The number of Faradays F required to produce | vedclass

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(B) The reduction reaction for calcium is: $Ca^{2+} + 2e^{-} \rightarrow Ca_{(s)}$The valence factor ($n$-factor) for $Ca^{2+}$ is $2$.According to Fa

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$1$

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(B) The reduction reaction for calcium is: $Ca^{2+} + 2e^{-} ightarrow Ca_{(s)}$The valence factor ($n$-factor) for $Ca^{2+}$ is $2$.According to Faraday's $1^{st}$ law of electrolysis,the number of Faradays required is equal to the number of gram equivalents of the product formed.Number of moles of $Ca = \frac{	ext{mass}}{	ext{molar mass}} = \frac{20 \ g}{40 \ g \ mol^{-1}} = 0.5 \ mol$.Number of gram equivalents = $	ext{moles} 	imes n	ext{-factor} = 0.5 	imes 2 = 1 \ F$.

The number of Faradays $F$ required to produce $20 \ g$ of calcium from molten $CaCl_2$ (Atomic mass of $Ca = 40 \ g \ mol^{-1}$) is

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