Purpose Of The Salt Bridge

This explanation makes reference the Zn/Cu2+ cell shown above.

The cell cannot operate unless the circuit is complete. The oxidation half-cell originally contains a neutral solution of Zn2+ and SO42- ions, but as Zn atoms in the bar lose electrons, the solution would develop a net positive charge from the Zn2+ ions entering. Similarly, in the reduction half-cell, the neutral solution of Cu2+ and SO42- ions would develop a net negative charge as Cu2+ ions leave the solution to form Cu atoms. A charge imbalance would arise and stop cell operation if the half-cells were not neutral. To avoid this situation and enable the cell to operate, the two half-cells are joined by a salt bridge, which acts as a "liquid wire," allowing ions to flow through both compartments and complete the circuit. The salt bridge shown in the diagram is an inverted U tube containing a solution of the nonreacting ions Na+ and SO42 - in a gel. The solution cannot pour out, but ions can diffuse through it into and out of the half-cells.

To maintain neutrality in the reduction half-cell (right; cathode compartment) as Cu2+ ions change to Cu atoms, Na+ ions move from the salt bridge into the solution (and some SO42- ions move from the solution into the salt bridge). Similarly, to maintain neutrality in the oxidation half-cell (left; anode compartment) as Zn atoms change to Zn2+ ions, SO42- ions move from the salt bridge into that solution (and some Zn2+ ions move from the solution into the salt bridge). Thus, as the diagram shows, the circuit is completed as electrons move left to right through the wire, while anions move right to left and cations move left to right through the salt bridge.

Partition Coefficient - What's The Numerator?

The process of a solute dissolved in one solvent being pulled out, or “extracted” into a new solvent actually involves an equilibrium process. At the time of initial contact, the solute will move from the original solvent to the extracting solvent at a particular rate, but, after a time, it will begin to move back to the original solvent at a particular rate. When the two rates are equal, we have equilibrium. We can thus write the following:

Aorig ⇌ Aext

in which A refers to analyte and orig and ext refer to original solvent and extracting solvent, respectively. If the analyte is more soluble in the extracting solvent than in the original solvent, then, at equilibrium, a greater percentage will be found in the extracting solvent and less in the original solvent. If the analyte is more soluble in the original solvent, then the greater percentage of analyte will be found in the original solvent. Thus, the amount that gets extracted depends on the relative distribution between the two layers, which, in turn, depends on the solubilities in the two layers. A distribution coefficient analogous to an equilibrium constant (also called the partition coefficient) can be defined as follows:

Often, the value of K is approximately equal to the ratio of the solubilities of A in the two solvents. If the value of K is very large, the transfer of solute to the extracting solvent is considered to be quantitative. A value around 1.0 would indicate equal distribution and a small value would indicate very little transfer. Uses of the distribution coefficient include:

1.the calculation of the amount of a solute that is extracted in a single extraction step,

2.the determination of the weight of the solute in the original solute (important if you are quantitating the solute in this solvent),

3.the calculation of the optimum volumes of both the extracting solvent and the original solution to be used,

4.the number of extractions needed to obtain a particular quantity or concentration in the extracting solvent, and

5.the percent extracted.

The following expansion of the previous equation is useful for these:

Electronic Configuration – d-block Elements and Ions

The electronic configurations of the first row d-block elements are given in the table below, along with those for the M2+ and M3+ ions. Because the 3d orbitals are all of the same energy, they are each filled with a single electron first.  Only after each of the five 3d orbitals is singly filled (3d5), do the electrons start pairing up, from 3d6 onwards.

For a free atom, the 4s orbital is normally filled before the 3d orbitals. This means that most first row d-block elements have the electronic configuration 3dn 4s2. Chromium (3d5  4s1) and copper (3d10  4s1) are exceptions to this.

The 4s orbital is more diffuse than the 3d orbitals and is affected more by the presence of other atoms or by the charge on the metal. As a result, in an ion or a compound, the 4s orbital is higher in energy than the 3d orbitals. This means that, when electrons are lost to form ions, it is the 4s electrons that are lost first. This is reflected in the electronic configurations of the d-block elements in ions and compounds , as these never contain 4s electrons  unless the d orbitals are full. For example, the electronic configuration of V2+ is 3d3 not 3d1 4s2, and the electronic configuration of Cr(0) in a compound is 3d6, not 3d5 4s1 as it is in atomic chromium.

IMPORTANT: In a compound of a first row d-block element, the 4s orbital is higher in energy than the 3d orbitals. For oxidation states of +2 and higher, the electronic configuration can be found by removing the 4s electrons, plus the appropriate number of 3d electrons, but for lower oxidation states the 4s electrons must be transferred into 3d orbitals before removing electrons.

The video below gives examples of how to obtain electronic configurations for:

•Fe3+,

•Fe(0), which is different from atomic iron,

•Ni2+,

•Ni(0),

•Co, and,

•Co+.

Click Here For Video:

Determining the electron configuration of a metal