For the cell prepared from electrode A and B; Electrode A∶${\rm{C}}{{\rm{r}}_2}{\rm{O}}_7^{2 - }|{\rm{C}}{{\rm{r}}^{3 + }},E_{{\rm{red}}}^ \circ = + 1.33{\rm{V}}$ and Electrode B : ${\rm{F}}{{\rm{e}}^{3 + }}|{\rm{F}}{{\rm{e}}^{2 + }},E_{{\rm{red}}}^ \circ = 0.77{\rm{V}}$. Which of the following statement is correct?

The electrons will flow from B to A when connection is made

The e.m.f. of the cell will be 0.56 V

Chlorine gas is passed into a solution containing KF, KI, and KBr and ${\rm{CHC}}{{\rm{l}}_3}$ is added. The initial colour in ${\rm{CHC}}{{\rm{l}}_3}$ layer is :

Violet due to formation of ${{\rm{I}}_2}$

Orange due to formation of ${\rm{B}}{{\rm{r}}_2}$

Colourless due to formation of ${{\rm{F}}_2}$

No colour change due to no reaction

${\rm{\Lambda }}_{{\rm{ClC}}{{\rm{H}}_2}{\rm{COONa}}}^\infty = 224{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$ ${\rm{\Lambda }}_{NaCl}^\infty = 38.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$ ${\rm{\Lambda }}_{HCl}^\infty = 203{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$ What is the value of ${\lambda _{{\rm{ClC}}{{\rm{H}}_2}{\rm{COOH}}}} = ?$

$388.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$

$288.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$

$289.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$

$59.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{c}}{{\rm{m}}^2}{\kern 1pt} {\kern 1pt}{\rm{g}}{\kern 1pt} {\kern 1pt} {\rm{equi}}{{\rm{v}}^{ - 1}}$

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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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Showing 1-30 of 145 questions

1 Medium

1 of 145

Remembering +4 | -1

Which are used as secondary reference electrodes?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because Calomel electrode and Ag/AgCl electrode are commonly used as secondary reference electrodes due to their stability and ease of use.

2 Medium

2 of 145

Understanding +4 | -1

Which one is correct?

View answer and explanation

Correct option: B

Zn displaces iron from its solution

Option B is correct because zinc is more reactive than iron, meaning it has a greater tendency to lose electrons and displace iron from its solution. This is based on the electrochemical series where Zinc is above Iron.

3 Medium

3 of 145

Application +4 | -1

The equivalent weight and electrochemical equivalent of copper deposited from cupric salt.

View answer and explanation

Correct option: A

31.75,\,\,3.29 \times {10^{ - 4}}\,\,g\,\,cou{l^{ - 1}}

Option A (

31.75,\,\,3.29 \times {10^{ - 4}}\,\,g\,\,cou{l^{ - 1}}

) is correct because it represents the equivalent weight and electrochemical equivalent of copper deposited from cupric salt.

4 Medium

4 of 145

Understanding +4 | -1

When A lead storage battery is discharged

View answer and explanation

Correct option: D

Sulphuric acid is consumed

Option D (Sulphuric acid is consumed) is correct because during the discharge of a lead storage battery, sulphuric acid is used up to form lead sulphate at both the anode and the cathode.

5 Medium

5 of 145

Application +4 | -1

How many faraday are needed to reduce a mole of

MnO_4^ -

M{n^{2 + }}

View answer and explanation

Correct option: B

Option B (5) is correct because the reduction of

MnO_4^-

Mn^{2+}

involves a change in oxidation state of 5, requiring 5 moles of electrons, hence 5 Faradays.

6 Medium

6 of 145

Application +4 | -1

When 1 faraday of electricity is passed through

\mathrm{CuS}{\mathrm{O}_4}

solution, number of atoms formed is :

View answer and explanation

Correct option: B

3.01 \times {10^{23}}

Option B (

3.01 \times {10^{23}}

) is correct because 1 Faraday deposits 0.5 moles of Cu atoms, which is

3.01 \times {10^{23}}

atoms.

