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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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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 301-330 of 396 questions

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301 Easy
301 of 396
Analyzing +4 | -1
What will be pH of aqueous solution of electrolyte in electrolytic cell during electrolysis of CuSO4(aq)\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right) between graphite electrodes?
View answer and explanation
Correct option: C
pH < 7.0
Option C (pH < 7.0) is correct because the electrolysis of aqueous CuSO4 produces H2SO4, making the solution acidic.
302 Medium
302 of 396
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.
303 Easy
303 of 396
Understanding +4 | -1
Rusting of iron is catalysed by which of the following?
View answer and explanation
Correct option: D
H+{\mathrm{H}^ + }
The correct answer is D (H+{\mathrm{H}^ + }) because hydrogen ions catalyze the rusting process of iron.
304 Hard
304 of 396
Application +4 | -1
A 0.200 M KOH0.200 \mathrm{~M} \mathrm{~KOH} solution is electrolysed for 1.5 h1.5 \mathrm{~h} using a current of 8.00 A8.00 \mathrm{~A}. How many moles of O2\mathrm{O}_2 were produced at the anode?
View answer and explanation
Correct option: A
0.112
Option A (0.112) is correct because it calculates the moles of O2 produced during electrolysis using Faraday's laws and the given current and time.
305 Medium
305 of 396
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.
306 Hard
306 of 396
Analyzing +4 | -1
Small quantities of solutions of compounds TX,TYTX,TY and TZTZ are put into separate test tubes containing X, Y and Z solutions. TXTX does not react with any of these. TYTY reacts with both X and Z. TZTZ reacts with X. The decreasing order of ease of oxidation of the anions X−,Y−,Z−{X^ - },{Y^ - },{Z^ - } is
View answer and explanation
Correct option: A
Y−,Z−,X−{Y^ - },{Z^ - },{X^ - }
Option A (Y−,Z−,X−{Y^ - },{Z^ - },{X^ - }) is correct because the reactivity series indicates that Y−{Y^-} is most easily oxidized, followed by Z−{Z^-}, and then X−{X^-}.
307 Medium
307 of 396
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).
308 Medium
308 of 396
Application +4 | -1
The Λ∞{\mathrm{\Lambda }^\infty } of NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH} at infinite dilution is &hellip;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.
309 Medium
309 of 396
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.
310 Easy
310 of 396
Remembering +4 | -1
In an electrolytic cell, the anode and cathode are respectively represented as :
View answer and explanation
Correct option: A
Positive electrode, negative electrode
Option A (Positive electrode, negative electrode) is correct because in an electrolytic cell, the anode is the positive electrode and the cathode is the negative electrode.
311 Medium
311 of 396
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.
312 Hard
312 of 396
Application +4 | -1
How much charge (in F\mathrm{F}) must flow through solution during electrolysis of aq. Na2SO4\mathrm{Na}_2 \mathrm{SO}_4 at 0∘C0^{\circ} \mathrm{C} and 1 atm to produce 33.6 L of product gases at 50%50 \% current efficiency?
View answer and explanation
Correct option: A
4 F4 \mathrm{~F}
The correct answer is A (4 F) because the electrolysis of aq. Na2SO4 produces H2 and O2, and the charge required to produce 33.6 L of product gases at 50% efficiency is 4F.
313 Medium
313 of 396
Application +4 | -1
Find weight of Zinc in Zinc sulphate electrolysis i=0.015 A,t=15\mathrm{i}=0.015 \mathrm{~A,t} =15 minutes
View answer and explanation
Correct option: A
0.00460.0046
Option A (0.0046) is correct because it is calculated using Faraday's laws of electrolysis, considering the current, time, and the equivalent weight of Zinc.
314 Easy
314 of 396
Analyzing +4 | -1
Which of the following statements is/are incorrect?
View answer and explanation
Correct option: D
ΔG=−nEeell \Delta G=-n E_{\text {eell }}
Option D is correct because the correct formula is ΔrG∘=−nFEcell∘\Delta_r G^{\circ}=-n F E^{\circ}_{cell}, which includes the Faraday constant.
315 Medium
315 of 396
Application +4 | -1
Find the emf of the cell in which the following reaction takes place at 298 K
Ni(s)+2Ag+(0.001M)→Ni2+(0.001M)+2Ag(s)Ni(s) + 2A{g^ + }(0.001M) \to N{i^{2 + }}(0.001M) + 2Ag(s)
(Given that Ecell∘=10.5 V, 2.303 RTF=0.059E_{cell}^ \circ = 10.5\,V,\,\frac{{2.303\,RT}}{F} = 0.059 at 298 K)
View answer and explanation
Correct option: D
0.9615 V
Option B (0.36 kg) is correct because it is the mass of glucose required to produce the specified amount of energy, calculated using the heat of combustion.
316 Easy
316 of 396
Application +4 | -1
Consider the following cell representation:
Pt(s)∣Br2(l)∣Br⊖(0.010)∣∣H⊕(0.030 M)∣H2(g)(1 bar)∣Pt(s)\mathrm{Pt}(\mathrm{s})\left|\mathrm{Br}_2({l})\right| \mathrm{Br}^{\ominus}(0.010)|| \mathrm{H}^{\oplus}(0.030 \mathrm{~M})\left|\mathrm{H}_2(\mathrm{g})(1 \mathrm{~bar})\right| \mathrm{Pt}(\mathrm{s})
The EMF of the above cell at 298 K298 \mathrm{~K} is −xV-\mathrm{xV}. Find 3x3 \mathrm{x} (nearest integer)

