When 0.04 F of electricity is passed through a solution of CaSO 4 , then the weight of Ca 2 + metal deposited at the cathode is

Ca 2 + + 2 e - → Ca E Ca = 4 0 2 = 2 0 w Ca = E Ca × No. of faradays = 2 0 × 0.04 = 0.8 g .

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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

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When 0.04 F of electricity is passed through a solution of CaSO₄, then the weight of Ca²⁺metal deposited at the cathode is

Answer:

Solution:

${Ca}^{2 +} + 2 e^{-} \to Ca $

$E_{Ca} = \frac{40}{2} = 20$

$w_{Ca} = E_{Ca} \times No. of faradays = 20 \times 0.04 = 0.8 g$ .

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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