7 Medium

7 of 145

Analyzing +4 | -1

A correct electrochemical series can be obtained from K, Ca, Na, Al, Mg, Zn. Fe, Pb, H, Cu, Hg, Ag, Au by interchanging :

View answer and explanation

Correct option: A

Al and Mg

Option A (Al and Mg) is correct because the electrochemical series is determined by the standard reduction potentials, and Al and Mg are often misplaced in introductory series.

8 Medium

8 of 145

Understanding +4 | -1

Which metal does not give the following reaction?
M + water or steam ⟶ oxide +

{H_2}

↑

View answer and explanation

Correct option: C

Mercury

Option C (Mercury) is correct because mercury is a noble metal and does not react with water or steam to produce oxide and hydrogen.

9 Medium

9 of 145

Understanding +4 | -1

If an iron rod is dipped in

CuS{O_4}

solution, then :

View answer and explanation

Correct option: B

Brown layer is deposited on iron rod

Option B (Brown layer is deposited on iron rod) is correct because iron is more reactive than copper and will displace copper ions from the solution, resulting in the deposition of copper metal on the iron rod.

10 Medium

10 of 145

Understanding +4 | -1

Ionic mobility is equal to:

View answer and explanation

Correct option: C

Speed of ions under unit potential gradient

Option C (Speed of ions under unit potential gradient) is correct because ionic mobility is defined as the drift velocity of an ion per unit electric field strength.

11 Medium

11 of 145

Analyzing +4 | -1

Red hot carbon will remove oxygen from the oxides XO, and YO but not from ZO. Y will remove oxygen from XO. Use this evidence to reduce the order of activity of the three metals X, Y and Z putting the most reactive first:

View answer and explanation

Correct option: B

Z, Y, X

Option B (Z, Y, X) is correct because Z does not react with carbon, while Y reacts with XO, indicating the order of reactivity.

12 Medium

12 of 145

Remembering +4 | -1

The expression connecting

\mathrm{E}_{\mathrm{cell}}^0

and

\mathrm{K}_{\mathrm{eq}}

of the reaction occurring in a Galvanic cell is:

View answer and explanation

Correct option: C

\ln \mathrm{K}_{\mathrm{eq}}=\mathrm{nFE^0} / \mathrm{RT}

Option C (

\ln \mathrm{K}_{\mathrm{eq}}=\mathrm{nFE^0} / \mathrm{RT}

) is correct because it represents the correct relationship between the equilibrium constant and the standard cell potential.

13 Medium

13 of 145

Analyzing +4 | -1

Two electrochemical cells are assembled in which the following reactions occur:

\begin{aligned} & \mathrm{V}^{2+}+\mathrm{VO}^{2+}+2 \mathrm{H}^{+} \rightarrow 2 \mathrm{V}^{3+}+\mathrm{H}_2 \mathrm{O} ; E^{\circ}_{\text {Cell }}= 0.616 \mathrm{~V} \\ & \mathrm{V}^{3+}+\mathrm{Ag}^{+}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{VO}^{2+}+\mathrm{Ag}(\mathrm{s})+2 \mathrm{H}^{+} ; E^{\circ}_\text {Cell} =0.439 \mathrm{~V} \end{aligned}

E_{\mathrm{Ag}^{+} \mid \mathrm{Ag}^{\circ}}^{\circ}=0.799 \mathrm{~V}

, what is

E_{\mathrm{V}^{3+} \mid \mathrm{V}^{2+}}^{\circ}?

View answer and explanation

Correct option: A

-0.256 \mathrm{~V}

Option A (

-0.256 \mathrm{~V}

) is correct because it is calculated by manipulating the given cell potentials and the standard reduction potential of Ag+/Ag to find the standard reduction potential of V3+/V2+.

14 Medium

14 of 145

Application +4 | -1

The emf of the cell

\mathrm{Mg}|\mathrm{M}{\mathrm{g}^{2 + }}\left( {0.01\mathrm{M}} \right)|\left| {\mathrm{S}{\mathrm{n}^{2 + }}\left( {0.1\mathrm{M}} \right)} \right|\mathrm{Sn}

at 298 K is (Given,

E_{M{g^{2 + }},Mg}^ \circ = - 2.34\mathrm{V},E_{S{n^{2 + }},Sn}^ \circ = - 0.14\mathrm{V}

View answer and explanation

Correct option: A

2.23 V

Option A is correct. The EMF of the cell is calculated using the Nernst equation, considering the standard electrode potentials and the concentrations of the ions.

