tendency to be reduced is measured against the standard hydrogen electrode (SHE), defined as 0.00 V: platinum in 1.00 mol dm^{-3} H^+ with H_2 at 100 kPa, all at 298 K.

The standard electrode potential, E^\circ, is the potential of a half-cell under these standard conditions relative to the SHE, always written as a reduction:

• \mathrm{Zn^{2+} + 2e^- \rightleftharpoons Zn}, E^\circ = -0.76 V
• \mathrm{Cu^{2+} + 2e^- \rightleftharpoons Cu}, E^\circ = +0.34 V

The more positive E^\circ, the stronger the oxidising agent on the left.

Cell EMF

In conventional representation, \mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,||\,Cu^{2+}(aq)\,|\,Cu(s)}, the oxidation half-cell is on the left. Here E^\circ_\text{cell} = 0.34 - (-0.76) = +1.10 V.

Predicting feasibility

A reaction is feasible if E^\circ_\text{cell} is positive for the cell written in that direction: Cu^{2+} will oxidise Zn but Zn^{2+} will not oxidise Cu. Predictions may fail because:

• the reaction may be too slow (high activation energy);
• conditions are not standard; changing concentrations shifts the electrode potentials;
• the cell may have a high internal resistance.

Commercial cells

Lithium-ion cells are rechargeable because the electrode reactions can be reversed by an external supply. Fuel cells are fed continuously with fuel and oxidant; in an alkaline hydrogen cell, \mathrm{O_2 + 2H_2O + 4e^- \rightarrow 4OH^-} occurs at the positive electrode.

Electrical work and Gibbs energy

A reversible cell delivers the maximum non-expansion work, which equals -\Delta_\text{r}G. Moving n moles of electrons through a potential difference E does work nFE, where F = 96\,485 C mol^{-1}, so

A positive EMF corresponds to a spontaneous cell reaction. For the Daniell cell (n = 2, E^\circ = 1.10 V), \Delta_\text{r}G^\circ = -212 kJ mol^{-1} and \ln K = 2 \times 96\,485 \times 1.10/(8.314 \times 298) \approx 85.7, so K \approx 10^{37}: the reaction runs essentially to completion.

Combining half-reactions

Potentials are intensive and do not add directly when half-reactions are combined into another half-reaction; Gibbs energies do. For example, from \mathrm{Fe^{3+}/Fe^{2+}} (+0.77 V, n=1) and \mathrm{Fe^{2+}/Fe} (-0.44 V, n=2): E^\circ(\mathrm{Fe^{3+}/Fe}) = [1(0.77) + 2(-0.44)]/3 = -0.04 V. Latimer and Frost diagrams organise this bookkeeping and reveal species prone to disproportionation.

Thermodynamics from EMF measurements

Because (\partial G/\partial T)_p = -S,

Precise cell measurements over a range of temperatures therefore give \Delta G, \Delta S and \Delta H without calorimetry.

Faraday's laws of electrolysis

The amount of substance changed at an electrode is proportional to the charge passed, Q = It:

where z is the number of electrons per formula unit. Worked example: 2.00 A for 30.0 min through CuSO_4(aq) passes 3600 C, i.e. 0.0373 mol e^-, depositing 0.0187 mol Cu, or 1.19 g. Real processes have a current efficiency below 100% when side reactions (often hydrogen evolution) consume charge.

Conventions and practicalities

IUPAC defines E_\text{cell} = E_\text{R} - E_\text{L} for the cell diagram as written, with all electrode potentials as reduction potentials. A liquid junction between different electrolytes adds a junction potential; a salt bridge of concentrated KCl minimises it because K^+ and Cl^- have similar mobilities. In electrolysis the applied voltage must exceed the thermodynamic cell potential by the overpotentials at each electrode plus the IR drop, which is why electrolysis of water needs well above 1.23 V in practice.