The standard oxidation potentials of Zn and Ag in water at ${25^ \circ }C$ are, ${\rm{Zn}}\left( s \right) \to {\rm{Z}}{{\rm{n}}^{2 + }} + 2e;{E^ \circ } = 0.76{\rm{V}}$ ${\rm{Ag}}\left( s \right) \to {\rm{A}}{{\rm{g}}^ + } + e;{E^ \circ } = - 0.80{\rm{V}}$ Which reaction actually takes place?

${\rm{Zn}}\left( s \right) + 2{\rm{A}}{{\rm{g}}^ + }\left( {aq} \right) \to {\rm{Z}}{{\rm{n}}^{2 + }}{\rm{ + Ag}}\left( s \right)$

${\rm{Z}}{{\rm{n}}^{2 + }} + 2{\rm{A}}{{\rm{g}}^ + }\left( s \right) \to 2{\rm{A}}{{\rm{g}}^ + }\left( {aq} \right) + {\rm{Zn}}\left( s \right)$

${\rm{Zn}}\left( s \right) + 2{\rm{Ag}}\left( s \right) \to {\rm{Z}}{{\rm{n}}^{2 + }}\left( {aq} \right) + {\rm{A}}{{\rm{g}}^ + }\left( {aq} \right)$

${\rm{Z}}{{\rm{n}}^{2 + }}\left( {aq} \right) + {\rm{A}}{{\rm{g}}^ + }\left( {aq} \right) \to {\rm{Zn}}\left( s \right) + {\rm{Ag}}\left( s \right)$

The equivalent conductivity of a solution containing 2.54 g of ${\rm{CuS}}{{\rm{O}}_4}{\kern 1pt} {\kern 1pt} {\rm{per L is}}{\kern 1pt} {\kern 1pt} {\kern 1pt} 91.0{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}{\rm{e}}{{\rm{q}}^{ - 1}}$. Its conductivity would be

$2.9 \times {10^{ - 3}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

$1.8 \times {10^{ - 2}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

$2.4 \times {10^{ - 4}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

$3.6 \times {10^{ - 3}}{{\rm{\Omega }}^{ - 1}}{\rm{c}}{{\rm{m}}^2}$

Standard reduction potentials of the half reactions are given below : ${{\rm{F}}_2}\left( {\rm{g}} \right) + 2{e^ - } \to 2{{\rm{F}}^ - }\left( {aq} \right);{E^ \circ } = + 2.85{\rm{V}}$ ${\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right) + 2{e^ - } \to 2{\rm{C}}{{\rm{l}}^ - }\left( {aq} \right);{E^ \circ } = + 1.36{\rm{V}}$ ${\rm{B}}{{\rm{r}}_2}\left( l \right) + 2{e^ - } \to 2{\rm{B}}{{\rm{r}}^ - }\left( {aq} \right);{E^ \circ } = + 1.06{\rm{V}}$ ${{\rm{I}}_2}\left( s \right) + 2{{\rm{e}}^ - } \to 2{{\rm{I}}^ - }\left( {aq} \right);{E^ \circ } = + 0.53V$ The strongest oxidising and reducing agents respectively are :

${{\rm{F}}_2}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {\kern 1pt} {{\rm{I}}^ - }$

${\rm{B}}{{\rm{r}}_2}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {\rm{C}}{{\rm{l}}^ - }$

${\rm{C}}{{\rm{l}}_2}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {\rm{B}}{{\rm{r}}^ - }$

${\rm{C}}{{\rm{l}}_2}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} {{\rm{I}}_2}$

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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Easy

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Showing 1-30 of 138 questions

1 Easy

1 of 138

Understanding +4 | -1

The correct order of molar conductivity at infinite dilution of

\mathrm{LiCl},{\rm{NaCl {\kern 1pt}{\kern 1pt}and {\kern 1pt}{\kern 1pt}KCl}}

View answer and explanation

Correct option: B

\mathrm{KCl} > NaCl > LiCl

Option B (

\mathrm{KCl} > NaCl > LiCl

) is correct because ionic mobility decreases with increasing hydrated size, and Li+ has the largest hydrated size among these ions.

2 Easy

2 of 138

Application +4 | -1

A solution of copper sulphate

\left(\mathrm{CuSO}_4\right)

is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at the cathode (at. mass of Cu = 63 u) is

View answer and explanation

Correct option: B

0.2938 g

Option B is correct. The mass of copper deposited is calculated using Faraday's laws of electrolysis, considering the current, time, and the electrochemical equivalent of copper.

