equivalent weights of oxidants and reductants

[satrohraj 10]

how do we calculate the equivalents weights of different oxidants and reductants!

I know how to calculate the EW of acids, basis and salts

 

if possible give an example

 

What is the difference between EW and GEW (gram equivalent weight)?

 

thank you

[woelen 18]

If you can do this for acids and bases, then you can also do it for reductors and oxidizers. With acids and bases, you are equalling H(+) ions and/or OH(-) ions on both side of the equation (exactly neutralizing). Now you do the same for electrons, transferred in a redox reaction.

 

Example: redox between sodium sulfite and potassium permanganate in sufficiently strong acidic medium.

 

KMnO4 : Oxidizer, Mn going from +7 to +2 oxidation state, 5 electrons gained.

Na2SO3: Reductor, S going from +4 to +6 oxidation state, 2 electrons lost.

 

In order to have electron balance on both sides, you need 2 'molecules' of KMnO4 for 5 'molecules' of Na2SO3. The rest of the task you can do yourself, just proceed as with acids/bases.

 

I'm not sure about GEW. It is an old-fashioned way of expressing things, IIRC it stands for normality (number of mols of electrons per unit solution, notation N), e.g. 1 M KMnO4 is 5N KMnO4.

[satrohraj 10]

that's not I am talking about

 

EW_acids = molecular weight or formula weight of the acid / basicity of the acid

EW_Bases = " " " " " " base / acidity of the base

EW_salts = " " " " " " salt / total number of positive charges on cation or negative charges of anion

EW_ oxidant = ?/?

 

***Here "/" means 'divided by'

[John Cuthber 3775]

EW_ oxidant = molecular weight or formula weight of the oxidant/ number of electrons transfered

This is a bit awkward because the number of electrons transfered depends on the conditions. For example KMnO4 can transfer 5 electrons in acid conditions to give a Mn++ salt or 3 in near neutral conditions to give MnO2

or, in rather strong alkali it can accept just one electron and give the green MnO4-- ion.

That means the equivalent weight of KMnO4 is 158.03/ 5 or 158.03/3 or 158.03 depending on the conditions.

It's no wonder that equivalent weight and Normality have been largely replaced by Molecular weight and molarity.

A solution of 158.03 g of KMnO4 in a litre of water (I'm not sure if it's that soluble- but that's not the point) could be labeled as "N", "0.2N" or "0.3333N" depending on what was going to be done with it.

However it would be one molar no matter what.