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NEET CHEMISTRYElectrochemistryEasy

Question

Standard electrode potential for Sn4+/Sn2+\text{Sn}^{4+}/\text{Sn}^{2+} couple is +0.15 V+0.15 \text{ V} and that for Cr3+/Cr\text{Cr}^{3+}/\text{Cr} couple is 0.74 V-0.74 \text{ V}. These two couples in their standard state are connected to make a cell. The cell potential will be:

A

+0.89 V

B

+0.18 V

C

+1.83 V

D

+1.199 V

Step-by-Step Solution

The standard reduction potentials are given as: ESn4+/Sn2+=+0.15 VE^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} = +0.15 \text{ V} ECr3+/Cr=0.74 VE^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \text{ V}

The species with the higher (more positive) standard reduction potential will act as the cathode (undergo reduction), and the one with the lower (more negative) standard reduction potential will act as the anode (undergo oxidation). Thus, the Sn4+/Sn2+\text{Sn}^{4+}/\text{Sn}^{2+} couple acts as the cathode and the Cr3+/Cr\text{Cr}^{3+}/\text{Cr} couple acts as the anode.

The standard cell potential (EcellE^\circ_{\text{cell}}) is calculated using the formula: Ecell=EcathodeEanodeE^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} Ecell=+0.15 V(0.74 V)E^\circ_{\text{cell}} = +0.15 \text{ V} - (-0.74 \text{ V}) Ecell=+0.15 V+0.74 V=+0.89 VE^\circ_{\text{cell}} = +0.15 \text{ V} + 0.74 \text{ V} = +0.89 \text{ V}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Electrochemistry. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYElectrochemistrystandardelectrodepotentialtextsntextsncouple

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