3 Easy

3 of 138

Application +4 | -1

The desired amount of charge for obtaining one mole of Al from

\mathrm{A}{\mathrm{l}^{3 + }}

View answer and explanation

Correct option: C

3 \times 96500\mathrm{C}

Option C (

3 \times 96500\mathrm{C}

) is correct because the reduction of one mole of

Al^{3+}

to Al requires 3 moles of electrons, and each mole of electrons requires 96500 Coulombs of charge.

4 Easy

4 of 138

Understanding +4 | -1

When an aqueous solution of lithium chloride is electrolysed using graphite electrodes :

View answer and explanation

Correct option: A

pH of the resulting solution increases

Option A (pH of the resulting solution increases) is correct because during electrolysis of aqueous lithium chloride, hydrogen gas is produced at the cathode, leading to an increase in hydroxide ion concentration and thus an increase in pH.

5 Easy

5 of 138

Understanding +4 | -1

The correct value of e.m.f. of cell is given by :

i)

{E_{cell}} = {E_{OP}}{\kern 1pt} anode - {E_{RP}}{\kern 1pt} cathode

ii)

{E_{cell}} = {E_{OP}}{\kern 1pt} anode + {E_{RP}}{\kern 1pt} cathode

iii)

{E_{cell}} = {E_{RP}}{\kern 1pt} anode + {E_{RP}}{\kern 1pt} cathode

iv)

{E_{cell}} = {E_{OP}}{\kern 1pt} anode - {E_{OP}}{\kern 1pt} cathode

View answer and explanation

Correct option: D

(ii) and (iv)

Option D ((ii) and (iv)) is correct because

{E_{cell}} = {E_{OP}}{\kern 1pt} anode + {E_{RP}}{\kern 1pt} cathode

and

{E_{cell}} = {E_{OP}}{\kern 1pt} anode - {E_{OP}}{\kern 1pt} cathode

are both valid ways to calculate cell potential.

6 Easy

6 of 138

Remembering +4 | -1

In electrolysis, oxidation takes place at:

View answer and explanation

Correct option: A

Anode

Option A is correct because oxidation, which is the loss of electrons, always occurs at the anode in any electrochemical cell, including during electrolysis.

7 Easy

7 of 138

Analyzing +4 | -1

Which one of the following has the highest molar conductivity?

View answer and explanation

Correct option: C

Potassium hexacyanoferrate (II)

Option C (Potassium hexacyanoferrate (II)) is correct because it dissociates into the largest number of ions, leading to the highest molar conductivity.

8 Easy

8 of 138

Remembering +4 | -1

Kohlrausch’s law states that at :

View answer and explanation

Correct option: C

Infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte

The correct answer is C because Kohlrausch's Law states that at infinite dilution, each ion contributes independently to the equivalent conductance, regardless of the other ion present.

9 Easy

9 of 138

Understanding +4 | -1

Which is correct representation for a cell at equilibrium?

View answer and explanation

Correct option: D

All of the above.

Option D (All of the above) is correct because all three equations represent the relationship between Gibbs free energy, equilibrium constant, and standard cell potential at equilibrium.

10 Easy

10 of 138

Application +4 | -1

Number of faraday required to liberate 8 g of

{\mathrm{H}_2}

is :

View answer and explanation

Correct option: A

Option A (8) is correct because 8g of H2 is 4 moles, and each mole requires 2 Faradays, so 4 moles require 8 Faradays.

11 Easy

11 of 138

Understanding +4 | -1

On passing electricity through dilute

{H_2}S{O_4}

between

pt

electrodes, the gas evolved at anode and cathode are

View answer and explanation

Correct option: C

{O_2},{H_2}

Option C (

{O_2},{H_2}

) is correct because during the electrolysis of dilute

{H_2}S{O_4}

, oxygen gas is evolved at the anode and hydrogen gas is evolved at the cathode.

12 Easy

12 of 138

Remembering +4 | -1

The cell constant is

View answer and explanation

Correct option: A

\frac{l}{a}

Option A (

\frac{l}{a}

) is correct because the cell constant is defined as the ratio of the distance between the electrodes (l) to the area of cross-section of the electrode (a).

