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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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Showing 1-30 of 381 questions

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1 Hard
1 of 381
Application +4 | -1
4.5 g of Al (at. mass 27 amu) is deposited at cathode from Al3+A{l^{3 + }} solution by a certain quantity of charge. The volume of H2{H_2} produced at STP from H+{H^ + } ions in solution by the same quantity of charge will be :
View answer and explanation
Correct option: C
5.6 L
Option C (5.6 L) is correct because it correctly calculates the volume of H2 produced at STP using Faraday's laws of electrolysis and stoichiometry.
2 Easy
2 of 381
Understanding +4 | -1
The correct order of molar conductivity at infinite dilution of LiCl,NaClandKCl\mathrm{LiCl},{\rm{NaCl {\kern 1pt}{\kern 1pt}and {\kern 1pt}{\kern 1pt}KCl}} is
View answer and explanation
Correct option: B
KCl>NaCl>LiCl\mathrm{KCl} > NaCl > LiCl
Option B (KCl>NaCl>LiCl\mathrm{KCl} > NaCl > LiCl) is correct because ionic mobility decreases with increasing hydrated size, and Li+ has the largest hydrated size among these ions.
3 Easy
3 of 381
Application +4 | -1
A solution of copper sulphate (CuSO4)\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.
4 Easy
4 of 381
Application +4 | -1
The desired amount of charge for obtaining one mole of Al from Al3+\mathrm{A}{\mathrm{l}^{3 + }} is
View answer and explanation
Correct option: C
3×96500C3 \times 96500\mathrm{C}
Option C (3×96500C3 \times 96500\mathrm{C}) is correct because the reduction of one mole of Al3+Al^{3+} to Al requires 3 moles of electrons, and each mole of electrons requires 96500 Coulombs of charge.
5 Easy
5 of 381
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.
6 Medium
6 of 381
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.
7 Medium
7 of 381
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.
8 Easy
8 of 381
Understanding +4 | -1
The correct value of e.m.f. of cell is given by :

i) Ecell=EOPanode−ERPcathode{E_{cell}} = {E_{OP}}{\kern 1pt} anode - {E_{RP}}{\kern 1pt} cathode

ii) Ecell=EOPanode+ERPcathode{E_{cell}} = {E_{OP}}{\kern 1pt} anode + {E_{RP}}{\kern 1pt} cathode

iii) Ecell=ERPanode+ERPcathode{E_{cell}} = {E_{RP}}{\kern 1pt} anode + {E_{RP}}{\kern 1pt} cathode

iv) Ecell=EOPanode−EOPcathode{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 Ecell=EOPanode+ERPcathode{E_{cell}} = {E_{OP}}{\kern 1pt} anode + {E_{RP}}{\kern 1pt} cathode and Ecell=EOPanode−EOPcathode{E_{cell}} = {E_{OP}}{\kern 1pt} anode - {E_{OP}}{\kern 1pt} cathode are both valid ways to calculate cell potential.
9 Easy
9 of 381
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.
10 Medium
10 of 381
Application +4 | -1
The equivalent weight and electrochemical equivalent of copper deposited from cupric salt.
View answer and explanation
Correct option: A
31.75,  3.29×10−4  g  coul−131.75,\,\,3.29 \times {10^{ - 4}}\,\,g\,\,cou{l^{ - 1}}
Option A (31.75,  3.29×10−4  g  coul−131.75,\,\,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.
11 Medium
11 of 381
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.
12 Medium
12 of 381
Application +4 | -1
How many faraday are needed to reduce a mole of MnO4−MnO_4^ - of Mn2+M{n^{2 + }} ?
View answer and explanation
Correct option: B
5
Option B (5) is correct because the reduction of MnO4−MnO_4^- to Mn2+Mn^{2+} involves a change in oxidation state of 5, requiring 5 moles of electrons, hence 5 Faradays.
13 Medium
13 of 381
Application +4 | -1
When 1 faraday of electricity is passed through CuSO4\mathrm{CuS}{\mathrm{O}_4} solution, number of atoms formed is :
View answer and explanation
Correct option: B
3.01×10233.01 \times {10^{23}}
Option B (3.01×10233.01 \times {10^{23}}) is correct because 1 Faraday deposits 0.5 moles of Cu atoms, which is 3.01×10233.01 \times {10^{23}} atoms.
14 Easy
14 of 381
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.
15 Easy
15 of 381
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.

