Stannous sulphate `(SnSO_4)` and potassium permanganate are used as oxidising agents in acidic medium for oxidation of ferrous ammonium sulphate to ferric sulphate. The ratio of number of moles of stannous sulphate required per mole of ferrous ammounium sulphate to the number of moles of `KMnO_4` required per mole of ferrous ammonium sulphate, is :

Stannous sulphate (SnSO_(4)) and potassium permanganate are used as oxidising agents in acidic medium for oxidation of ferrrous ammnium sulphate to ferric sulphate. The ration of number of moles of stannous sulphate required per mole of ferrous ammonium sulphate to the number of moles of KMnO_(4) required per mole of ferrous ammonium sulphate, is:

How will K_2Cr_2O_7 oxidise: Ferrous sulphate to ferric sulphate.

How will ozone oxidise the following : Acidified ferrous sulphate to ferric sulphate.

Why nitric acid acts as an oxididing agent? How it oxidises: Ferrous Sulphate to Ferric sulphate acid.

When potassium permanganate is titrated against ferrous ammonium sulphate in acidic medium, the equivalent mass potassium permanganate is ,

When an acidified solution of ferrous ammonium sulphate is treated with potassium permanganate solution, the ion which is oxidized is

Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is

1 mole of equimolar mixture of ferric oxalate and ferrous oxalate requires x mole of KMnO_4 in acidic medium for complete oxidation. x is :

Why does ozone ( 0_3 ) acts as a powerful oxidising agent ? Give examples acidified ferrous sulphate.

"Equivalent mass" =("Molecular mass/Atomic mass" )/("n-factor") n-factor is very important in redox as well as non-redox reactions. With the help of n-factor we can predict the molar ratio of the reactant species taking part in reactions. The reciprocal of n-factor's ratio of the reactions is the molar ratio of the reactants. In general n-factor of acid/base is number of moles of H^(+)//OH^(-) furnished per mole of acid/base n-factor of a reactant is number of moles electrons lost or gained per mole of reactant. Example 1: (1)In acidic medium : KMnO_(4) (n=5)to Mn^(2+) (2) In neutral medium : KMnO_(4)(n=3) to Mn^(2+) (3) In basic medium : KMnO_(4)(n=1) to Mn^(6+) Example 2 : FeC_(2)O_(4)to Fe^(3+)+2CO_(2) Total number of moles e^(-) lost by 1 mole of FeC_(2)O_(4) =1+1xx2 implies 3 n-factor of Ba(MNO_(4))_(2) in acidic medium is :