Half reactions and conventions

A half-cell contains a metal in two oxidation states. Inside an isolated half-cell, there is an oxidation-reduction (redox) reaction that is in chemical equilibrium, a condition written symbolically as follows (here, "M" represents a metal cation, an atom that has a charge imbalance due to the loss of "n" electrons):

Mⁿ⁺ (oxidized species) + ne⁻ ⇌ M (reduced species)

A galvanic cell consists of two half-cells, such that the electrode of one half-cell is composed of metal A, and the electrode of the other half-cell is composed of metal B; the redox reactions for the two separate half-cells are thus:

Aⁿ⁺ + ne⁻ ⇌ A

B^m+ + me⁻ ⇌ B

The overall balanced reaction is:

m A + n B^m+ ⇌ n B + m Aⁿ⁺

In other words, the metal atoms of one half-cell are oxidized while the metal cations of the other half-cell are reduced. By separating the metals in two half-cells, their reaction can be controlled in a way that forces transfer of electrons through the external circuit where they can do useful work.

• The electrodes are connected with a metal wire in order to conduct the electrons that participate in the reaction.

In one half-cell, dissolved metal-B cations combine with the free electrons that are available at the interface between the solution and the metal-B electrode; these cations are thereby neutralized, causing them to precipitate from solution as deposits on the metal-B electrode, a process known as plating.

This reduction reaction causes the free electrons throughout the metal-B electrode, the wire, and the metal-A electrode to be pulled into the metal-B electrode. Consequently, electrons are wrestled away from some of the atoms of the metal-A electrode, as though the metal-B cations were reacting directly with them; those metal-A atoms become cations that dissolve into the surrounding solution.

As this reaction continues, the half-cell with the metal-A electrode develops a positively charged solution (because the metal-A cations dissolve into it), while the other half-cell develops a negatively charged solution (because the metal-B cations precipitate out of it, leaving behind the anions); unabated, this imbalance in charge would stop the reaction. The solutions of the half-cells are connected by a salt bridge or a porous plate that allows ions to pass from one solution to the other, which balances the charges of the solutions and allows the reaction to continue.

By definition:

• The anode is the electrode where oxidation (loss of electrons) takes place (metal-A electrode); in a galvanic cell, it is the negative electrode, because when oxidation occurs, electrons are left behind on the electrode.^[10] These electrons then flow through the external circuit to the cathode (positive electrode) (while in electrolysis, an electric current drives electron flow in the opposite direction and the anode is the positive electrode).
• The cathode is the electrode where reduction (gain of electrons) takes place (metal-B electrode); in a galvanic cell, it is the positive electrode, as ions get reduced by taking up electrons from the electrode and plate out (while in electrolysis, the cathode is the negative terminal and attracts positive ions from the solution). In both cases, the statement 'the cathode attracts cations' is true.

Galvanic cells, by their nature, produce direct current. The Weston cell has an anode composed of cadmium mercury amalgam, and a cathode composed of pure mercury. The electrolyte is a (saturated) solution of cadmium sulfate. The depolarizer is a paste of mercurous sulfate. When the electrolyte solution is saturated, the voltage of the cell is very reproducible; hence, in 1911, it was adopted as an international standard for voltage.

A battery is a set of galvanic cells that are connected together to form a single source of voltage. For instance, a typical 12V lead–acid battery has six galvanic cells connected in series with the anodes composed of lead and cathodes composed of lead dioxide, both immersed in sulfuric acid. Large battery rooms, for instance in a telephone exchange providing central office power to user's telephones, may have cells connected in both series and parallel.

Cell voltage

The voltage (electromotive force E^o) produced by a galvanic cell can be estimated from the standard Gibbs free energy change in the electrochemical reaction according to:

$${\displaystyle E_{\text{cell}}^{o}=-\Delta _{r}G^{o}/(\nu _{e}F)}$$

where ν_e is the number of electrons transferred in the balanced half reactions, and F is Faraday's constant. However, it can be determined more conveniently by the use of a standard potential table for the two half cells involved. The first step is to identify the two metals and their ions reacting in the cell. Then one looks up the standard electrode potential, E^o, in volts, for each of the two half reactions. The standard potential of the cell is equal to the more positive E^o value minus the more negative E^o value.

For example, in the figure above the solutions are CuSO₄ and ZnSO₄. Each solution has a corresponding metal strip in it, and a salt bridge or porous disk connecting the two solutions and allowing SO²⁻
₄ ions to flow freely between the copper and zinc solutions. To calculate the standard potential one looks up copper and zinc's half reactions and finds:

Cu²⁺ + 2
e⁻
⇌ Cu   E^o = +0.34 V

Zn²⁺ + 2
e⁻
⇌ Zn   E^o = −0.76 V

Thus the overall reaction is:

Cu²⁺ + Zn ⇌ Cu + Zn²⁺

The standard potential for the reaction is then +0.34 V − (−0.76 V) = 1.10 V. The polarity of the cell is determined as follows. Zinc metal is more strongly reducing than copper metal because the standard (reduction) potential for zinc is more negative than that of copper. Thus, zinc metal will lose electrons to copper ions and develop a positive electrical charge. The equilibrium constant, K, for the cell is given by:

$${\displaystyle \ln K={\frac {\nu _{e}FE_{\text{cell}}^{o}}{RT}}}$$

where F is the Faraday constant, R is the gas constant and T is the temperature in kelvins. For the Daniell cell K is approximately equal to 1.5×10³⁷. Thus, at equilibrium, a few electrons are transferred, enough to cause the electrodes to be charged.^[11]

Actual half-cell potentials must be calculated by using the Nernst equation as the solutes are unlikely to be in their standard states:

$${\displaystyle E_{\text{half-cell}}=E^{o}-{\frac {RT}{\nu _{e}F}}\ln _{e}Q}$$

where Q is the reaction quotient. When the charges of the ions in the reaction are equal, this simplifies to:

$${\displaystyle E_{\text{half-cell}}=E^{o}-2.303{\frac {RT}{\nu _{e}F}}\log _{10}\left\{{\text{M}}^{n+}\right\}}$$

where {Mⁿ⁺} is the activity of the metal ion in solution. In practice concentration in mol/L is used in place of activity. The metal electrode is in its standard state so by definition has unit activity. The potential of the whole cell is obtained as the difference between the potentials for the two half-cells, so it depends on the concentrations of both dissolved metal ions. If the concentrations are the same, ${\displaystyle E_{\text{cell}}=E_{\text{cell}}^{o}}$and the Nernst equation is not needed under the conditions assumed here.

The value of 2.303R/F is 1.9845×10⁻⁴ V/K, so at 25 °C (298.15 K) the half-cell potential will change by only 0.05918 V/ν_e if the concentration of a metal ion is increased or decreased by a factor of 10.

$${\displaystyle E_{\text{half-cell}}=E^{o}-{\frac {0.05918\ {\text{V}}}{\nu _{e}}}\log _{10}\left[{\text{M}}^{n+}\right]}$$

These calculations are based on the assumption that all chemical reactions are in equilibrium. When a current flows in the circuit, equilibrium conditions are not achieved and the cell voltage will usually be reduced by various mechanisms, such as the development of overpotentials.^[12] Also, since chemical reactions occur when the cell is producing power, the electrolyte concentrations change and the cell voltage is reduced. A consequence of the temperature dependency of standard potentials is that the voltage produced by a galvanic cell is also temperature dependent.