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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 241-270 of 396 questions

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241 Medium
241 of 396
Application +4 | -1
The E∘{E^ \circ } of Fe2+/Fe\mathrm{F}{\mathrm{e}^{2 + }}/\mathrm{Fe} and Sn2+/Sn\mathrm{S}{\mathrm{n}^{2 + }}/\mathrm{Sn} are -0.44 V and -0.14 V respectively. If cell reaction is

Fe+Sn2+→Fe2++Sn\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
#ERROR!
Option A (+0.30 V) is correct because the cell emf is calculated by subtracting the anode's standard reduction potential from the cathode's.
242 Medium
242 of 396
Application +4 | -1
A 4.0 molar aqueous solution of NaCl\mathrm{NaCl} is prepared and 500 mL500 \mathrm{~mL} of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes. If the cathode is a Hg\mathrm{Hg} electrode, the maximum weight (g)(\mathrm{g}) of amalgam (Na−Hg)(\mathrm{Na}-\mathrm{Hg}) formed from this solution is : (atomic mass: Na=23,Hg=200\mathrm{Na}=23, \mathrm{Hg}=200 )
View answer and explanation
Correct option: A
446
Option A (446) is correct because it is the mass of the Na-Hg amalgam formed based on the moles of Na+ reduced at the cathode.
243 Easy
243 of 396
Remembering +4 | -1
A cell in which electric current is produced by net oxidation and reduction process is called :
View answer and explanation
Correct option: A
Voltaic cell
Option A (Voltaic cell) is correct because a voltaic cell produces electric current through spontaneous oxidation-reduction reactions.
244 Hard
244 of 396
Understanding +4 | -1
Statement-I: The voltage of mercury cell remains constant for long period of time.

Statement - II: The net cell reaction does not involve active species(ions) .
View answer and explanation
Correct option: A
Both Statements-I and II are correct and Statement - II is correct explanation for Statement - I.
Option A (Both Statements-I and II are correct and Statement - II is correct explanation for Statement - I.) is correct because the voltage of a mercury cell remains constant due to the net cell reaction not involving ions in solution.
245 Easy
245 of 396
Understanding +4 | -1
Among the following correct statements

