Q92 · CSIR-NET Chemistry, December 2011

Paper: CSIR-NET December 2011 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Nernst Cells · Marks: 2 · Difficulty: Medium

For a potentiometric titration, in the curve of emf (E) vs volume (V) of the titrant added, the equivalence point is indicated by

(a)$|\mathrm{dE/dV}|=0 ;\ |\mathrm{d^2/d^2V}|=0$

(b)$|\mathrm{dE/dV}|=0 ;\ |\mathrm{d^2E/d^2V}|>0$

(c)$|\mathrm{dE/dV}|>0 ;\ |\mathrm{d^2E/d^2V}|=0$

(d)$|\mathrm{dE/dV}|>0 ;\ |\mathrm{d^2E/d^2V}|>0$

Answer

Answer: C ✓ checked by 4AB · confidence high

Explanation

At the equivalence point the potential changes fastest: |dE/dV| is a maximum (> 0) and the second derivative passes through zero.

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