[Given E12Br2∣Br−∘ (Reduction potential of Br2)=+1.08 V]\begin{aligned} & \text {[Given } \mathrm{E}^{\circ}_{\frac{1}{2}\mathrm{Br}_2 \mid \mathrm{Br}^{-}} \text{ (Reduction potential of } \mathrm{Br}_2)=+1.08 \mathrm{~V} \text {]} \end{aligned}
View answer and explanation
Correct option: A
4
Option A (4) is correct because the EMF of the cell is calculated using the Nernst equation, and then multiplied by 3 as requested.
317 Hard
317 of 396
Application +4 | -1
The cell, Zn∣Zn2+(1M)∣∣Cu2+(1M)∣Cu\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\left( {1\mathrm{M}} \right)} \right|\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {1\mathrm{M}} \right)} \right|\mathrm{Cu}

(Ecell∘=E_{\mathrm{cell}}^ \circ =1.10 V), was allowed to be completely discharged at 298 K. The relative concentration of

Zn2+\mathrm{Z}{\mathrm{n}^{2 + }} to Cu2+([Zn2+][Cu2+])\mathrm{C}{\mathrm{u}^{2 + }}\left( {\frac{{\left[ {\mathrm{Z}{\mathrm{n}^{2 + }}} \right]}}{{\left[ {\mathrm{C}{\mathrm{u}^{2 + }}} \right]}}} \right) is
View answer and explanation
Correct option: B
Antilog (37.3)
Option C (1037.3{10^{37.3}}) is the closest to the correct answer because it is derived from the Nernst equation at equilibrium, where the cell potential is zero.
318 Easy
318 of 396
Remembering +4 | -1
Electrode potential of hydrogen electrode is …….. Volt.
View answer and explanation
Correct option: A
0
Option A (0) is correct because the standard hydrogen electrode (SHE) is defined to have a standard electrode potential of 0 volts.
319 Medium
319 of 396
Application +4 | -1
If the following half cells have the E∘E^{\circ} values as Fe3++e−→Fe2+;E∘=+0.77 V\mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} ; E^{\circ}=+0.77 \mathrm{~V} and Fe2++2e−→Fe;E∘=−0.44 V;\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} ; E^{\circ}=-0.44 \mathrm{~V}; the E∘E^{\circ} of the half-cell Fe3++3e−→Fe\mathrm{Fe}^{3+}+3 e^{-} \rightarrow \mathrm{Fe} will be
View answer and explanation
Correct option: C
−0.036 V-0.036 \mathrm{~V}
Option C (-0.036 V) is correct because the standard reduction potential for the half-cell Fe3++3e−→Fe\mathrm{Fe}^{3+}+3 e^{-} \rightarrow \mathrm{Fe} is calculated using the given E∘E^{\circ} values and the relationship between Gibbs free energy and standard potential.
320 Easy
320 of 396
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 to have a standard electrode potential of 0 V.
321 Medium
321 of 396
Remembering +4 | -1
An electronic conductor is
View answer and explanation
Correct option: C
AgAg
Option C (AgAg) is correct because silver (AgAg) is a metal and metals are electronic conductors due to the presence of delocalized electrons.
322 Medium
322 of 396
Understanding +4 | -1
The SRP values of the elements A, B and C are 0.68, -2.50 and 0.50V respectively. The order of their reducing power is
View answer and explanation
Correct option: D
B > C > A
Option D (B > C > A) is correct because reducing power is inversely proportional to the standard reduction potential (SRP).
323 Hard
323 of 396
Application +4 | -1
Find the solubility product of a saturated solution of Ag2CrO4\mathrm{Ag}_2 \mathrm{CrO}_4 in water at 298 K298 \mathrm{~K} if the emf of the cell Ag∣Ag+(satd. Ag2CrO4 soln.)∥Ag+(0.1 M)∣Ag\mathrm{Ag} | \mathrm{Ag}^{+} \text{(satd. }\mathrm{Ag}_2 \mathrm{CrO}_4 \text{ soln.)}\| \mathrm{Ag}{ }^{+}(0.1 \mathrm{~M}) | \mathrm{Ag} is 0.164 V0.164 \mathrm{~V} at 298 K.298 \mathrm{~K}.
View answer and explanation
Correct option: A
2.287×10−12{2.287} \times 10^{-12}
Option A (2.287 × 10-12) is correct because it's the solubility product calculated using the Nernst equation and the given EMF.
324 Hard
324 of 396
Understanding +4 | -1
Which colourless gas evolves when NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} reacts with zinc in A dry cell battery?
View answer and explanation
Correct option: C
H2{\mathrm{H}_2}
Option C is correct because hydrogen gas (H2) is produced when zinc reacts with ammonium chloride in a dry cell battery.
325 Medium
325 of 396
Application +4 | -1
Without losing its concentration ZnCl2{ZnCl}_2 solution cannot be kept in contact with
View answer and explanation
Correct option: B
Al{Al}
Option B (2×10−82 \times 10^{-8}) is correct because it is the approximate partial pressure of SO2 at equilibrium, calculated using the given Kp and initial partial pressures.
326 Medium
326 of 396
Understanding +4 | -1
Molar conductance of an electrolyte increases with dilution according to the equation.

