frequently asked questions

Manganese (VI) is usually in the form of the MnO4(2-) ion. This ion

disproportionated in acid solution but not in base. What is going on here?

The Mn(VI) state is not stable under acid conditions because it CAN disproportionate (notice that this is NOT a half-equation - it is a reaction):

3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O

The Mn(VI) state cannot do this under alkaline conditions, and the alkaline disproportionation reaction (theoretical) would be:

3MnO4(2-) + 2H2O –> MnO2 + 2MnO4(-) + 4OH-

This is now in direct competition with the reverse reaction (as all reactions are) which we know DOES occur.

However, what we are really comparing is the Gibbs free energy of the two disproportionation equations. In the case of acid conditions:

3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O

The Gibbs Free energy makes the forward reaction feasible. In the case of alkaline conditions however, Gibbs Free energy makes the backward reaction equally feasible:

3MnO4(2-) + 2H2O <– MnO2 + 2MnO4(-) + 4OH-

This can be predicted by reference to the electrode potential of the half equations involved. If you break down each disporportionation into two half-equations and compare the redox potentials using E = E(red) - E(ox).

OK, let’s do it!
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Acidic conditions:

3MnO4(2-) + 4H+ –> MnO2 + 2MnO4(-) + 2H2O

This disproportionation can be though of as two half equations:
1. MnO4(2-) + 4H+ + 2e –> MnO2 + 2H2O ……….. Eº = +2.26V ( the Manganate ion is reduced)
2. MnO4(2-) –> MnO4(-) + 1e ……………………….Eº = +0.56V ( the Manganate ion is oxidised)

Calculating Eº = E(reduced state) - E(oxidised state) = +2.26V - +0.56V = + 1.70V
This is a large positive value (remember that any Eº value greater than 0.3v means that the reaction is spontaneous as shown) so the forward reaction proceeds.

Alkaline conditions:

3MnO4(2-) + 2H2O –> MnO2 + 2MnO4(-) + 4OH-

This disproportionation can be though of as two half equations:
MnO4(2-) + 2H2O + 2e –> MnO2 + 4OH- ……….. Eº = +0.67V (reduction)
MnO4(2-) –> MnO4(-) + 1e…………………………Eº = +0.56V (oxidation)

Calculating Eº = E(red) - E(ox) = 0.67 - 0.56 =  +0.11V

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Hardness is a term to describe water that doesn’t allow soaping action, in other words no bubbles (lather) and no cleaning.  The term ’soaping action’ refers to the formation of micelles by negative ions joined to long carbon chains that dissolve in the fat or grease leaving the negative charges outside the micelle structure. These can then bond to the water molecules. In this way the hydrophobic grease is made to dissolve and lifts off the fabric or skin.

Hardness is due to Ca2+ or Mg2+ ions dissolved in the water. These ions react with the long negative ions of soap (stearate ions C17H35COO-) to make an insoluble grey compound which we call ’scum’. As the reaction removes the necessary soap ions from the water it prevents lathering and the soap cannot perform its function.
Hardness can be removed by various means depending on the other ion that is dissolved with the Mg2+ or Ca2+ ions.

There are two subcategories of hardness:
1. Temporary
2. Permanent

Temporary hardness is due to dissolved calcium (or magnesium) hydrogen carbonate Ca(HCO3)2, it can be removed by boiling the water, when the hydrogen carbonate ions decompose forming insoluble calcium (or magnesium) carbonate:

Ca(HCO3)2 –> CaCO3 + CO2 + H20

Permanent hardness (due to dissolved calcium and magnesium sulphates and chlorides) must be removed by another method either:

1. ion exchange using resins etc.
2. chemical precipitation

Ca2+ + CO3(2-) –> CaCO3(ppt)

use of substances such as washing soda (sodium carbonate) do just that.

Water of crystallisation is the term given to the molecules of water that are used to build up a crystal lattice in some ionic compounds.

In copper II sulphate the blue crystals would be impossible without using water molecules to act as ’scaffolding’ within the structure CuSO4.5H2O. When this is heated the water molecules are driven off and the blue crystals become a white powder.

