Calculate the quantity of electricity require | vedclass

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

Question

(D) The reduction reaction at the cathode is: $Ag^+ + e^- \rightarrow Ag(s)$.According to Faraday's law of electrolysis,the mass deposited $(W)$ is gi

Vedclass · Accepted Answer

$375.3$

Vedclass · Answer

(D) The reduction reaction at the cathode is: $Ag^+ + e^- ightarrow Ag(s)$.According to Faraday's law of electrolysis,the mass deposited $(W)$ is given by $W = \frac{M 	imes Q}{n 	imes F}$,where $M$ is the molar mass,$Q$ is the charge,$n$ is the number of electrons involved ($n=1$ for $Ag^+$),and $F$ is Faraday's constant $(96500 \ C \ mol^{-1})$.Substituting the given values: $0.42 = \frac{108 	imes Q}{1 	imes 96500}$.Solving for $Q$: $Q = \frac{0.42 	imes 96500}{108}$.$Q = 375.277 \ C \approx 375.3 \ C$.

Calculate the quantity of electricity required to produce $0.42 \ g$ of $Ag$ at the cathode during the electrolysis of an $AgNO_3$ solution. (Molar mass of $Ag = 108 \ g \ mol^{-1}$) (in $C$)

Similar Questions

The $SI$ unit of electrochemical equivalent is

When $1 \ C$ of charge is passed through an electrolyte solution,the mass deposited is equal to:

$5 \ A$ current is passed through $AgNO_3$ solution for $193 \ s$. If $0.972 \ g$ of $Ag$ is deposited,calculate the current efficiency (cell capacity) of the electrolytic cell. $[Ag = 108 \ u]$ (in $\%$)

Silver plating is done on a silver cup by passing a current of $965 \ A$ for $1 \ s$. How many grams of $Ag$ are deposited? $(Ag = 107.87 \ g/mol)$

When $10 \,A$ current is passed for $80 \,min$, the volume of hydrogen gas liberated at $STP$ is: (in $\,L$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module