How many atoms of calcium will be deposited f | vedclass

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Question

(A) The reaction for the deposition of calcium is: $Ca^{2+} + 2e^- ightarrow Ca(s)$.Quantity of electricity passed $(Q)$ $= I 	imes t = 25 	imes 1

Vedclass · Accepted Answer

$4.68 	imes 10^{18}$

Vedclass · Answer

(A) The reaction for the deposition of calcium is: $Ca^{2+} + 2e^- ightarrow Ca(s)$.Quantity of electricity passed $(Q)$ $= I 	imes t = 25 	imes 10^{-3} \, A 	imes 60 \, s = 1.5 \, C$.According to Faraday's law,$2 \, mol$ of electrons $(2 	imes 96500 \, C)$ deposit $1 \, mol$ of $Ca$ atoms ($6.022 	imes 10^{23}$ atoms).Number of $Ca$ atoms deposited $= \frac{1.5 \, C}{2 	imes 96500 \, C \, mol^{-1}} 	imes 6.022 	imes 10^{23} \, 	ext{atoms} \, mol^{-1}$.Number of $Ca$ atoms $= 4.68 	imes 10^{18}$ atoms.

How many atoms of calcium will be deposited from molten $CaCl_2$ by a current of $25 \, mA$ flowing for $60 \, s$?

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