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CHEMICAL BONDING & MOLECULAR STRUCTURE 595 MCQs THERMODYNAMICS 583 MCQs STRUCTURE OF ATOM 499 MCQs CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES 429 MCQs Equilibrium 412 MCQs STATES OF MATTER 410 MCQs
16 Medium
16 of 381
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.
17 Hard
17 of 381
Remembering +4 | -1
The reference cell used in the determination of the E.M.F of a half–cell is
View answer and explanation
Correct option: C
Calomel electrode
Option C (Calomel electrode) is the correct answer as it is a commonly used reference electrode with a stable and well-defined potential.
18 Medium
18 of 381
Understanding +4 | -1
Which metal does not give the following reaction?
M + water or steam ⟶ oxide + H2{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.
19 Easy
19 of 381
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.
20 Easy
20 of 381
Application +4 | -1
Number of faraday required to liberate 8 g of H2{\mathrm{H}_2} is :
View answer and explanation
Correct option: A
8
Option A (8) is correct because 8g of H2 is 4 moles, and each mole requires 2 Faradays, so 4 moles require 8 Faradays.
21 Medium
21 of 381
Understanding +4 | -1
If an iron rod is dipped in CuSO4CuS{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.
22 Medium
22 of 381
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.
23 Medium
23 of 381
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.
24 Medium
24 of 381
Remembering +4 | -1
The expression connecting Ecell0\mathrm{E}_{\mathrm{cell}}^0 and Keq\mathrm{K}_{\mathrm{eq}} of the reaction occurring in a Galvanic cell is:
View answer and explanation
Correct option: C
ln⁡Keq=nFE0/RT\ln \mathrm{K}_{\mathrm{eq}}=\mathrm{nFE^0} / \mathrm{RT}
Option C (ln⁡Keq=nFE0/RT\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.
25 Medium
25 of 381
Analyzing +4 | -1
Two electrochemical cells are assembled in which the following reactions occur:

V2++VO2++2H+→2V3++H2O;ECell ∘=0.616 VV3++Ag++H2O→VO2++Ag(s)+2H+;ECell∘=0.439 V\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}

If EAg+∣Ag∘∘=0.799 VE_{\mathrm{Ag}^{+} \mid \mathrm{Ag}^{\circ}}^{\circ}=0.799 \mathrm{~V}, what is EV3+∣V2+∘?E_{\mathrm{V}^{3+} \mid \mathrm{V}^{2+}}^{\circ}?
View answer and explanation
Correct option: A
−0.256 V-0.256 \mathrm{~V}
Option A (−0.256 V-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+.
26 Hard
26 of 381
Analyzing +4 | -1
What is EoE^{\mathrm{o}} for the following half reaction?

MX2(s)+2e−→M(s)+2X−;E∘=?M2+(s)+2e−→M(s);E∘=0.100 VMX2(s)→M2++2X−;Ksp=e−10\begin{aligned} & \mathrm{MX}_2(\mathrm{s})+2 e^{-} \rightarrow \mathrm{M}(\mathrm{s})+2 \mathrm{X}^{-} ; E^{\circ}=? \\ & \mathrm{M}^{2+}(\mathrm{s})+2 e^{-} \rightarrow \mathrm{M}(\mathrm{s}) ; E^\circ=0.100 \mathrm{~V} \\ & \mathrm{MX}_2(\mathrm{s}) \rightarrow \mathrm{M}^{2+}+2 \mathrm{X}^{-} ; K_{\mathrm{sp}}=e^{-10} \end{aligned}
View answer and explanation
Correct option: B
E∘=0.100−0.05922log⁡1KspE^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}
Option B (E∘=0.100−0.05922log⁡1KspE^{\circ}=0.100-\frac{0.0592}{2} \log \frac{1}{K_{\text {sp}}}) is correct because it correctly applies the Nernst equation to relate the standard potential to the solubility product.
27 Medium
27 of 381
Application +4 | -1
The emf of the cell Mg∣Mg2+(0.01M)∣∣Sn2+(0.1M)∣Sn\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, EMg2+,Mg∘=−2.34V,ESn2+,Sn∘=−0.14VE_{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.
28 Medium
28 of 381
Application +4 | -1
Aluminium displaces hydrogen from dilute HCl whereas silver does not. The emf of A cell prepared by combining Al/Al3+\mathrm{Al}/\mathrm{A}{\mathrm{l}^{3 + }} and Ag/Ag+\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.
29 Medium
29 of 381
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.
30 Easy
30 of 381
Understanding +4 | -1
On passing electricity through dilute H2SO4{H_2}S{O_4} between ptpt electrodes, the gas evolved at anode and cathode are
View answer and explanation
Correct option: C
O2,H2{O_2},{H_2}
Option C (O2,H2{O_2},{H_2}) is correct because during the electrolysis of dilute H2SO4{H_2}S{O_4}, oxygen gas is evolved at the anode and hydrogen gas is evolved at the cathode.
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