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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minPYQ_2006

ChemistryHardsingle choice

The charge required for the reduction of $1 mol$ of $MnO_{4}^{-}$ to $MnO_{2}$ is

Answer:

Solution:

$MnO_{4}^{-} + 2 H_{2} O + 3 e^{-} \to MnO_{2} + 4 OH^{-}$ Charge required for the reduction of $1 mol_{of MnO_{4}^{-}}$ to $MnO_{2} = 3 F$ .

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

⏱ 2mℹ️ Source: PYQ_2006

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