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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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91 Medium
91 of 381
Remembering +4 | -1
An electronic conductor is
View answer and explanation
Correct option: C
AgAg
Option C (AgAg) is the correct answer because silver is a metal and conducts electricity through the movement of electrons.
92 Medium
92 of 381
Remembering +4 | -1
Cathode used in Leclanche cell is
View answer and explanation
Correct option: B
Graphite rod surrounded by MnO2Mn{O_2} and carbon
Option B (Graphite rod surrounded by MnO2Mn{O_2} and carbon) is correct because this is the standard cathode composition in a Leclanche cell.
93 Easy
93 of 381
Remembering +4 | -1
E∘{E^ \circ } of an electrode is :
View answer and explanation
Correct option: D
Intensive property
Option D (Intensive property) is correct because the standard electrode potential is independent of the amount of substance and depends only on the nature of the electrode and the temperature.
94 Medium
94 of 381
Application +4 | -1
If the standard electrode potential for a cell A(aq)2++2B(s)→A(s)+2B(aq)+A_{\left( {aq} \right)}^{2 + } + 2{B_{\left( s \right)}} \to {A_{\left( s \right)}} + 2B_{\left( {aq} \right)}^ + is 1.9 V, then the standard Gibbs energy for the reaction is
View answer and explanation
Correct option: C
-3.8 F J/mol
Option C is correct because the standard Gibbs energy is calculated using the formula ΔG=−nFE∘\Delta G = -nFE^\circ, where n=2 and E° = 1.9 V.
95 Easy
95 of 381
Understanding +4 | -1
In a cell containing zinc electrode and standard hydrogen electrode(SHE), the zinc electrode acts as :
View answer and explanation
Correct option: A
Anode
Option A is correct because zinc has a higher tendency to lose electrons than hydrogen, so it undergoes oxidation at the anode, while hydrogen ions get reduced at the cathode (SHE).
96 Easy
96 of 381
Understanding +4 | -1
Which ion has exceptionally higher Λ∞{\mathrm{\Lambda }^\infty } values?
View answer and explanation
Correct option: A
H+{\mathrm{H}^ + }
Option A (H+{\mathrm{H}^ + }) is correct because hydrogen ions have exceptionally high molar conductivity due to their unique mechanism of transport in water.
97 Medium
97 of 381
Analyzing +4 | -1
Statement I: 96,500C96,500 \mathrm{C} charge is required for the reduction of one mole of silver ions.

