The number of electrons required to reduce 4. | vedclass

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

Question

(B) The reduction reaction is: $Al^{3+} + 3e^- ightarrow Al$. Moles of $Al = \frac{	ext{Mass}}{	ext{Atomic mass}} = \frac{4.5 	imes 10^{-5} 	ext

Vedclass · Accepted Answer

$3.01 	imes 10^{18}$

Vedclass · Answer

(B) The reduction reaction is: $Al^{3+} + 3e^- ightarrow Al$. Moles of $Al = \frac{	ext{Mass}}{	ext{Atomic mass}} = \frac{4.5 	imes 10^{-5} 	ext{ g}}{27 	ext{ g/mol}} = 1.666 	imes 10^{-6} 	ext{ mol}$. Moles of electrons required $= 3 	imes 	ext{moles of } Al = 3 	imes 1.666 	imes 10^{-6} = 5 	imes 10^{-6} 	ext{ mol}$. Number of electrons $= 	ext{moles of electrons} 	imes N_A = 5 	imes 10^{-6} 	imes 6.022 	imes 10^{23} \approx 3.011 	imes 10^{18}$.

The number of electrons required to reduce $4.5 \times 10^{-5} \text{ g}$ of $Al^{3+}$ to $Al$ is:

Similar Questions

Which metal cannot be obtained by the electrolysis of its aqueous salt solution?

The number of electrons involved in the reaction when $1 \ Faraday$ of electricity is passed through an electrolytic solution is?

If a current of $0.5 \ A$ flows through a metallic wire for $2 \ hours$,then how many electrons would flow through the wire?

Which of the following is used as an electrolyte in a dry cell?

The main reason for not using a mercury electrolytic cell in $NaOH$ manufacture is that

Vedclass Test Series

Exam Paper Generator

Online Exam Module