As most crystals are made by evaporation from an aqueous solution, it makes sense that water molecules can incorporate themselves into the ionic crystal structure. Water, after all, has an oxygen atom with two lone pairs capable of behaving as a Lewis base; it bonds easily to metal ions as evidenced by complex ions in transition metal chemistry. The metal ions and the water molecules are bonded by dative coordinate bonds from the oxygen atom of the water.

To calculate the RMM you must take into account the water of crystallisation when weighing hydrated crystals out, after all the water molecules are acually present in the crysals and contribute to the mass of the solid.

For example, if you wish to weigh out 0.1 moles of copper II sulphate crystals then you have to weigh out 0.1 x the mass of the hydrated salt.

CuSO4.5H2O has a relative mass = CuSO4 + (5 x H2O) = 249.5

therefore 0.1 moles = 24.95g

Colour is caused by selective absorption of certain wavelengths and reflection of the rest. The most familiar example would be that of phosphorus, which has three allotropes each of which have a different colour; white, red, and black, and several variations within these three basic forms. The structure of the red form is not certain, although one form of red phosphorus has been elucidated there is no clear evidence that the other red forms are the same.

All of the allotropes have molecular structures with corresponding molecular orbitals. It is the electrons within these orbitals that absorb light that promote electrons to higher molecular orbitals giving the charcteristic colour by subtraction from white light.

The term oxidation state is a convenient way of describing the effective condition of an atom in a compound in terms of the charge that it would have if it were ionic. It is not intended to suggest that atoms have a specific charge, but it is a useful way of describing the state or condition that an atom of an element is in within a compound; useful for the purposes of redox chemistry.

Whether an element is assigned a specific oxidation state, for example, +1 or -1 oxidation state depends on what it is bonded to. THIS IS IMPORTANT

If an atom of a specific element is bonded to an atom that is MORE electronegative, then it is assigned a positive oxidation state, and vice versa.

In reality this assignment is for descriptive terms only - it doesn’t really imply anything about the atoms or the bonds apart from the fact that one atom will tend to attract electrons more than the other.

Take the following nitrogen example:

In ammonia, NH3, nitrogen is more electronegative than hydrogen and so is assigned a negative oxidation state of 3- (assuming that each hydrogen is +1)

When bonded to oxygen however, as is N2O, the nitrogen is assigned a positive oxidation state of +1 (as oxygen is -2 (usually except for in peroxides) when attached to something less electronegative)

The extreme case that highlights this is F2O.

What is the oxidation state of oxygen here? well since oxygen is LESS electronegative that fluorine, and fluorine always has an oxidation state of -1,  it must be +2.

In summary, the most important factor to establish is which is the more electronegative of the two attached atoms. This atom takes the negative oxidation state, which can then be calculate from the valencies of the atoms in question.

There are several misconceptions about electrolysis, such as the idea that electricity flows across the electrolysis cell.

The conditions required for electrolysis are an electrolyte with ions that are free to move. They may be in solution or in the form of a molten salt. There must then be an electrical potential applied by means of electrodes across the electrolyte. This electrical potential creates a large electrostatic force field that attracts the charged ions.

The actual charges on the ions are what attracts them to the electrodes (that are charged electrically) the positive ions migrate to the negative electrode (the cathode) and the negative ions migrate to the positive electrode (the anode).

When a negative ion arrives at the positive electrode it has its extra electron stripped off by the electrode. This electron continues on to the the battery just like normal electricity (all electrons are, of course, indistinguishable)

Cl-      –>      Cl + 1e

When a positive ion arrives at the negative electrode it finds billions of electrons just waiting to fill up its positive ‘hole’. One of these extra electrons jumps onto the positive ion cancelling it out.

Na+    +     1e      –>      Na

The overall effect of these two processes, or electrode reactions, is that one electron has left the cathode and one electron has arrived at the anode. From the point of view of the battery and external circuit one electron has left the negative side and has arrived at the positive side. i.e. a current has flowed around the circuit.

Nothing actually flows across the electrolysis cell, the electrons are absorbed by one electrode and produced at the other by two different reactions. For these reactions to be able to occur the ions in the cell must be able to move, i.e. in solution or molten.