15 Medium

15 of 145

Application +4 | -1

Aluminium displaces hydrogen from dilute HCl whereas silver does not. The emf of A cell prepared by combining

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

and

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

is 2.46 V. The reduction potential of silver electrode is +0.80 V. The reduction potential of aluminium electrode is

View answer and explanation

Correct option: D

-1.66 V

Option D (-1.66 V) is correct because the cell potential is the difference between the reduction potentials of the cathode and anode, and aluminium is the anode in this case.

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16 Medium

16 of 145

Understanding +4 | -1

Dipping iron article into a strongly alkaline solution of sodium phosphate

View answer and explanation

Correct option: A

Does not affect the article

Option A (Does not affect the article) is correct because sodium phosphate in a strongly alkaline solution passivates the iron surface, preventing further reaction or corrosion.

17 Medium

17 of 145

Analyzing +4 | -1

For the cell prepared from electrode A and B; Electrode A∶

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }|\mathrm{C}{\mathrm{r}^{3 + }},E_{\mathrm{red}}^ \circ = + 1.33\mathrm{V}

and Electrode B :

\mathrm{F}{\mathrm{e}^{3 + }}|\mathrm{F}{\mathrm{e}^{2 + }},E_{\mathrm{red}}^ \circ = 0.77\mathrm{V}

. Which of the following statement is correct?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because, based on the reduction potentials, electrons flow from B to A, the EMF is 0.56V, and A is the positive electrode.

18 Medium

18 of 145

Understanding +4 | -1

Chlorine gas is passed into a solution containing KF, KI, and KBr and

\mathrm{CHC}{\mathrm{l}_3}

is added. The initial colour in

\mathrm{CHC}{\mathrm{l}_3}

layer is :

View answer and explanation

Correct option: A

Violet due to formation of

{\mathrm{I}_2}

Option A (Violet due to formation of

I_2

) is correct because chlorine will oxidize iodide ions to iodine, which dissolves in chloroform to give a violet color.

19 Medium

19 of 145

Analyzing +4 | -1

Which of the following statement is incorrect for dry cell?

View answer and explanation

Correct option: B

Manganese is reduced from +5 to +4 state

Option B is the correct answer because in a dry cell, manganese is reduced from +4 to +3 state, not +5 to +4.

20 Medium

20 of 145

Understanding +4 | -1

In an aqueous solution, hydrogen (

{H_2}

) will not reduce :

View answer and explanation

Correct option: C

Z{n^{2 + }}

Option C (

Z{n^{2 + }}

) is correct because hydrogen cannot reduce ions of metals that are more electropositive than itself (i.e., have a more negative standard reduction potential).

21 Medium

21 of 145

Application +4 | -1

The

{E^ \circ }

\mathrm{F}{\mathrm{e}^{2 + }}/\mathrm{Fe}

and

\mathrm{S}{\mathrm{n}^{2 + }}/\mathrm{Sn}

are -0.44 V and -0.14 V respectively. If cell reaction is

\mathrm{Fe} + \mathrm{S}{\mathrm{n}^{2 + }} \to \mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{Sn}

then emf of the cell is

View answer and explanation

Correct option: A

+0.30 V

Option A is correct because the cell emf is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode: E°cell = E°cathode - E°anode.

22 Medium

22 of 145

Remembering +4 | -1

The composition of rust is

View answer and explanation

Correct option: A

F{e_2}{O_3} \cdot x{H_2}O

Option A (

F{e_2}{O_3} \cdot x{H_2}O

) is correct because rust is hydrated iron(III) oxide, where 'x' represents a variable number of water molecules.

23 Medium

23 of 145

Remembering +4 | -1

In mercury cells, cathode used is

View answer and explanation

Correct option: A

Paste of

\mathrm{HgO}

and carbon

Option A (Paste of

\mathrm{HgO}

and carbon) is correct because this mixture is used as the cathode in mercury cells.

24 Medium

24 of 145

Understanding +4 | -1

At cathode the electrolysis of aqueous

Na_2SO_4

gives

View answer and explanation

Correct option: B

H_2

Option B (

H_2

) is correct because during the electrolysis of aqueous

Na_2SO_4

, water is reduced at the cathode to produce hydrogen gas.

