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NEET CHEMISTRYElectrochemistryMedium

Question

Two half cell reactions are given below: Co3++eCo2+ ;\ECo2+/Co3+=1.81 VCo^{3+} + e^- \rightarrow Co^{2+} \ ; \E^{\circ}_{Co^{2+}/Co^{3+}} = -1.81\text{ V} 2Al3++6e2Al(s) ;\EAl/Al3+=+1.66 V2Al^{3+} + 6e^- \rightarrow 2Al(s) \ ; \E^{\circ}_{Al/Al^{3+}} = +1.66\text{ V} The standard EMF of a cell with feasible redox reaction will be:

A

+7.09 V

B

+0.15 V

C

+3.47 V

D

–3.47 V

Step-by-Step Solution

For a feasible cell reaction, the standard cell potential (EcellE^{\circ}_{cell}) must be positive. The given potentials are standard oxidation potentials (as inferred from the subscript notation): ECo2+/Co3+=1.81 VE^{\circ}_{Co^{2+}/Co^{3+}} = -1.81\text{ V}, so the standard reduction potential is ECo3+/Co2+=+1.81 VE^{\circ}_{Co^{3+}/Co^{2+}} = +1.81\text{ V}. EAl/Al3+=+1.66 VE^{\circ}_{Al/Al^{3+}} = +1.66\text{ V}, so the standard reduction potential is EAl3+/Al=1.66 VE^{\circ}_{Al^{3+}/Al} = -1.66\text{ V}. Since ECo3+/Co2+>EAl3+/AlE^{\circ}_{Co^{3+}/Co^{2+}} > E^{\circ}_{Al^{3+}/Al}, the Co3+/Co2+Co^{3+}/Co^{2+} half-cell has a higher tendency to get reduced and will act as the cathode, while the Al/Al3+Al/Al^{3+} half-cell will act as the anode. The standard EMF of the cell is: Ecell=EcathodeEanodeE^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} Ecell=1.81 V(1.66 V)=1.81+1.66=+3.47 VE^{\circ}_{cell} = 1.81\text{ V} - (-1.66\text{ V}) = 1.81 + 1.66 = +3.47\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.

CHEMISTRYElectrochemistryreactionsrightarrowecirccocorightarrowecircalal

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