Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

Using the data given below: $E_{Ce^{4 +} / Ce^{3 +}}^{\circ} = 1.61 V$ , $E_{Br_{2} / Br^{-}}^{\circ} = 1.09 V$ , $E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = 0.77 V$ , $E_{Zn^{2 +} / Zn}^{\circ} = - 0.76 V$ .

Find out which of the following is the strongest oxidising agent.

Options:

Answer:

Solution:

Oxidising power rises with the reduction potential. $Ce^{4 +} / Ce^{3 +}$ is highest (1.61 V), so $Ce^{4 +}$ is the strongest oxidising agent.

Answer: (C).

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: QB

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