I) Oxidising power ∝ SRP ∝1SOP\propto \,\mathrm{SRP}\, \propto \frac{1}{{SOP}}

II) Oxidising power ∝\propto SOP ∝1SOP\propto \frac{1}{{SOP}}

III) Reducing power ∝\propto SOP ∝  SOP ∝ 1SRP\propto \,\,\mathrm{SOP}\, \propto \,\frac{1}{{SRP}}
IV) Reducing power ∝\propto SRP ∝1SOP\propto \frac{1}{{SOP}}
View answer and explanation
Correct option: D
I & III
Option D (I & III) is correct because oxidizing power is directly proportional to SRP and inversely proportional to SOP, while reducing power is directly proportional to SOP and inversely proportional to SRP.
246 Medium
246 of 396
Application +4 | -1
A solution is 1 molar in each of NaCl,CdCl2,ZnCl2{NaCl}, {CdCl}_2,{ZnCl}_2 and PbCl2{PbCl}_2. To this Sn{Sn} metal is added, which of the following is true?
[ Given E∘(Pb2+/Pb=−0.126 V)ESn2/Sn∘=−0.136 V,ECd2+/Cd∘=−0.40 VEZn2+/Zn∘=−0.763 V,ENa+/Na∘=−2.71 V]\left[\begin{array}{l} \text { Given } \quad {E}^{\circ}\left({Pb}^{2+} /{Pb}=-0.126 \mathrm{~V}\right) \\ {E}_{{Sn}^2 /{Sn}}^{\circ}=-0.136 \mathrm{~V},{E}_{{Cd}^{2+} / {Cd}}^{\circ}=-0.40 \mathrm{~V} \\ {E}_{{Zn}^{2+} / {Zn}}^{\circ}=-0.763 \mathrm{~V}, {E}_{{Na}^{+} /{Na}}^{\circ}=-2.71 \mathrm{~V} \end{array}\right]
View answer and explanation
Correct option: D
Sn{Sn} can reduce Pb2+{Pb}^{2+} to Pb{Pb}
Option D (Sn{Sn} can reduce Pb2+{Pb}^{2+} to Pb{Pb}) is correct because the standard reduction potential of Sn is lower than that of Pb, making Sn a stronger reducing agent.
247 Easy
247 of 396
Understanding +4 | -1
The correct order of electro chemical equivalent (e) of Sodium, Magnesium and Aluminium is
View answer and explanation
Correct option: A
Na > Mg > Al
The correct answer is A. Electrochemical equivalent is directly proportional to equivalent weight, so Na > Mg > Al.
248 Medium
248 of 396
Application +4 | -1
The equivalent conductivity of 0.1 M weak acid is 100 times less than that at infinite dilution. The degree of dissociation of weak electrolyte at 0.1 M is :
View answer and explanation
Correct option: C
0.01
Option A (333) is correct because it is calculated using Kohlrausch's law of independent migration of ions.
249 Medium
249 of 396
Application +4 | -1
A certain quantity electricity is passed through aqueous solutions of Al2(SO4)3A{l_2}{(S{O_4})_3} and CuSO4CuS{O_4} solutions connected in series. 0.09 g of AlAl is deposited on cathode during electrolysis. The amount of copper deposited on cathode in grams is (Atomic weight of Al=27Al = 27, Atomic weight of Cu=63.6Cu = 63.6)
View answer and explanation
Correct option: B
0.318
Option B (0.318) is correct because it is the amount of copper deposited calculated using Faraday's second law of electrolysis.
250 Medium
250 of 396
Application +4 | -1
The resistance of 2.5 N2.5 \mathrm{~N} solution of acetic acid is 230 ohm230 \mathrm{~ohm}, when measured in a cell having a cell constant of 1.15 cm−11.15 \mathrm{~cm}^{-1}. The equivalent conductance is ohm−1 cm2\mathrm{ohm}^{-1} \mathrm{~cm}^2 equiv −1{ }^{-1}
View answer and explanation
Correct option: A
2
Option A (2) is correct because the equivalent conductance is calculated using the formula Λeq=K×1000normality\Lambda_{eq} = \frac{{K \times 1000}}{{\text{{normality}}}} where K is conductivity.
251 Hard
251 of 396
Application +4 | -1
Specific conductance of 10−4 Mn10^{-4} \mathrm{~M} \mathrm{n}-Butyric acid aqueous solution is 1.9×10−9 S m−1.1.9 \times 10^{-9} \mathrm{~S} \mathrm{~m}^{-1}. If molar conductance of n\text{n}-Butyric acid at infinite dilution is 380×10−Sm2 mol1,380 \times 10^{-} \mathrm{S} \mathrm{m}^2 \mathrm{~mol}^1, then Ka\mathrm{K}_\text{a} for n\mathrm{n}-Butyric acid is:
View answer and explanation
Correct option: A
5×10−5M5 \times 10^{-5} \mathrm{M}
Option A (5×10−5M5 \times 10^{-5} \mathrm{M}) is correct because it is calculated using the formula for KaK_a after finding the degree of dissociation.
252 Hard
252 of 396
Analyzing +4 | -1
In which case (Ecell −Ecell ∘)\left(E_{\text {cell }}-E_{\text {cell }}^{\circ}\right) is zero
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because in both cases, the reaction quotient Q equals the equilibrium constant K, making the log term in the Nernst equation zero, thus Ecell - E°cell = 0.
253 Easy
253 of 396
Remembering +4 | -1
Hydrogen-oxygen fuel cells are used in spacecraft to supply :
View answer and explanation
Correct option: A
Power for heat and light
Option A (Power for heat and light) is correct because hydrogen-oxygen fuel cells in spacecraft provide power for heat and light, as well as drinking water.
254 Medium
254 of 396
Application +4 | -1
The standard reduction potential of EBi33+/Bi\mathrm{E}_{\mathrm{Bi}^3}^{3+} / \mathrm{Bi} and ECu2+/Cu\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}} are 0.226 V0.226 \mathrm{~V} and 0.344 V0.344 \mathrm{~V} respectively A\mathrm{A} mixture of salts of Bi\mathrm{Bi} and Cu\mathrm{Cu} at unit concentration each is electrolysed at 25∘C25^{\circ} \mathrm{C}. To what value can [Cu2+]\left[\mathrm{Cu}^{2+}\right] be brought down before bismuth starts to deposit during electrolysis.
View answer and explanation
Correct option: A
10−410^{-4}
Option A (10−410^{-4}) is correct because it is the concentration of Cu2+\mathrm{Cu}^{2+} at which the reduction potential of copper equals that of bismuth, initiating bismuth deposition.
255 Easy
255 of 396
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 the EMF of a cell can be calculated as Ecell = EOP anode + ERP cathode or Ecell = EOP anode - EOP cathode.
256 Medium
256 of 396
Application +4 | -1
During the electrolysis of 0.1 M−CuSO40.1 \mathrm{~M}-\mathrm{CuSO}_4 solution using copper electrodes, a depletion of Cu2+\mathrm{Cu}^{2+} occurs near the cathode with a corresponding excess near the anode, owing to inefficient stirring of the solution. If the local concentration of Cu2+\mathrm{Cu}^{2+} near the anode and cathode are, respectively, 0.12 M0.12 \mathrm{~M} and 0.08 M,0.08 \mathrm{~M}, the back EMF developed at 298 K298 \mathrm{~K} is (log⁡1.5=0.18,2.303RT/F=0.06)(\log 1.5=0.18,2.303 R T / F=0.06)
View answer and explanation
Correct option: B
5.4 mV5.4 \mathrm{~mV}
Option B (5.4 mV5.4 \mathrm{~mV}) is correct because the back EMF is calculated using the Nernst equation with the given concentrations.
257 Medium
257 of 396
Application +4 | -1
Resistivity of a material is equal to resistance when cell
constant is
View answer and explanation
Correct option: A
1 cm−11 \mathrm{~cm}^{-1}
Option A (1 cm−11 \mathrm{~cm}^{-1}) is correct because resistivity is equal to resistance when the cell constant is 1.
258 Medium
258 of 396
Remembering +4 | -1
Which of the following statements is correct? Galvanic cell converts:
View answer and explanation
Correct option: A
Chemical energy into electrical energy
Option A (Chemical energy into electrical energy) is correct because a galvanic cell (also known as a voltaic cell) uses spontaneous chemical reactions to generate electricity.
259 Easy
259 of 396
Application +4 | -1
100 g Al100 \mathrm{~g} \mathrm{~Al} is to be deposited on the cathode of the electrolytic cell containing Al2O3\mathrm{Al}_2 \mathrm{O}_3 in the solvent Na3AlF6\mathrm{Na}_3 \mathrm{AlF}_6 by a current of 125 A125 \mathrm{~A}. If at the cathode, reaction is