25 Medium

25 of 145

Application +4 | -1

\mathrm{\Lambda }_{\mathrm{ClC}{\mathrm{H}_2}\mathrm{COONa}}^\infty = 224{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{g}{\kern 1pt} {\kern 1pt} \mathrm{equi}{\mathrm{v}^{ - 1}}

\mathrm{\Lambda }_{NaCl}^\infty = 38.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{g}{\kern 1pt} {\kern 1pt} \mathrm{equi}{\mathrm{v}^{ - 1}}

\mathrm{\Lambda }_{HCl}^\infty = 203{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{g}{\kern 1pt} {\kern 1pt} \mathrm{equi}{\mathrm{v}^{ - 1}}

What is the value of

{\lambda _{\mathrm{ClC}{\mathrm{H}_2}\mathrm{COOH}}} = ?

View answer and explanation

Correct option: C

388.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{g}{\kern 1pt} \mathrm{equi}{\mathrm{v}^{ - 1}}

Option C (

388.5{\Omega ^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{g}{\kern 1pt} {\kern 1pt} \mathrm{equi}{\mathrm{v}^{ - 1}}

) is correct because it applies Kohlrausch's law of independent migration of ions.

26 Medium

26 of 145

Application +4 | -1

\mathrm{Ni}/\mathrm{N}{\mathrm{i}^{2 + }}\left[ {1.0\mathrm{M}} \right]||\mathrm{A}{\mathrm{u}^{3 + }}\left[ {1.0\mathrm{M}} \right]/\mathrm{Au}

where

{E^ \circ }

\mathrm{for}{\kern 1pt} \mathrm{N}{\mathrm{i}^{2 + }}/{\rm{Ni {\kern 1pt} {\kern 1pt} is}} - 0.250\mathrm{V}

; and

{E^ \circ }

for

\mathrm{A}{\mathrm{u}^{3 + }}/\mathrm{Au}

is 0.150 V. The emf of the cell is

View answer and explanation

Correct option: D

+0.4 V

Option D (+0.4 V) is correct because it represents the standard EMF of the cell calculated from the difference in standard reduction potentials.

27 Medium

27 of 145

Understanding +4 | -1

The molecular conductivity and equivalent conductivity are same for the solution of :

View answer and explanation

Correct option: A

1M{\kern 1pt} {\kern 1pt} \mathrm{NaCl}

Option A (

1M{\kern 1pt} {\kern 1pt} \mathrm{NaCl}

) is correct because NaCl dissociates into one cation and one anion, so its molar mass and equivalent mass are the same.

28 Medium

28 of 145

Analyzing +4 | -1

The standard electrode potential of

Zn,\,Ag

and

Cu

are

- 0.76,0.80

and

0.34V

resp. then,

View answer and explanation

Correct option: C

Zn

can reduce

A{g^ + }

and

C{u^{2 + }}

Option C (

Zn

can reduce

A{g^ + }

and

C{u^{2 + }}

) is correct because Zn has a more negative standard electrode potential than both Ag and Cu, making it a stronger reducing agent.

29 Medium

29 of 145

Application +4 | -1

The equilibrium constant for the reaction given below at 298 K is :

\mathrm{Zn}\left( s \right) + \mathrm{F}{\mathrm{e}^{2 + }}\left( {aq} \right) \to \mathrm{Z}{\mathrm{n}^{2 + }}\left( {aq} \right) + \mathrm{Fe}\left( s \right)

;

E_{cell}^ \circ

= 2905 V at 298 K

View answer and explanation

Correct option: D

{10^{0.32/0.295}}

Option D is correct because it uses the relationship between the standard cell potential and the equilibrium constant:

E_{cell}^\circ = \frac{{0.0591}}{n}logK

at 298 K.

30 Medium

30 of 145

Application +4 | -1

How many coloumbs of electricity are required for the oxidation of 1 mole of

{H_2}O

{O_2}

View answer and explanation

Correct option: C

1.93 \times {10^5}C

Option C (

1.93 \times {10^5}C

) is correct because the oxidation of 1 mole of

H_2O

O_2

requires 4 Faradays, which is equal to

4 \times 96500

Coulombs.

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