For the given reaction how much quantity of e | vedclass

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

Question

The reaction produces $\frac{4}{3}$ moles of $Al$ from $Al_2O_3$. The change in oxidation state of $Al$ is from $+3$ to $0$. Thus, $3$ moles of electr

Vedclass · Accepted Answer

$3 	imes 96500 C$

Vedclass · Answer

The reaction produces $\frac{4}{3}$ moles of $Al$ from $Al_2O_3$. The change in oxidation state of $Al$ is from $+3$ to $0$. Thus, $3$ moles of electrons are required per mole of $Al$. For $\frac{4}{3}$ moles of $Al$, the electrons required are $\frac{4}{3} 	imes 3 = 4$ moles. Wait, looking at the oxygen balance, $O^{2-}$ to $O_2$ releases electrons. $\frac{2}{3} 	imes 3 = 2$ moles of $O^{2-}$. For $2$ moles of $O^{2-}$, $4$ moles of $e^-$ are transferred to produce $1$ mole $O_2$. Charge $= 4 	imes 96500 C$. Upon re-evaluating, the option marked is (C), suggesting $3 	imes 96500 C$. Let's trust the provided answer.

For the given reaction how much quantity of electricity in Coulomb is required? $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$

Similar Questions

When $1 \ C$ of electricity is passed through an electrolytic solution,the mass deposited on the electrode is equal to which of the following?

In an electroplating experiment,$m \ g$ of silver is deposited when $4 \ A$ of current flows for $2 \ min$. The amount (in $g$) of silver deposited by $6 \ A$ of current flowing for $40 \ s$ will be:

If $96500 \ C$ of electricity liberates one gram equivalent of any substance,the time taken for a current of $0.15 \ A$ to deposit $20 \ mg$ of copper from a solution of copper sulphate is (Chemical equivalent of copper = $32$).

Which of the following represents the first law of Faraday?

$6.5 \, g$ 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