Statement II: The amount of electricity (or) charge required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
View answer and explanation
Correct option: B
If both Statement I and Statement II are correct and Statement II is not a correct explanation of Statement I
Option B is correct. Statement I is correct because the reduction of one mole of silver ions (Ag+) requires one Faraday of charge (96,500 C). Statement II is also correct, but it doesn't explain statement I.
98 Hard
98 of 381
Application +4 | -1
If the ΔG\Delta G of a cell reaction AgCl+e−→Ag+Cl−\mathrm{AgCl} + {e^ - } \to \mathrm{Ag} + \mathrm{C}{\mathrm{l}^ - } is −21.20- 21.20 kJ, the standard emf of cell is
View answer and explanation
Correct option: B
0.220 V
Option B (0.220 V) is correct because it is calculated using the relationship between Gibbs free energy and standard EMF: E0=−ΔGnFE^0 = -\frac{{\Delta G}}{{nF}}.
99 Easy
99 of 381
Remembering +4 | -1
A standard hydrogen electrode has Zero electrode potential because
View answer and explanation
Correct option: B
This electrode potential is assumed to be zero
Option B (This electrode potential is assumed to be zero) is correct because the standard hydrogen electrode (SHE) is defined as having a zero potential at all temperatures.
100 Medium
100 of 381
Remembering +4 | -1
The equation representing kohlrausch law from the following is (V+={V^ + } = No. of cations, V−={V^ - } = No. of anions)
View answer and explanation
Correct option: B
λmo=V+λ+o+V−λ−o\lambda _m^\mathrm{o} = {V^ + }\lambda _ + ^\mathrm{o} + {V^ - }\lambda _ - ^\mathrm{o}
Option B is the correct representation of Kohlrausch's law, stating that the limiting molar conductivity of an electrolyte is the sum of the contributions of the individual ions.
101 Medium
101 of 381
Analyzing +4 | -1
The reaction occurring at anode when the electrolysis of an aqueous solution containing Na2SO4\mathrm{Na}_2 \mathrm{SO}_4 and CuSO4\mathrm{CuSO}_4 is done using Pt\mathrm {Pt} electrode is
View answer and explanation
Correct option: C
2H2O⟶O2+4H++4e2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}
Option C (2H2O⟶O2+4H++4e2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 \mathrm{e}) is correct because in the electrolysis of aqueous Na2SO4 and CuSO4 using Pt electrodes, water is oxidized at the anode to produce oxygen, hydrogen ions, and electrons.
102 Medium
102 of 381
Remembering +4 | -1
The unit of equivalent conductivity is
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Correct option: D
ohm−1  cm2 (g  equivalent)−1\mathrm{oh}{\mathrm{m}^{ - 1}}\,\,\mathrm{c}{\mathrm{m}^\mathrm{2}}\,{\mathrm{(g}\,\,\mathrm{equivalent})^{ - 1}}
Option D is the correct unit for equivalent conductivity, representing conductance per gram equivalent per unit area.
103 Easy
103 of 381
Application +4 | -1
Passage of 96500 coulomb of electricity liberates ….litre of O2{\mathrm{O}_2} at NTP during electrolysis.
View answer and explanation
Correct option: A
5.6
Option A is correct because 96500 Coulombs (1 Faraday) liberates 0.5 mole of O2, which occupies 5.6 liters at NTP.
104 Easy
104 of 381
Remembering +4 | -1
The amount of substance deposited by the passage of 1 A of current for 1 s is equal to
View answer and explanation
Correct option: C
Electrochemical equivalent
Option C is correct because the electrochemical equivalent is defined as the mass of a substance deposited or liberated by the passage of 1 coulomb of charge (1 Ampere for 1 second).
105 Easy
105 of 381
Remembering +4 | -1
The fraction of the total current carried by an ion is known as:
View answer and explanation
Correct option: A
Transport number of that ion
Option A is correct because the transport number represents the fraction of the total current carried by a specific ion in an electrolytic solution.
106 Medium
106 of 381
Understanding +4 | -1
Which one of the following nitrates will leave behind A metal on strong heating?
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Correct option: D
Silver nitrate
Option D (Silver nitrate) is correct because silver is a relatively unreactive metal, and its nitrate decomposes upon heating to leave behind metallic silver.
107 Easy
107 of 381
Understanding +4 | -1
During the electrolysis of an electrolyte, the number of ions produced, is directly proportional to the
View answer and explanation
Correct option: C
Quantity of electricity passed
Option C (Quantity of electricity passed) is correct because Faraday's laws of electrolysis state that the amount of substance produced is directly proportional to the quantity of electricity passed.
108 Hard
108 of 381
Application +4 | -1
If the H+{H^ + } concentration is decreased from 1 M to 10−4{10^{ - 4}} M at 25∘C{25^ \circ }C  for the couple MnO4−/Mn(2+)MnO_4^ - /M{n^{(2 + )}}, then the oxidising power of the MnO4−/Mn(2+)MnO_4^ - /M{n^{(2 + )}} couple decreases by
View answer and explanation
Correct option: C
0.38 V
Option C (0.38 V) is correct because it is the calculated decrease in oxidizing power using the Nernst equation.
109 Easy
109 of 381
Understanding +4 | -1
The product obtained at anode when 50% H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} aqueous solution is electrolysed using platinum electrodes is
View answer and explanation
Correct option: B
H2S2O8{\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8}
Option B (H2S2O8{\mathrm{H}_2}{\mathrm{S}_2}{\mathrm{O}_8}) is correct because at high concentrations of sulfuric acid, peroxydisulfuric acid (H2S2O8H_2S_2O_8) is formed at the anode.
110 Easy
110 of 381
Analyzing +4 | -1
Two platinum electrodes were immersed in a solution of cupric sulphate and electric current passed through the solution. After sometime, it was found that the colour of copper sulphate disappeared with evolution of gas at the electrode. The colourless solution contains:
View answer and explanation
Correct option: D
Sulphuric acid
Option D (Sulphuric acid) is correct because the electrolysis of cupric sulphate solution leads to the formation of sulphuric acid as the copper ions are deposited and water is oxidized.
111 Easy
111 of 381
Remembering +4 | -1
Quantity of charge is measured in :
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Correct option: A
ampere-sec
Option A (ampere-sec) is correct because quantity of charge (Q) is defined as the product of current (I) and time (t), where current is measured in amperes and time in seconds.
112 Medium
112 of 381
Remembering +4 | -1
Cell constant of a conductivity cell is usually derived by using a solution of :
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Correct option: A
KCl\mathrm{KCl}
Option A (KCl\mathrm{KCl}) is correct because potassium chloride (KCl) solutions are commonly used to determine the cell constant due to their well-established conductivity values at various concentrations.
113 Easy
113 of 381
Remembering +4 | -1
Hydrogen-oxygen fuel cells are used in spacecraft to supply :
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Correct option: A
Power for heat and light
Option A (Power for heat and light) is correct because hydrogen-oxygen fuel cells produce electricity, which can be used to generate heat and light in spacecraft.
114 Hard
114 of 381
Understanding +4 | -1
Read the following statements carefully.

