The number of Faradays F required to produce | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(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

Similar Questions

The number of $Faradays$ needed to reduce $4 \ g$ equivalents of $Cu^{++}$ to $Cu$ metal will be:

$1$ mole of electrons corresponds to how much Faraday,how much Coulomb,and how many electrons?

Two electrolytic cells,one containing acidified ferrous chloride $(FeCl_2)$ and the other containing acidic ferric chloride $(FeCl_3)$,are connected in series. When an electric current is passed through the cells,what is the ratio of iron deposited at the cathodes in the two cells?

Which equation represents Faraday's first law of electrolysis? ($m \rightarrow$ mass,$c \rightarrow$ current,$t \rightarrow$ time,$z \rightarrow$ electrochemical equivalent)

$6.5 \, g$ of $Mg$ is deposited during the electrolysis of fused $MgCl_2$. What will be the volume of chlorine gas (in $L$) discharged at the electrode at $STP$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module