The net result for the electrolyte is that the sum of the two electrode reactions has happened in the cell.

2Cl-     –>      Cl2 + 2e
2Na+     +     2e      –>        2Na
———————————————–

2Cl-     +      2Na+       –>         2Na      +           Cl2

Or

NaCl(l)     —->   2Na(l)    +           Cl2(g)

The chlorine reaction was doubled up to show that chlorine is released as Cl2 molecules. The sodium is double up to equalise the charge. Under the conditions of electrolysis of molten sodium chloride the sodium is released as a liquid.

All liquids contain particles in which the energy distribution is govered by the laws of statistics. This energy distribution may be plotted as a curve, called the Maxwell Boltzmann graph that shows some particles with very little energy, some with large amounts of energy and the bulk of the particles with energy somwhere in between.  Those particles with large amounts of energy can escape from the body of the liquid producing a gas formed of liquid particles above the surface. This is called the vapour and the pressure that it exerts is called the vapour pressure.

The boiling point of a liquid is the temperature at which the vapour pressure equals the outside (atmospheric pressure). At this point bubbles of vapour form in the body of the liquid at this temperature - we call this “boiling”

The bubbles of vapour forming in the liquid can literally push the water apart as they have the same pressure as the atmosphere.

If the external pressure decreases then the vapour pressure will be able to equal it at a lower temperature. i.e. the liquid boils at a lower temperature.

To detect the presence of the OH group regardless there are several simple tests that can be done, however they all have the disadvantage of interference by water.

A small piece of sodium metal can be placed in the alcohol and a steady stream of hydrogen bubbles gives a positive indication. Remember that the presence of water will also cause hydrogen to be eveloved.

The suspected alcohol may be mixed with some PCl5 and the evolution of misty HCl gas is a positive indication of an OH group. Once again water interferes.

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Primary, secondary and tertiary alcohols 

There are three different types of alcohol depending on what is attached to the carbon holding the OH group.

• Primary - one alkyl group attached to the carbon holding the -OH
• Secondary - two alkyl groups attached to the carbon holding the -OH
• Tertiary alcohols - three alkyl groups attached to the carbon holding the -OH

To differentiate between the three it is possible to use oxidation with sodium dichromate in dilute sulphuric acid and heat.

The 1º alcohol oxidises to an aldehyde (alkanal) and then to a carboxylic acid (alkanoic acid)

The 2º alcohol oxidises to a ketone (alkanone) and stops there.

The 3º alcohol cannot be oxidised under these conditions.

Another (infrequently used) test is Lucas’ test for 1º, 2º, 3º alcohols.

It involves shaking the alcohol with ZnCl2 and dilute HCl. The 3º alcohol goes cloudy almost at once, the 2º alcohol goes cloudy after a few minutes whereas the 1º alcohol needs concentrated HCl to go cloudy. The cloudiness is due to the formation of the haloalkane that is immiscible with the aqueous solution of the zinc chloride and forms tiny droplets of an organic phase within the aqueous phase.

Hydrogen peroxide decomposes according to the equation: 2H2O2 –> 2H2O + O2

If you start with 100 volume H2O2 then 1 dm3 of solution will release 100 dm3 oxygen (at RTP)

This means that 1 litre of H2O2 solution releases 100/24 moles of oxygen and as 2 moles of H2O2 decompose to release 1 mole of oxygen then there must be 2 x 100/24 moles of H2O2 in the solution

This is equal to 8.33 mol/dm3

SO2 seems to be a little confusing as there are several different descriptions circulating. Some books claim that it has an expanded octet although there is no real evidence to support this. It can be simply described without octet expansion in that there are only 8 electrons around the central sulphur atom.

On a simplistic Lewis structure basis this can be achieved with one lone pair, one dative covalent bond and one double bond. According to bond measurements, however, the two S-O bonds are in resonance and have the equivalence of one and a half bonds each (three electrons per bond). Both of the S-O bonds are of equal length and energy as predicted by this model. The bond angle is that expected from three regions of electron density giving a slightly distorted trigonal planar electronic arrangement and an O-S-O bond angle of 119º.