Q63 · CSIR-NET Chemistry, June 2016
Paper: CSIR-NET June 2016 · Subject: Physical Chemistry · Chapter: Electrochemistry · Topic: Electrochemistry – General · Marks: 2 · Difficulty: Easy
In a potentiometric titration, the end point is characterised by
(a)$\frac{\mathrm{dE}}{\mathrm{dV}}=0 ; \frac{\mathrm{d}^{2} \mathrm{E}}{\mathrm{d}^{2} \mathrm{V}}=0$
(b)$\frac{\mathrm{dE}}{\mathrm{dV}} \neq 0 ; \frac{\mathrm{d}^{2} \mathrm{E}}{\mathrm{d}^{2} \mathrm{V}}=0$
(c)$\frac{\mathrm{dE}}{\mathrm{dV}}=0 ; \frac{\mathrm{d}^{2} \mathrm{E}}{\mathrm{d}^{2} \mathrm{V}} \neq 0$
(d)$\frac{\mathrm{dE}}{\mathrm{dV}} \neq 0 ; \frac{\mathrm{d}^{2} \mathrm{E}}{\mathrm{d}^{2} \mathrm{V}} \neq 0$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
At the end point of a potentiometric titration, dE/dV is maximum (non-zero) and d²E/dV² = 0 (b).
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