Λm=Λm∘−AC\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{A} \sqrt{\mathrm{C}}

Which of the following statement is/are true:
View answer and explanation
Correct option: B
Value of the constant A depends upon the nature of the solvent
Option B is correct because the constant A in the Debye-Huckel-Onsager equation depends on the nature of the solvent due to factors like dielectric constant and viscosity.
327 Hard
327 of 396
Application +4 | -1
A certain quantity of electricity is passed through aqueous solution of AgNO3\mathrm{AgN}{\mathrm{O}_3} and CuSO4\mathrm{CuS}{\mathrm{O}_4} connected in series. If Ag (at.wt.108) deposited at the cathode is 1.08 g then Cu deposited at the cathode is (at. wt. of Cu is 63.53 ):
View answer and explanation
Correct option: B
0.317 g
Option B (0.317 g) is correct because the mass of Cu deposited is calculated using Faraday's laws of electrolysis, equating the equivalents of Ag and Cu.
328 Medium
328 of 396
Understanding +4 | -1
The molecular conductivity and equivalent conductivity are same for the solution of :
View answer and explanation
Correct option: A
1MNaCl1M{\kern 1pt} {\kern 1pt} \mathrm{NaCl}
Option A (1MNaCl1M{\kern 1pt} {\kern 1pt} \mathrm{NaCl}) is correct because NaCl has a valency factor of 1, making its molar and equivalent conductivities equal.
329 Medium
329 of 396
Application +4 | -1
Calculate the standard Gibbs free energy of

Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s)\mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s)
View answer and explanation
Correct option: B
−2FEcell -2 F E_{\text {cell }}
Option B (−2FEcell-2FE_{cell}) is correct because the standard Gibbs free energy change is calculated using the formula ΔrG=−nFEcell\Delta_r G = -nFE_{cell}, where n is the number of moles of electrons transferred, which is 2 in this case.
330 Medium
330 of 396
Application +4 | -1
The cell reaction of A cell is
Mg(s)+Cu2+(aq)→Cu(s)+Mg2+(aq)\mathrm{Mg}\left( s \right) + \mathrm{C}{\mathrm{u}^{2 + }}\left( {aq} \right) \to \mathrm{Cu}\left( s \right) + \mathrm{M}{\mathrm{g}^{2 + }}\left( {aq} \right)
If the standard reduction potentials of Mg and Cu are &ndash; 2.37 and + 0.34 V respectively. The emf of the cell is
View answer and explanation
Correct option: C
#ERROR!
Option C (+2.71 V) is correct because the emf of the cell is calculated by subtracting the standard reduction potential of the anode from that of the cathode.
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