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(A) The total charge passed in Coulombs is $Q = I 	imes t = (25 	imes 10^{-3} \ A) 	imes (60 \ s) = 1.5 \ C$. The reduction reaction at the cathode

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$4.68 	imes 10^{18}$

Vedclass · Answer

(A) The total charge passed in Coulombs is $Q = I 	imes t = (25 	imes 10^{-3} \ A) 	imes (60 \ s) = 1.5 \ C$. The reduction reaction at the cathode is $Ca^{2+} + 2e^{-} ightarrow Ca$. According to the reaction,$2 \ F$ (where $1 \ F = 96500 \ C$) of charge deposits $1 \ 	ext{mole}$ of $Ca$ atoms,which is $6.022 	imes 10^{23}$ atoms. Therefore,the number of atoms deposited by $1.5 \ C$ is: $	ext{Number of atoms} = \frac{1.5 	imes 6.022 	imes 10^{23}}{2 	imes 96500} = 4.68 	imes 10^{18}$ atoms.

How many atoms of calcium will be deposited at the cathode when a current of $25 \ mA$ is passed through a $CaCl_2$ solution for $60 \ s$?

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