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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 151-180 of 230 questions

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151 Medium
151 of 230
Understanding +4 | -1
Which statement is not correct
View answer and explanation
Correct option: B
Conductance of an electrote solution decreases with dilution
Option B (Conductance of an electrote solution decreases with dilution) is correct because as dilution increases, the number of ions per unit volume decreases, leading to a decrease in conductance.
152 Medium
152 of 230
Remembering +4 | -1
In mercury cell electrolyte used is
View answer and explanation
Correct option: B
Paste of KOH{KOH} and ZnO{ZnO}
Option B (Paste of KOH{KOH} and ZnO{ZnO}) is correct because the electrolyte used in a mercury cell is a paste of potassium hydroxide (KOH{KOH}) and zinc oxide (ZnO{ZnO}).
153 Medium
153 of 230
Understanding +4 | -1
Which of the following metals cannot be used as sacrificial electrode for iron?
View answer and explanation
Correct option: A
Cu{Cu}
Option A (Cu{Cu}) is correct because Copper has a lower oxidation potential than Iron, and therefore cannot act as a sacrificial electrode.
154 Medium
154 of 230
Analyzing +4 | -1
Assertion: pH\mathrm{pH} of solution increases during electrolysis of aqueous solution of NaCl\mathrm{NaCl}.

Reason: Electrolysis is non-spontaneous process.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B is correct because during the electrolysis of aqueous NaCl, NaOH is formed, which increases the pH of the solution. The reason is true but not the correct explanation.
155 Medium
155 of 230
Application +4 | -1
In following cell reaction

Mg(s)+2Ag+(0.001M)⟶Mg2+(0.20M)+2Ag(s){Mg}({s})+2 {Ag}^{+}(0.001 {M}) \longrightarrow {Mg}^{2+}(0.20 {M})+2 {Ag}({s})

Calculate Ecell E_{\text {cell }} for the reaction [E0=3.17 V]\left[E^0=3.17 {~V}\right]
View answer and explanation
Correct option: B
3.04 V3.04 {~V}
Option B (3.04 V3.04 {~V}) is correct because it is the result of applying the Nernst equation with the given concentrations and standard potential.
156 Medium
156 of 230
Analyzing +4 | -1
Four successive members of the first row transition elements listed below with atomic numbers. Which one of them is expected to have the highest EM3+/M2+∘E_{{M^{3 + }}/{M^{2 + }}}^ \circ value is?
View answer and explanation
Correct option: D
Co (Z=27)Co\,(Z = 27)
Option D is correct because Co (Z = 27) has the highest EM3+/M2+∘E_{M^{3+}/M^{2+}}^\circ value due to the stability of Co²⁺.
157 Medium
157 of 230
Application +4 | -1
If the molar conductivity of 10−3 M HA10^{-3} \mathrm{~M} \mathrm{~HA} is 39 S cm2 mol−1,39 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, then its dissociation constant will be