Al3++3e−⟶Al\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}

then time required for the passage of current is (Al=27)(\mathrm{Al}=27)
View answer and explanation
Correct option: A
143 min
Option A (143 min) is correct because it is the time required to deposit 100g of Al, calculated using Faraday's laws of electrolysis.
260 Medium
260 of 396
Analyzing +4 | -1
Assertion (A) Ecell E_{\text {cell }} increases with increase in concentration of Ag+\mathrm{Ag}^{+}ions.

Reason (R) Ecell E_{\text {cell }} has positive value.
View answer and explanation
Correct option: B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and the reason are true, but the reason does not explain the assertion.
261 Easy
261 of 396
Understanding +4 | -1
Pure water does not conduct electricity because it is
View answer and explanation
Correct option: B
Almost not ionised
Option B (Almost not ionised) is correct because pure water has a very low concentration of ions (H+ and OH-) due to its very small degree of self-ionization.
262 Medium
262 of 396
Application +4 | -1
12.3 g nitrobenzene is reduced into aniline by electrolysis, if current efficiency is 25% calculate charge required in coulombs
View answer and explanation
Correct option: C
24×965024 \times 9650 coulombs
The correct answer is C (24×965024 \times 9650 coulombs) because it is calculated by considering the number of moles of nitrobenzene, the number of electrons involved in the reduction, and the current efficiency.
263 Easy
263 of 396
Remembering +4 | -1
In electrolysis, oxidation takes place at:
View answer and explanation
Correct option: A
Anode
Option A (Anode) is correct because oxidation, which is the loss of electrons, always occurs at the anode during electrolysis.
264 Medium
264 of 396
Understanding +4 | -1
Assertion: In electrolytic refining, electrolytic cell consists of pure and impure Cu electrodes dipped in CuSO4(ag)\mathrm{CuSO}_4(\mathrm{ag}).

Reason: Pure cu is obtained at anode.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C (If Assertion is True but the Reason is False) is correct because in electrolytic refining, pure copper is obtained at the cathode, not the anode.
265 Medium
265 of 396
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.
266 Medium
266 of 396
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}).
267 Medium
267 of 396
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.
268 Easy
268 of 396
Application +4 | -1
The amount of electricity that can deposit 108 g of
silver from silver nitrate solution is
View answer and explanation
Correct option: C
1 faraday
Option C (1 faraday) is correct because 1 mole of silver (108 g) requires 1 mole of electrons (1 faraday) for deposition.
269 Easy
269 of 396
Remembering +4 | -1
The charge required to liberate one gram equivalent of an element is
View answer and explanation
Correct option: B
1 F
Option B (1 F) is correct because one Faraday (1 F) is the charge required to liberate one gram equivalent of an element.
270 Medium
270 of 396
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.
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