13 Easy

13 of 138

Application +4 | -1

When electricity is passed through a solution of

AlC{l_3},13.5g\,Al

are deposited. The number of Faradays must be

View answer and explanation

Correct option: C

1.5

Option C (1.5) is correct because the deposition of 1 mole of Al requires 3 Faradays, and 13.5g of Al is 0.5 moles, thus requiring 1.5 Faradays.

14 Easy

14 of 138

Analyzing +4 | -1

Which of the following displacement does not occur.

View answer and explanation

Correct option: C

Cu + F{e^{2 + }} \to C{u^{2 + }} + Fe \downarrow

Option C (

Cu + F{e^{2 + }} \to C{u^{2 + }} + Fe \downarrow

) is correct because copper cannot displace iron from its solution due to copper having a lower reduction potential than iron.

15 Easy

15 of 138

Understanding +4 | -1

If the current is passed into the solution of an electrolyte:

View answer and explanation

Correct option: A

Anions move towards anode, cations towards cathode

Option A (Anions move towards anode, cations towards cathode) is correct because anions are negatively charged and move towards the positive electrode (anode), while cations are positively charged and move towards the negative electrode (cathode).

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16 Easy

16 of 138

Remembering +4 | -1

The charge required to liberate one gram equivalent of an element is

View answer and explanation

Correct option: B

1 F

Option C (Temperature) is correct because according to the Nernst equation, electrode potential is directly dependent on temperature.

17 Easy

17 of 138

Remembering +4 | -1

The approximate e.m.f. of a dry cell is :

View answer and explanation

Correct option: D

1.5 V

Option D is correct because the standard electromotive force (e.m.f.) of a dry cell is approximately 1.5 V.

18 Easy

18 of 138

Remembering +4 | -1

E.m.f. of a cell in terms of reduction potential of its left and right electrode is :

View answer and explanation

Correct option: C

E = {E_R} - {E_L}

Option C (

E = {E_R} - {E_L}

) is correct because the EMF of a cell is the difference between the reduction potential of the right-hand electrode (cathode) and the left-hand electrode (anode).

19 Easy

19 of 138

Analyzing +4 | -1

The standard oxidation potentials of Zn and Ag in water at

{25^ \circ }C

are,

\mathrm{Zn}\left( s \right) \to \mathrm{Z}{\mathrm{n}^{2 + }} + 2e;{E^ \circ } = 0.76\mathrm{V}

\mathrm{Ag}\left( s \right) \to \mathrm{A}{\mathrm{g}^ + } + e;{E^ \circ } = - 0.80\mathrm{V}

Which reaction actually takes place?

View answer and explanation

Correct option: A

\mathrm{Zn}\left( s \right) + 2\mathrm{A}{\mathrm{g}^ + }\left( {aq} \right) \to \mathrm{Z}{\mathrm{n}^{2 + }}\mathrm{ + Ag}\left( s \right)

The correct answer is A because Zinc has a positive oxidation potential and Silver has a negative oxidation potential, making Zinc oxidation and Silver reduction favorable.

20 Easy

20 of 138

Application +4 | -1

\mathrm{M}{\mathrm{g}^{2 + }} + 2e \to \mathrm{Mg}\left( s \right);E = - 2.37\mathrm{V}

\mathrm{C}{\mathrm{u}^{2 + }} + 2e \to \mathrm{Cu}\left( s \right);E = + 0.34\mathrm{V}

?

The e.m.f. of the cell

\mathrm{Mg}\left| {\mathrm{M}{\mathrm{g}^{2 + }}} \right|\left| {\mathrm{C}{\mathrm{u}^{2 + }}} \right|\mathrm{Cu}

is:

View answer and explanation

Correct option: A

2.71 V

Option A (2.71 V) is correct because the EMF of the cell is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.

21 Easy

21 of 138

Analyzing +4 | -1

Which one of the following solutions has highest conductance power?

View answer and explanation

Correct option: D

0.1M{\kern 1pt} {\kern 1pt} \mathrm{HCl}

Option D (

0.1M{\kern 1pt} {\kern 1pt} \mathrm{HCl}

) is correct because HCl is a strong acid and dissociates completely, leading to a higher concentration of ions and thus higher conductance.

22 Easy

22 of 138

Application +4 | -1

By passing 9.65 A current for 16 min 40 s , the volume of

{\mathrm{O}_2}

liberated at STP will be

View answer and explanation

Correct option: C

1120 mL

Option C is correct. The volume of

{O_2}

liberated at STP is calculated using Faraday's laws of electrolysis, considering the number of moles of electrons and the stoichiometry of the oxygen evolution reaction.

