Above plot represents the variation of molar conductivity against C (where, C = molar concentration of the electrolyte). Select the incorrect options among the following.

Curve I is for weak electrolyte and curve II is for strong electrolyte.

Curve I is for strong electrolyte and curve II is for weak electrolyte.

Above plot represents the variation of molar conductivity against C (where, C = molar concentration of the electrolyte). Select the incorrect options among the following.

Curve I is for weak electrolyte and curve II is for strong electrolyte.

A current of 2 Ampere is passed for one hour between nickel electrodes in 0 . 5 L of 2 M Ni NO 3 2 solution. What will be the molarity of the solution at the end of the electrolysis?

The electrolysis of Ni NO 3 2 is as follows: At cathode: Ni + 2 + 2 e - → Ni At anode: Ni → Ni + 2 + 2 e - The amount of nickel deposited at the cathode is equal to the amount of nickel dissolved at the anode. Given: i = 2 Ampere , 0 . 5 L of 2 M Ni NO 3 2 solution, Ni electrode. At the cathode, reduction of Ni + 2 ion into Ni s takes place and at the anode, oxidation of Ni s into Ni + 2 ion takes place. So, there is no change in the concentration of Ni + 2 and NO 3 - ion in the solution and the molarity of the solution at the end of electrolysis will be 2 M Ni NO 3 2 .

One faraday of electricity deposits one mole of Na from the molten salt but 1 3 mole of Al from an aluminium salt. Why?

The electrode reactions for deposition of sodium and aluminium are as follows: Na + aq + e - → Na s Al + 3 aq + 3 e - → Al s One mole electron or one faraday can deposit one mole of sodium but three moles of electrons or three faraday are required to deposit one mole of aluminium. According to faraday's II law, when equal amount of electricity is passed through different electrolytic solutions, it results in deposition of one equivalent mass of the element at the respective electrode. So, 1 faraday of electricity deposits 1 equivalent of Na and 1 equivalent of Al metal. Valence factor ( n - factor) of Na is 1 and Al is 3 . Number of Moles = No . of equivalents n - factor Number of moles of Na = 1 1 = 1 mole Number of moles of Al = 1 3 moles

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Above plot represents the variation of molar conductivity against $\sqrt{C}$ (where, $C =$ molar concentration of the electrolyte). Select the incorrect options among the following.

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Important Questions on Electrochemistry

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 1

An aqueous solution of copper sulphate is electrolysed using platinum electrodes in one case and copper electrodes in another case. Will the products of electrolysis be the same or different. Give reason?

EASY

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 2

State the products of electrolysis obtained on the cathode and the anode in the following case:
A dilute solution of

H_{2} {SO}_{4}

with platinum electrodes.

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 2

State the products of electrolysis obtained on the cathode and the anode in the following case:
An aqueous solution of

{AgNO}_{3}

with silver electrodes.

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JEE Main

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 3

A current of

2

Ampere is passed for one hour between nickel electrodes in

0.5 L

2 M Ni {({NO}_{3})}_{2}

solution. What will be the molarity of the solution at the end of the electrolysis?

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Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 4

On electrolysis of an aqueous solution of

NaCl

, why is

H_{2}

and not

Na

liberated at the cathode?

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 4

One faraday of electricity deposits one mole of

Na

from the molten salt but

\frac{1}{3}

mole of

Al

from an aluminium salt. Why?

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JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 5

In the electrolysis of an aqueous solution of sodium sulphate,

2.4 L

of oxygen at STP was liberated at anode. The volume of hydrogen at STP liberated at cathode would be:

HARD

JEE Main

IMPORTANT

Chapterwise/Topicwise Daily Practice Problems (DPP) Physical Chemistry Part - 2 JEE Main & Advanced>Chapter 3 - Electrochemistry>DPP 3.2>Q 6

What mass of silver could be plated out on a serving tray by electrolysis of a solution containing silver in

+ 1

oxidation state for a period of

8

hours with a current of

8.46 amperes

? What is the area of tray if thickness of the silver plating is

0.00254 cm

? Given density of the silver is

10.5 g {cm}^{- 3}

Above plot represents the variation of molar conductivity against C (where, C= molar concentration of the electrolyte). Select the incorrect options among the following.

Important Questions on Electrochemistry

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Above plot represents the variation of molar conductivity against $\sqrt{C}$ (where, $C =$ molar concentration of the electrolyte). Select the incorrect options among the following.