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

How much Faraday of electricity is required to reduce $1.5 \ mole$ of $KMnO_4$ into $MnO_2$ in a basic medium (in $F$)?

What is the number of moles of electrons passed when a current of $2 \text{ A}$ flows through an electrolyte solution for $10 \text{ minutes}$?

Faraday's laws of electrolysis will fail when

The quantity of electricity needed to separately electrolyze $1 \ M$ solution of $ZnSO_{4}$, $AlCl_{3}$ and $AgNO_{3}$ completely is in the ratio of

$0.066 \ g$ of metal was deposited when a current of $2 \ A$ is passed through a metal ion solution for $100 \ s$. What is the electrochemical equivalent (in $g \ C^{-1}$) of the metal?

Vedclass Test Series

Exam Paper Generator

Online Exam Module