Chemistry - Electrochemistry Question with Solution | TestHub
ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsEasy2 minPYQ_2020
ChemistryEasysingle choice
Consider the following relations for emf of an electrochemical cell
EMF of cell = (oxidation potential of anode) - (Reduction potential of cathode)
EMF of cell = (oxidation potential of anode) + (Reduction potential of cathode)
EMF of cell = (reduction potential of anode) + (oxidation potential of cathode)
EMF of cell = (oxidation potential of anode) - (oxidation potential of cathode)
Which of these expressions are correct?
Options:
Answer:
D
Solution:
EMF of a cell = reduction potential of cathode - reduction potential of anode
= reduction potential of cathode + oxidation potential of anode
= oxidation potential of anode - oxidation potential of cathode.
Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells
⏱ 2mℹ️ Source: PYQ_2020
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