ELECTROCHEMISTRY. During electrolysis positive ions (cations) move to negatively charged electrode (catode) and negative ions (anions) to positively charged.

ELECTROCHEMISTRY

During electrolysis positive ions (cations) move to negatively chargedelectrode (catode) and negative ions (anions) to positively chargedelectrode (anode)

For the case of NaCl we have:cathodic reduction:

Na + + electron = Na

andanodic oxidation:

Cl - - electron = Cl

Faraday laws:

1. The mass of product formed in an electrolysis is directly proportional to the electric charge moved during the process.

2. The masses of different compounds formed by the same electric charge are chemically equivalent.

The amount of electric charge needed for formation of 1 gramionof a substance:

N . e (N is Avogadro number, e electron charge)

N . e = F (Faraday constant, 1Faraday = 96 494 Coulomb)

Work produced by electric current: w = F .

Electrochemical cell is composed of two half-cells, realized e.g. asmetal electrode immersed in the solution of its salt.

The half-cells are conductively connected, e.g. by salt bridge.

Each half-cell contains oxidized and reduced component, which create a redox couple

p .... osmotic pressureP .... solvatation pressureP > p negative electrode chargeP < p positive electrode charge

+ -

Hydrogen electrode

Precious metals such as platinum or palladium absorb vigorouslyhydrogen. A solid solution is formed, analogical to the metal alloys.Here we have hydrogen present in its atomic form, not as a two atommolecule. Thus, in this state hydrogen has properties of a metal.

If we saturate a platinum electrode coated with platinum black by astream of hydrogen and immerse this electrode to the solution, protons will be released into the solution due to the solvatation pressure until they balance the proton osmotic pressure. This leadsto the generation of a potential, dependent on hydrogen partial pressure.

Standard hydrogen electrodeis realized under conditions of [ H+] = 1 (i.e. pH = 0) and hydrogenpressure 1 atm. By convention its potential = 0

By comparison of the potential of a half-cell, realized as a metalelectrode immersed in 1 N solution of its salt, with standard hydrogenelectrode we obtain electrochemical series.

Some examples:

Electrode Potential (Volt) Li/Li+ - 3, 02 K/K+ - 2, 92 Na/Na+ - 2, 71 Zn/Zn2+ - 0, 76 Fe/Fe2+ - 0, 43 Fe/Fe3+ - 0, 04 H/H+ 0, 00 Cu/Cu2+ + 0, 34 Cu/Cu+ + 0, 51 Ag/Ag+ + 0,80 Au/Au+ + 1, 50

Metals, placed above hydrogen in this table, have a tendency to formpositive cations and with distance from hydrogen, their electropositivity increases.

More electropositive metal displaces less electropositive metal fromthe solution.

Potential of a metal electrode dissolving metal cations into solutionis given by

Nernst equation:

E = - RT/nF . ln c

where R ...universal gas constant n ... number of electrons representing the difference between the metal and its ion c ... concentration of the ions in solution

We can express the amount of energy released in electrochemicalprocess as:

G = - nFE

Under standard conditions (concentration 1 M, pressure 1 atm) we get:

G0 = - nFE0

where E0 is the standard potential of the cell

Standard potential of the cell can be calculated as a sum of standardpotentials of electrodes:

E0 = E0(anode) + E0(catode)

We can express the amount of energy released under standardconditions in a general form:G0 = - RT lnKAnother expression can be used for the electrochemicalprocess:G0 = - nFE0

If we consider a real process out of standard conditions we get:

G = G0 + RT lnQ

where Q corresponds to the actual ratio of products and reactants

For electrode potential we get:

-nFE = -nFE0 + RT lnQ and hence E = E0 - RT/nF ln Q

This is an important expression of Nernst equation

Concentration cell

E = E0 – RT/2F ln (c2/c1)

We can use Nernst equation for the calculation of a potential generated in redox reactions in the living cell. The reaction:

reductant + oxidant = oxidized reductant + reduced oxidant

can be simplified:

electron donor = electron acceptor + electron

then acceptor) - (donor)

as G0 = - nF

we get:

E = E0 + RT/nF ln( [acceptor]/[donor] )

Membrane potential

To calculate membrane potential we first consider the amount ofenergy needed for the transport of a substance across the membrane.For the transport of 1 mol of a substance from the region of concentration c1 to the region of concentration c2 we get:

G = RT ln (c2/ c1)

When c2 is lower than c1, G is negative and the transport proceeds.Under equilibrium G = 0, concentrations are equal and transport is stopped.

If we have an ion with a charge Z, the change of Gibbs function during its transport will contain two components – a concentration part and a part describing charge movement:

G = RT ln (c2/ c1) + ZF

where is the membrane potential in Volts