[Given Λm0(HA)=390 S cm2 mol−1]\Lambda_{\mathrm{m}}^0(\mathrm{HA})=390 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}]
View answer and explanation
Correct option: D
10−5M10^{-5} \mathrm{M}
The correct answer is D (10-5 M) because it is calculated using the formula Ka = Cα^2, where α is the degree of dissociation and C is the concentration.
158 Medium
158 of 230
Application +4 | -1
In acidic medium MnO4−\mathrm{MnO}_4^ - is converted to Mn2+\mathrm{M}{\mathrm{n}^{2 + }} . The quantity of electricity in faraday required to reduce 0.5 mole of MnO4−\mathrm{MnO}_4^ - to Mn2+\mathrm{M}{\mathrm{n}^{2 + }} would be
View answer and explanation
Correct option: A
2.5
Option A (2.5) is correct because it is the quantity of electricity in faradays required to reduce 0.5 mole of MnO₄⁻ to Mn²⁺, calculated based on the change in oxidation state.
159 Medium
159 of 230
Application +4 | -1
During the electrolysis of a solution of AgNO3,9650C\mathrm{AgN}{\mathrm{O}_3},9650\mathrm{C} of charge is passed through the electroplating bath.
The mass of silver deposited at the cathode will be
View answer and explanation
Correct option: B
10.8 g
Option B (10.8 g) is correct because it calculates the mass of silver deposited using Faraday's laws of electrolysis.
160 Medium
160 of 230
Application +4 | -1
The equivalent conductance of an aqueous solution of 1.0283×10−3 g1.0283 \times 10^{-3} \mathrm{~g} equivalent acetic acid per litre is 48.15π−1 cm248.15 \pi^{-1} \mathrm{~cm}^2 equiv −1^{-1} at 25∘C25^{\circ} \mathrm{C}. At infinite dilution value is 390.7π−1 cm2390.7 \pi^{-1} \mathrm{~cm}^2 equiv −1{ }^{-1}. Calculate the degree of ionisation and ionisation constant of acetic acid.
View answer and explanation
Correct option: A
0.1232,1.78×10−50.1232,1.78 \times 10^{-5}
Option A (0.1232, 1.78×10−51.78 \times 10^{-5}) is correct because it accurately calculates the degree of ionization and ionization constant using the given equivalent conductance values.
161 Medium
161 of 230
Remembering +4 | -1
Which of the following electrolytes is used in lead storage battery?
View answer and explanation
Correct option: C
38 % solution of H2SO4{H_2}S{O_4}
Option C (38 % solution of H2SO4{H_2}S{O_4}) is correct because a 38% solution of sulfuric acid is used as the electrolyte in lead storage batteries.
162 Medium
162 of 230
Application +4 | -1
On the basis of the information available from the 43Al+O2⟶23Al2O3,ΔG=−827 kJ mol−1\frac{4}{3} \mathrm{Al}+\mathrm{O}_2 \longrightarrow \frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3, \Delta G=-827 \mathrm{~kJ} \mathrm{~mol}^{-1} of O2\mathrm{O}_2, the minimum emf required to carry out an electrolysis of Al2O3\mathrm{Al}_2 \mathrm{O}_3 is : (F=96500 C mol−1\left(\mathrm{F}=96500 \mathrm{~C} \mathrm{~mol}^{-1}\right. )
View answer and explanation
Correct option: B
2.14 V2.14 \mathrm{~V}
Option B (2.14 V2.14 \mathrm{~V}) is correct because the minimum EMF required is calculated from the Gibbs free energy change and the number of electrons transferred in the reaction.
163 Medium
163 of 230
Application +4 | -1
In a fuel cell hydrogen and oxygen react to produce electricity. In the process hydrogen gas is oxidised at the anode and oxygen at the cathode. If 67.2 L67.2 \mathrm{~L} of H2\mathrm{H}_2 at STP react in 15 minutes, what is the average current produced ? If the entire current is used for electrodeposition of copper from copper (II) solution, how many grams of copper will be deposited ?

Anode reaction : H2+2OH−⟶2H2O+2e−\mathrm{H}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-}

Cathode reaction: O2+2H2+2e−⟶4OH−\mathrm{O}_2+2 \mathrm{H}_2+2 \mathrm{e}^{-} \longrightarrow 4 \mathrm{OH}^{-}
View answer and explanation
Correct option: A
190
Option A (190) is correct because it is the approximate mass of copper deposited, calculated based on the stoichiometry of the reaction and Faraday's laws of electrolysis.
164 Medium
164 of 230
Analyzing +4 | -1
Select the correct statement(s) for the given reaction.

