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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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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 181-210 of 230 questions

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181 Medium
181 of 230
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.
182 Medium
182 of 230
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.
183 Medium
183 of 230
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.
184 Medium
184 of 230
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.
185 Medium
185 of 230
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).
186 Medium
186 of 230
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.
187 Medium
187 of 230
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.
188 Medium
188 of 230
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.
189 Medium
189 of 230
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.
190 Medium
190 of 230
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 – 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.
191 Medium
191 of 230
Application +4 | -1
Charge required to liberate 11.5 g11.5 \mathrm{~g} sodium is
View answer and explanation
Correct option: A
0.5 F0.5 \mathrm{~F}
Option A (0.5 F) is correct because it is calculated using Faraday's law of electrolysis, where the charge required is proportional to the mass of sodium liberated.
192 Medium
192 of 230
Analyzing +4 | -1
Standard potentials (E∘)\left(E^{\circ}\right) for some half-reactions are given below:

(i) Sn4++2e−→Sn2+;E∘=+0.15 V{Sn}^{4+}+2 {e}^{-} \rightarrow {Sn}^{2+} ; E^{\circ}=+0.15 {~V}

(ii) 2Hg2++2e−→Hg22+;E∘=+0.92 V2 {Hg}^{2+}+2 {e}^{-} \rightarrow {Hg}_2{ }^{2+} ; E^{\circ}=+0.92 {~V}

(iii) PbO2+4H++2e−→Pb2++2H2O;E∘=+1.45 V{PbO}_2+4 {H}^{+}+2 {e}^{-} \rightarrow {Pb}^{2+}+2 {H}_2 {O} ; E^{\circ}=+1.45 {~V}

Based on the above, which one of the following statements is correct?
View answer and explanation
Correct option: B
Sn2+{Sn}^{2+} is a stronger reducing agent than Hg22+{Hg}_2{ }^{2+}
Option B (Sn2+{Sn}^{2+} is a stronger reducing agent than Hg22+{Hg}_2{ }^{2+}) is correct because a more negative standard reduction potential indicates a stronger reducing agent.
193 Medium
193 of 230
Application +4 | -1
An electrochemical cell is constructed as:

Zn∣Zn2+(0.1M)∣∣H+(XM)∣(H2)Pt1 bar; EZn2+∣Zn0=−0.76  V{\text{Zn}}\left| {{\text{Z}}{{\text{n}}^{2 + }}(0.1{\text{M}})||{{\text{H}}^ + }({\text{XM}})} \right|\mathop {\left( {{{\text{H}}_2}} \right){\text{Pt}}}\limits_{1\,{\text{bar}}} {\text{; E}}_{{\text{Z}}{{\text{n}}^{2 + }}\mid {\text{Zn}}}^0 = - 0.76\;{\text{V}}

If emf of cell is 0.17 V0.17 \mathrm{~V} then pH\mathrm{pH} of cathodic compartment is:
View answer and explanation
Correct option: C
11.511.5
Option C (11.5) is correct because the pH of the cathodic compartment is calculated using the Nernst equation and the given cell emf.
194 Medium
194 of 230
Understanding +4 | -1
Assertion:The electrode potential of SHE is zero.