23 Easy

23 of 138

Remembering +4 | -1

A depolariser used in dry cell batteries is :

View answer and explanation

Correct option: B

Manganese dioxide

Option B is correct because Manganese dioxide acts as a depolarizer by oxidizing hydrogen gas produced at the cathode, preventing polarization and maintaining the cell's voltage.

24 Easy

24 of 138

Remembering +4 | -1

Galvanic cell is A device in which

View answer and explanation

Correct option: A

Chemical energy is converted into electrical energy.

Option A is correct because a galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions.

25 Easy

25 of 138

Understanding +4 | -1

In electrolysis of aqueous copper sulphate, the gas at anode and cathode are

View answer and explanation

Correct option: A

{\mathrm{O}_2}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {\mathrm{H}_2}

Option A (

{\mathrm{O}_2}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {\mathrm{H}_2}

) is correct because during the electrolysis of aqueous copper sulfate, oxygen gas is evolved at the anode and hydrogen gas is evolved at the cathode.

26 Easy

26 of 138

Understanding +4 | -1

The oxidation potential of Mg and Al are + 2.37 and + 1.66 volt respectively. The Mg in chemical reactions :

View answer and explanation

Correct option: B

Will replace Al

Option B (Will replace Al) is correct because Mg has a higher oxidation potential than Al, meaning it is more easily oxidized and can displace Al from its compounds.

27 Easy

27 of 138

Application +4 | -1

The quantity of charge required to obtain one mole of aluminium from

A{l_2}{O_3}

View answer and explanation

Correct option: C

3 F

Option C (3 F) is correct because the reduction of

Al^{3+}

to Al requires 3 moles of electrons, which is equivalent to 3 Faradays.

28 Easy

28 of 138

Application +4 | -1

Ionic mobility of

\mathrm{A}{\mathrm{g}^ + }

is
(

{\lambda _{\mathrm{A}{\mathrm{g}^ + }}} = 5 \times {10^{ - 1}}{\kern 1pt} {\kern 1pt}{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{equi}{\mathrm{v}^{ - 1}}

)

View answer and explanation

Correct option: A

5.2 \times {10^{ - 9}}

Option A is correct because ionic mobility is calculated using the formula: ionic mobility = (ionic conductance) / (Faraday's constant), which gives approximately 5.2 × 10^-9 m² V^-1 s^-1.

29 Easy

29 of 138

Application +4 | -1

The equivalent conductivity of a solution containing 2.54 g of

\mathrm{CuS}{\mathrm{O}_4}{\kern 1pt} {\kern 1pt} \mathrm{per L is}{\kern 1pt} {\kern 1pt} {\kern 1pt} 91.0{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}\mathrm{e}{\mathrm{q}^{ - 1}}

. Its conductivity would be

View answer and explanation

Correct option: A

2.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}

Option A (

2.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}

) is correct because conductivity is calculated by dividing equivalent conductivity by 1000 and multiplying by the concentration in normality.

30 Easy

30 of 138

Understanding +4 | -1

Standard reduction potentials of the half reactions are given below :

{\mathrm{F}_2}\left( \mathrm{g} \right) + 2{e^ - } \to 2{\mathrm{F}^ - }\left( {aq} \right);{E^ \circ } = + 2.85\mathrm{V}

\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) + 2{e^ - } \to 2\mathrm{C}{\mathrm{l}^ - }\left( {aq} \right);{E^ \circ } = + 1.36\mathrm{V}

\mathrm{B}{\mathrm{r}_2}\left( l \right) + 2{e^ - } \to 2\mathrm{B}{\mathrm{r}^ - }\left( {aq} \right);{E^ \circ } = + 1.06\mathrm{V}

{\mathrm{I}_2}\left( s \right) + 2{\mathrm{e}^ - } \to 2{\mathrm{I}^ - }\left( {aq} \right);{E^ \circ } = + 0.53V

The strongest oxidising and reducing agents respectively are :

View answer and explanation

Correct option: A

{\mathrm{F}_2}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\mathrm{I}^ - }

Option A (

{\mathrm{F}_2}{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} {\kern 1pt} {\mathrm{I}^ - }

) is correct because F2 has the highest reduction potential (strongest oxidizing agent) and I- is the easiest to oxidize (strongest reducing agent).

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