(i) According to a convention cell potential of hydrogen electrode (S.H.E.) is considered to be zero at all temperatures.

(ii) e.m.f. of the cell Pt(s)/H2( g,1bar)/H+(aq,1M){Pt}({s}) / {H}_2({~g}, 1 {bar}) / {H}^{+}({aq}, 1 {M}) ∥Zn2+(aq,1M)/Zn\| {Zn}^{2+}({aq}, 1 {M}) / {Zn} is -0.76 . This negative value indicates that Zn2+{Zn}^{2+} ion reduces less easily then H+{H}^{+} ions.

(iii) Copper does not dissolve in HCl{HCl} but dissolves in HNO3{HNO}_3 as in nitric acid it gets oxidised by nitrate ion.

(iv) Inert metals like PtPt or Au are used in certain electrodes i.e., these metals does not participate in reaction but provide surface for oxidation and reduction reactions.

(v) Fluorine has the highest electrode potential thereby making it strongest oxidising agent whereas lithium with lowest electrode potential is the weakest oxidising and strongest reducing agent.

Which of the following is the correct coding for the statements above.
View answer and explanation
Correct option: A
TTTTT
Option A (TTTTT) is correct because all the statements are true regarding electrochemical concepts.
115 Easy
115 of 381
Remembering +4 | -1
The calomel electrode is a :
View answer and explanation
Correct option: B
Reference electrode
Option D (AuCl3+NaCN\mathrm{AuC}{\mathrm{l}_3} + \mathrm{NaCN}) is the correct answer because the electrolytic bath for gold plating typically contains a gold salt (like AuCl3) and sodium cyanide (NaCN) to form a complex ion that facilitates smooth and even deposition of gold.
116 Medium
116 of 381
Understanding +4 | -1
The product obtained from the electrolytic oxidation of acidified sulphate solutions, is
View answer and explanation
Correct option: B
HO3SOOSO3HH{O_3}SOOS{O_3}H
Option D (HO3SOSO3HH{O_3}SOS{O_3}H) is correct because the electrolytic oxidation of acidified sulfate solutions results in the formation of peroxydisulfuric acid (H2S2O8H_2S_2O_8), which has the structure HO3SOSO3HH{O_3}SOS{O_3}H.
117 Medium
117 of 381
Remembering +4 | -1
Hydrogen peroxide is obtained by electrolysis of
View answer and explanation
Correct option: B
Sulphurie acid
Option B (Sulphuric acid) is correct because hydrogen peroxide can be obtained by the electrolysis of sulphuric acid.
118 Medium
118 of 381
Remembering +4 | -1
The electrolytic bath used in gold plating of copper articles contains :
View answer and explanation
Correct option: D
AuCl3+NaCN\mathrm{AuC}{\mathrm{l}_3} + \mathrm{NaCN}
Option D (AuCl3+NaCN\mathrm{AuC}{\mathrm{l}_3} + \mathrm{NaCN}) is correct because the addition of NaCN helps in forming a complex with gold, facilitating a smoother and more even plating process.
119 Hard
119 of 381
Understanding +4 | -1
Which of the following statement(s) is/an incorrect for corrosion of iron?

(i) Reaction occurring at anode is

O2(g)+4H+(aq)+4e−→2H2O(ℓ){O}_2(g)+4 {H}^{+}(a q)+4 e^{-} \rightarrow 2 {H}_2 {O}(\ell)

(ii) Reaction occurring at cathode is

2Fe(s)→2Fe2++4e−2 {Fe}({s}) \rightarrow 2 {Fe}^{2+}+4 e^{-}

(iii) Rust is Fe2O3⋅H2O{Fe}_2 {O}_3 \cdot {H}_2 {O}

(iv) H+{H}^{+}involved in corrosion reaction is provided from H2CO3{H}_2 {CO}_3 which is formed due to dissolution of carbon dioxide from air in to water.
View answer and explanation
Correct option: C
(i) and (ii)
Option C ((i) and (ii)) is correct because the reaction at the anode is the oxidation of Fe to Fe2+Fe^{2+}, and the reaction at the cathode is the reduction of O2O_2 to H2OH_2O. The given reactions are reversed.
120 Hard
120 of 381
Analyzing +4 | -1
Assertion (A) At the end of electrolysis using platinum electrodes, an aqueous solution of CuSO4\mathrm{CuSO}_4 turns colourless.

Reason (R) During the electrolysis CuSO4\mathrm{CuSO}_4 changes to Cu(OH)2\mathrm{Cu}(\mathrm{OH})_2 during electrolysis.
View answer and explanation
Correct option: C
Assertion is true but Reason is false.
Option C is correct because the assertion is true (the solution turns colorless), but the reason is false (CuSO4 does not change to Cu(OH)2). Copper is deposited at the cathode, and oxygen is evolved at the anode, leading to the formation of sulfuric acid.
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