A solution of copper sulphate is electrolysed | vedclass

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

Question

(A) The mass of the substance deposited during electrolysis is given by Faraday's law: $m = (I 	imes t 	imes M) / (n 	imes F)$.Here, the current $I

Vedclass · Accepted Answer

$(1)$ $0.2938 	ext{ g}$

Vedclass · Answer

(A) The mass of the substance deposited during electrolysis is given by Faraday's law: $m = (I 	imes t 	imes M) / (n 	imes F)$.Here, the current $I = 1.5 	ext{ A}$, time $t = 10 	ext{ minutes} = 600 	ext{ s}$, molar mass $M = 63 	ext{ g mol}^{-1}$, and $n = 2$ (since $Cu^{2+} + 2e^{-} ightarrow Cu$).Substituting the values: $m = (1.5 	imes 600 	imes 63) / (2 	imes 96487)$.$m = 56700 / 192974 \approx 0.2938 	ext{ g}$.Thus, the mass of copper deposited is $0.2938 	ext{ g}$.

$A$ solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at the cathode is:
(Given: Molar mass of $Cu = 63 \text{ g mol}^{-1}$; $1F = 96487 \text{ C mol}^{-1}$)

Similar Questions

What is the number of moles of electrons passed when a current of $5 \ A$ is passed through a solution of $FeCl_3$ for $20 \ minutes$?

How many electrons flow when a current of $5 \ A$ is passed through a cell for $200 \ s$?

Calculate the time required in seconds to deposit $6.35 \ g$ of copper from its salt solution by passing a $5 \ A$ current. [Molar mass of $Cu = 63.5 \ g \ mol^{-1}$]

According to the first law of Faraday,the weight of a substance discharged at the electrode is:

The amount of silver deposited on passing $2 \ F$ of electricity through an aqueous solution of $AgNO_3$ is: (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at the cathode is:(Given: Molar mass of $Cu = 63 \text{ g mol}^{-1}$; $1F = 96487 \text{ C mol}^{-1}$)