Cu(s)+2Ag+(aq)⟶Cu2+(aq)+2Ag(s)\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the provided statements regarding the electrochemical cell representation, cell potential, and electrode reactions are all incorrect for the given reaction.
165 Medium
165 of 230
Analyzing +4 | -1
The electrical conductivity of the flowing aqueous solutions is highest for:
View answer and explanation
Correct option: C
0.1 M CHF2COOH0.1 \mathrm{~M} \mathrm{~CHF}_2 \mathrm{COOH}
Option C (0.1 M CHF2COOH0.1 \mathrm{~M} \mathrm{~CHF}_2 \mathrm{COOH}) is correct because difluoroacetic acid is the strongest acid among the options due to the electron-withdrawing effect of the two fluorine atoms, leading to the highest electrical conductivity.
166 Medium
166 of 230
Understanding +4 | -1
KCl(aq)\mathrm{KCl}\left( {aq} \right) cannot be used as a salt bridge for the cell Cu(s)∣CuSO4(aq)∣∣AgNO3(aq)∣Ag(s)\mathrm{Cu}\left( s \right)\left| {\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right)} \right|\left| {\mathrm{AgN}{\mathrm{O}_3}\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right) because :
View answer and explanation
Correct option: C
AgCl\mathrm{AgCl} is precipitated
Option C (AgCl\mathrm{AgCl} is precipitated) is correct because AgCl\mathrm{AgCl} precipitates when KCl\mathrm{KCl} reacts with AgNO3\mathrm{AgN}{\mathrm{O}_3} in the cell.
167 Medium
167 of 230
Application +4 | -1
A galvanic cell : Cu∣Cu2+∣∣Ag+∣Ag\mathrm{Cu}\left|\mathrm{Cu}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag} initially contains 1 M Ag+1 \mathrm{~M}\mathrm{~Ag}^{+}and 1 M Cu2+1 \mathrm{~M} \mathrm{~Cu}^{2+} ions. The cell potential after the passage of 9.65 A9.65 \mathrm{~A} of current for 1 h1 \mathrm{~h} will be ___ V.
View answer and explanation
Correct option: A
0.010.01
Option A (0.01) is correct because the cell potential is calculated using the Nernst equation after determining the change in concentrations due to the passage of current.
168 Medium
168 of 230
Understanding +4 | -1
Assertion (A): Λm\Lambda_m for weak electrolytes shows a sharp increase, when the electrolytic solution is diluted.

Reason (R): For weak electrolytes degree of dissociation increases with dilution of solution.
View answer and explanation
Correct option: A
Assertion and Reason both are correct and Reason is the correct explanation of Assertion.
Option A is correct because as a weak electrolyte is diluted, its degree of dissociation increases, leading to a sharp increase in molar conductivity.
169 Medium
169 of 230
Application +4 | -1
2MnO4−+aH++bH2O2⟶2Mn+2+cH2O+dO22 \mathrm{MnO}_4^{-}+\mathrm{aH}^{+}+\mathrm{bH}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{Mn}^{+2}+\mathrm{cH}_2 \mathrm{O}+\mathrm{dO}_2. What are the values of a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} and d\mathrm{d}?
View answer and explanation
Correct option: B
6,5,8,56,5,8,5
Option B (6,5,8,56,5,8,5) is correct because it represents the coefficients that balance the given redox reaction.
170 Medium
170 of 230
Understanding +4 | -1
The correct order of molar conductance at infinite dilution of NaCl,KCl,RbCl{NaCl}, {KCl}, {RbCl} is
View answer and explanation
Correct option: B
RbCl>KCl>NaCl{RbCl}>{KCl}>{NaCl}
Option B is correct because ionic conductance increases with ionic mobility, which is inversely proportional to hydrated size.
171 Medium
171 of 230
Understanding +4 | -1
KCl(aq)\mathrm{KCl}\left( {aq} \right) cannot be used as a salt bridge for the cell Cu(s)∣CuSO4(aq)∣∣AgNO3(aq)∣Ag(s)\mathrm{Cu}\left( s \right)\left| {\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right)} \right|\left| {\mathrm{AgN}{\mathrm{O}_3}\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right) because :
View answer and explanation
Correct option: C
AgCl\mathrm{AgCl} is precipitated
Option C (AgCl is precipitated) is correct because Ag+ ions from AgNO3 react with Cl- ions from KCl to form insoluble AgCl, disrupting the salt bridge function.
172 Medium
172 of 230
Application +4 | -1
Following is the concentration cell in which electrode is
reversible with respect to anion
Pt(Cl2)1  bar∣Cl−(C1)∣∣Cl−(C2)∣Pt(Cl2)1  bar\mathop {Pt\left( {C{l_2}} \right)}\limits_{1\,\,{\text{bar}}} \left| {C{l^ - }\left( {{C_1}} \right)} \right|\left| {C{l^ - }\left( {{C_2}} \right)} \right|\mathop {Pt\left( {C{l_2}} \right)}\limits_{1\,\,{\text{bar}}}
The cell reaction is spontaneous, if
View answer and explanation
Correct option: A
C1>C2{C_1} > {C_2}
Option A (C1>C2{C_1} > {C_2}) is correct because the cell reaction is spontaneous when the concentration of the anion at the left electrode is greater than the concentration at the right electrode.
173 Medium
173 of 230
Analyzing +4 | -1
Assertion: The cell constant of a cell depends upon the nature of the material of the electrodes.

