By Sunil Bhardwaj


If in a titration current is measured as a function of volume of titratnt, to determine the end point. The titration is known as Amperometric titration. The graph of current Vs volume of titrant added is plotted and from the graph endpoint is calculated. The Nature of Amperometric titration curves can be explained with three different cases. when

1. Titrand (Reactant) is reducible species: e.g. titration of Pb+2 ions with sulphate (SO4-2) OR oxalate ions C2O4-2. In this case Pb+2 ions solution is taken in beaker is titrand and SO4-2 ions solution is filled in burette as titrant. As the titration proceeds, the concentration of Pb+2 decreases due to formation of PPt. of PbSO4. Hence diffusion current also decreases till it becomes minimum at the end point. The diffusion current remain constant beyond the end point, since the titrant does not yield a diffusion current, During the titration the potential applied to the cell is kept constant at -0.8 volt. (VS S.C.E.)$${ Pb }^{ 2+ }+{ SO }_{ 4 }^{ 2- }\longrightarrow PbS{ O }_{ 4 }(s)\downarrow $$


2. Titrant is reducible species: In this case, as the titration proceeds the current remains steady till the equivalence point. Since no species is reduced. However at the equivalence point the added titrant is reduced and the current is steadily increases.e.g. titration of non reducible Mg+2 ions with reducible 8 hydroxy quinoline (oxine) at a potential of -1.6 volt (Vs SCE)



3. Both titrand and titrant are reducible: The current first decreases due to the removal of the titrand (reactant) due to reaction with titrant. The current is minimum at the end point. On further addition of the titrant the current once again increases. When we plot a graph of current V/S Vol. of titrant. Thus V shaped curve is obtained.e.g. titration of Pb+2 ions V/S K2Cr2O7 at a potential of -0.8 Volt. Vol. of titrant (Vs SCE).$$ { Pb }^{ 2+ }+{ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }\longrightarrow Pb{ Cr }_{ 2 }{ O }_{ 7 }+2{ K }^{ + } $$


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