Reason: SHE is a standard reference electrode.
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 (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because while both the assertion and the reason are true, the reason doesn't explain why the electrode potential of SHE is zero.
195 Medium
195 of 230
Application +4 | -1
An aqueous solution of 0.02 M MgSO40.02 \, \mathrm{M} \, {MgSO}_4 electrolysed by passing a current of 9.65 A9.65 \mathrm{~A} for 30 min. Mass of product obtained at cathode is
View answer and explanation
Correct option: A
0.18 g0.18 \mathrm{~g}
Option A (0.18 g0.18 \mathrm{~g}) is correct because hydrogen gas is produced at the cathode, and its mass is calculated using Faraday's laws of electrolysis.
196 Medium
196 of 230
Application +4 | -1
The standard cell potential for the cell is : Zn∣Zn2+(1M)∣∣Cu2+(1M)∣Cu\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\left( {1M} \right)} \right|\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {1M} \right)} \right|\mathrm{Cu}
[E∘forZn2+/Zn=−0.76;E∘forCu2+/Cu=+0.34]\left[ {{E^ \circ }{\kern 1pt} {\kern 1pt}{\rm{for {\kern 1pt} {\kern 1pt}Z}}{\mathrm{n}^{2 + }}/\mathrm{Zn} = - 0.76;{E^ \circ }{\kern 1pt} {\kern 1pt}{\rm{for {\kern 1pt} {\kern 1pt}C}}{\mathrm{u}^{2 + }}/\mathrm{Cu} = + 0.34} \right]
View answer and explanation
Correct option: C
0.34−(−0.76)=+1.10V0.34 - \left( { - 0.76} \right) = + 1.10\mathrm{V}
Option C (0.34−(−0.76)=+1.10V0.34 - \left( { - 0.76} \right) = + 1.10\mathrm{V}) is correct because the standard cell potential is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.
197 Medium
197 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 the reduction of 0.5 mole of MnO4−\mathrm{MnO}_4^ - to Mn2+\mathrm{M}{\mathrm{n}^{2 + }} requires 2.5 Faraday of electricity.
198 Medium
198 of 230
Application +4 | -1
ZnZn amalgam is prepared by electrolysis of aqueous ZnCl2ZnCl_2 using HgHg cathode (9gm)(9 \mathrm{gm}). How much current is to be passed through ZnCl2ZnCl_2 solution for 1000 seconds to prepare a ZnZn Amalgam with 25%Zn25 \% Zn by wt. (Zn=65.4)(Zn=65.4)
View answer and explanation
Correct option: C
8.85amp8.85 \mathrm{amp}
Option C (8.85 amp) is correct because the current is calculated using Faraday's laws of electrolysis and the given weight percentage of zinc in the amalgam.
199 Medium
199 of 230
Application +4 | -1
On electrolysis of acidified water, if volume of hydrogen liberated is 5.6 cm35.6 \mathrm{~cm}^3, then the volume of oxygen liberated equal to:
View answer and explanation
Correct option: B
2.8 cm32.8 \mathrm{~cm}^3
Option B (2.8 cm32.8 \mathrm{~cm}^3) is correct because the volume of oxygen liberated is half the volume of hydrogen liberated during electrolysis of water.
200 Medium
200 of 230
Application +4 | -1
The potential of Hydrogen electrode (PH2=1atm,[H+]=O.1M)\left( {{P_{{H_2}}} = 1\mathrm{ atm },\mathrm{ }\left[ {{H^ + }} \right] = O.1M} \right) at 25oC{25^\mathrm{o}}C will be
View answer and explanation
Correct option: B
−0.059V- 0.059V
Option B (−0.059V- 0.059V) is correct because the Nernst equation for the hydrogen electrode gives a potential of -0.059V at 25°C when [H+] = 0.1M.
201 Medium
201 of 230
Application +4 | -1
If the pressure of H2(g)\mathrm{H}_2(\mathrm{g}) is increased from 1 atm1 \mathrm{~atm} to 100 atm100 \mathrm{~atm} keeping [H+]\left[\mathrm{H}^{+}\right] constant at 1M1 \mathrm{M}, the change in reduction potential of hydrogen half cell at 25∘C25^{\circ} \mathrm{C} will be:
View answer and explanation
Correct option: B
−0.059 V-0.059 \mathrm{~V}
Option B (-0.059 V) is correct because the change in reduction potential is calculated using the Nernst equation, considering the change in hydrogen gas pressure.
202 Medium
202 of 230
Application +4 | -1
By passage of 1F of electricity, through 1L of 1M solution
of an electrolyte which is not obtained
View answer and explanation
Correct option: A
1 mol1 \mathrm{~mol} of Cu\mathrm{Cu}
Option A (1 mol of Cu) is correct because 1F of electricity can deposit 0.5 mol of Cu from Cu²⁺ solution, not 1 mol.
203 Medium
203 of 230
Application +4 | -1
Passage of three faraday of charge through aqueous solution of AgNO3,CuSO4,Al(NO3)3\mathrm{AgN}{\mathrm{O}_3},\mathrm{CuS}{\mathrm{O}_4},\mathrm{Al}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_3} and NaCl will deposit metals at the cathode in the molar ratio of:
View answer and explanation
Correct option: C
6:3:0:06\mathrm{ }:\mathrm{ }3\mathrm{ }:\mathrm{ }0\mathrm{ }:\mathrm{ }0
Option C (6:3:0:06\mathrm{ }:\mathrm{ }3\mathrm{ }:\mathrm{ }0\mathrm{ }:\mathrm{ }0) is correct because Ag and Cu will deposit, while Na and Al will not deposit from their aqueous solutions.
204 Medium
204 of 230
Analyzing +4 | -1
Tell which of the following statements accurately describes the effect of adding CN−\mathrm{CN}^{-} to the cathode of a cell with a cell reaction:

Cd+2Ag+→2Ag+Cd2+,E∘=1.2 V\mathrm{Cd}+2 \mathrm{Ag}^{+} \rightarrow 2 \mathrm{Ag}+\mathrm{Cd}^{2+}, E^{\circ}=1.2 \mathrm{~V}
View answer and explanation
Correct option: D
E∘E^{\circ} decreases because Ag(CN)2−\mathrm{Ag}(\mathrm{CN})_2^{-} forms
Option D is correct because adding CN- to the cathode will form a complex with Ag+ ions, decreasing the concentration of Ag+ and thus decreasing the cell potential E°.
205 Medium
205 of 230
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 composition of the cathode in a Leclanche cell.
206 Medium
206 of 230
Application +4 | -1
Passage of electricity through dil H2SO4{H}_2 {SO}_4 for 1 hour liberates a total of 224 ml224\, \mathrm{ml} of H2{H}_2 at S.T.P. The magnitude of current would be :-
View answer and explanation
Correct option: B
0.54 amperes
Option B (0.54 amperes) is correct because it is calculated by using Faraday's laws of electrolysis to relate the amount of hydrogen liberated to the current passed.
207 Medium
207 of 230
Analyzing +4 | -1
Assertion: Molar conductivity of CH3COOH\mathrm{CH}_3 \mathrm{COOH} increases with dilution.

Reason: The graph between λmv/s(C)1/2\lambda \mathrm{m} \mathrm{v} / \mathrm{s}(\mathrm{C})^{1 / 2} is a straight line for CH3COOH\mathrm{CH}_3 \mathrm{COOH}.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because the molar conductivity of a weak electrolyte like CH3COOH\mathrm{CH}_3\mathrm{COOH} increases with dilution as the degree of dissociation increases, but the graph between λm\lambda_m vs (C)1/2(\mathrm{C})^{1/2} is not a straight line.
208 Medium
208 of 230
Analyzing +4 | -1
The conductivity of 0.1 (M) NaOH\mathrm{NaOH} solution is 0.0221 ohm−1 cm−1\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}. When an equal volume of 0.1 (M) HCl\mathrm{HCl} solution is added, the conductivity decreases to 0.0056 ohm−1 cm−1\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}.
A further addition of HCl\mathrm{HCl} solution, the volume of which is equal to that of the first portion added, the conductivity increases to 0.017ohm−1 cm−10.017 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}. What is the molar conductance at infinite dilution of H2O\mathrm{H}_2 \mathrm{O} in ohm −1{ }^{-1} cm2 mol−1\mathrm{cm}^2 \mathrm{~mol}^{-1} in nearest possible integers ?
View answer and explanation
Correct option: A
512512
Option A (512) is correct because it is calculated using the molar conductivities of NaOH, HCl, and NaCl at infinite dilution.
209 Medium
209 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.
210 Medium
210 of 230
Analyzing +4 | -1
Statement I: A metal having negative reduction
potential when dipped in the solution of its own
ions has a tendency to pass into the solution.

Statement II: Metals having negative reduction
potential have large hydration energy.
View answer and explanation
Correct option: B
If both statements are CORRECT, and Statement
II is NOT the CORRECT explanation of
Statement I
Option B is correct because a metal with a negative reduction potential tends to pass into solution, and metals with negative reduction potential have large hydration energy, but the hydration energy is not the direct explanation for the tendency to pass into solution.
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