Reason: The observed conductance depends upon the nature of the electrolyte and the concentration of the solution.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because the cell constant depends on the geometry of the cell, not the electrode material, and conductance depends on electrolyte nature and concentration.
174 Medium
174 of 230
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 is the most reactive as carbon cannot remove oxygen from ZO, Y is more reactive than X as Y removes oxygen from XO.
175 Medium
175 of 230
Application +4 | -1
The charge required to deposit 40.5 g40.5 \mathrm{~g} of Al\mathrm{Al} (atomic mass =27.0 g=27.0 \mathrm{~g}) from the fused Al2(SO4)3\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 is:
View answer and explanation
Correct option: B
4.34×105C4.34 \times 10^5 \mathrm{C}
Option B (4.34×105C4.34 \times 10^5 \mathrm{C}) is correct because it is the charge required to deposit 40.5 g of Al calculated using Faraday's laws of electrolysis.
176 Medium
176 of 230
Application +4 | -1
For complete reduction of 1 mole of nitrobenzene to aniline, charge needed will be ( FF represents Faraday's constant)
View answer and explanation
Correct option: D
6 F6 \mathrm{~F}
Option D (6 F) is correct because the complete reduction of nitrobenzene to aniline involves a change in oxidation number of 6, requiring 6 Faradays of charge.
177 Medium
177 of 230
Analyzing +4 | -1
Statement I: At the end of electrolysis using platinum electrodes an aqueous solution of copper sulphate turns colourless

Statement II: Copper in CuSO4\mathrm{CuSO}_4 is converted to Cu(OH)2\mathrm{Cu}(\mathrm{OH})_2 during electrolysis
View answer and explanation
Correct option: C
Statement I is correct but statement II is false
Option C is correct because Statement I is correct (aqueous CuSO4 turns colourless after electrolysis), but Statement II is false (CuSO4 is not converted to Cu(OH)2 during electrolysis).
178 Medium
178 of 230
Application +4 | -1
The Λ∞{\mathrm{\Lambda }^\infty } of NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH} at infinite dilution is …Scm2eq.−1S{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^2}\mathrm{eq}{\mathrm{.}^{ - 1}} .Given λOH−∞=174,λCl−∞=66\lambda _{\mathrm{O}{\mathrm{H}^ - }}^\infty = 174,\lambda _{\mathrm{C}{\mathrm{l}^ - }}^\infty = 66 and λNH4Cl∞=130Scm2eq−1:\lambda _{\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}}^\infty = 130{\kern 1pt} {\kern 1pt} {\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{e}{\mathrm{q}^{ - 1}}:
View answer and explanation
Correct option: A
238
Option A (238) is correct because it is calculated using Kohlrausch's law of independent migration of ions: ΛNH4OH∞=ΛNH4Cl∞+λOH−∞−λCl−∞\mathrm{\Lambda }_{\mathrm{N}{\mathrm{H}_4}\mathrm{OH}}^\infty = \mathrm{\Lambda }_{\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}}^\infty + \lambda _{\mathrm{O}{\mathrm{H}^ - }}^\infty - \lambda _{\mathrm{C}{\mathrm{l}^ - }}^\infty.
179 Medium
179 of 230
Application +4 | -1
Calculate the electrode potential of given electrode at 25∘C25^{\circ} {C}

Pt/H2/H+(pH=6){Pt} / {H}_2 / {H}^{+}({pH}=6)
View answer and explanation
Correct option: A
0.354  V0.354\, {~V}
The correct answer is A (0.354 V) because the electrode potential is calculated using the Nernst equation, considering the pH of the solution.
180 Medium
180 of 230
Understanding +4 | -1
Statement I: A negative value of standard reduction potential means that reduction takes place on this electrode with reference to standard hydrogen electrode.

Statement II: The standard electrode potential of a half-cell has fixed value.
View answer and explanation
Correct option: D
Statement I is false but Statement II is correct
Option D (Statement I is false but Statement II is correct) is correct because a negative standard reduction potential means the electrode is more likely to be oxidized, not reduced, and the standard electrode potential of a half-cell has a fixed value.
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