(A) At Anode: $^-OOC-CH_2-CH_2-COO^- \rightarrow CH_2=CH_2 + 2CO_2 + 2e^-$. For $2 \ F$ electricity,$3 \ moles$ of gas ($1 \ mole \ C_2H_4$ and $2 \ m

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

(A) At Anode: $^-OOC-CH_2-CH_2-COO^- \rightarrow CH_2=CH_2 + 2CO_2 + 2e^-$. For $2 \ F$ electricity,$3 \ moles$ of gas ($1 \ mole \ C_2H_4$ and $2 \ moles \ CO_2$) are produced.At Cathode: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. For $2 \ F$ electricity,$1 \ mole \ H_2$ gas is produced.Total moles of gas for $2 \ F = 3 + 1 = 4 \ moles$.Total moles of gas for $0.2 \ F = (4/2) \times 0.2 = 0.4 \ moles$.Total volume at $STP = 0.4 \times 22.4 = 8.96 \ L$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Difficult

Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing $0.2 \ F$ electricity through an aqueous solution of potassium succinate,the total volume of gases (at both cathode and anode) at $STP$ ($1 \ atm$ and $273 \ K$) is .............. $L$

JEE MAIN 2016 All JEE MAIN

A
$8.96$
B
$4.48$
C
$6.72$
D
$2.24$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Faraday’s law of electrolysis

Previous Year

JEE MAIN 2016 Paper All JEE MAIN years →

Faraday's laws of electrolysis are related to the

MediumIIT 1983

View Solution

How many amperes must be passed through a Down's cell to produce sodium metal at a rate of $30.0 \ kg/h$?

Medium

View Solution

When the same quantity of electricity is passed through aqueous solutions of $AgNO_3$ and $CuSO_4$,if the number of atoms of $Ag$ and $Cu$ deposited are $x$ and $y$ respectively,then determine the correct relationship.

Easy

View Solution

The laws of electrolysis were proposed by

Easy

View Solution

$A$ current of $10.0 \, A$ flows for $2.00 \, h$ through an electrolytic cell containing a molten salt of metal $X$. This results in the decomposition of $0.250 \, mol$ of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is $(F = 96,500 \, C \, mol^{-1})$ (in $+$)

DifficultJEE MAIN 2014

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing $0.2 \ F$ electricity through an aqueous solution of potassium succinate,the total volume of gases (at both cathode and anode) at $STP$ ($1 \ atm$ and $273 \ K$) is .............. $L$

Similar Questions

Faraday's laws of electrolysis are related to the

How many amperes must be passed through a Down's cell to produce sodium metal at a rate of $30.0 \ kg/h$?

When the same quantity of electricity is passed through aqueous solutions of $AgNO_3$ and $CuSO_4$,if the number of atoms of $Ag$ and $Cu$ deposited are $x$ and $y$ respectively,then determine the correct relationship.

The laws of electrolysis were proposed by

$A$ current of $10.0 \, A$ flows for $2.00 \, h$ through an electrolytic cell containing a molten salt of metal $X$. This results in the decomposition of $0.250 \, mol$ of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is $(F = 96,500 \, C \, mol^{-1})$ (in $+$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module