Skip to main content
Back to ELECTRO CHEMISTRY

Your Progress

0 / 396 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 361-390 of 396 questions

Preparing practice and saved progress…

361 Medium
361 of 396
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.
362 Medium
362 of 396
Application +4 | -1
The reduction potential of hydrogen half – cell will be negative if:
View answer and explanation
Correct option: C
P(H)2=2atm  and  [H+]=1.0 MP{\left( H \right)_2} = 2\mathrm{ atm\,\, and\,\, }\left[ {{H^ + }} \right] = 1.0\,M
Option C (P(H)2=2atm  and }[H+]=1.0 MP{\left( H \right)_2} = 2\mathrm{ atm\,\, and\,\}\left[ {{{{H^ + }}}} \right] = 1.0\,M}) is correct because, according to the Nernst equation, increasing the pressure of H2 and decreasing the concentration of H+ will result in a negative reduction potential.
363 Medium
363 of 396
Analyzing +4 | -1
A Galvanic cell is represented as: Pt,H2(0.1 bar)∣H+(aq)(2 M)∥M+(0.2 M)∣M(s)\mathrm{Pt}, \mathrm{H}_2(0.1 \mathrm{~bar})\left|\mathrm{H}^{+}(\mathrm{aq})(2 \mathrm{~M}) \| \mathrm{M}^{+}(0.2 \mathrm{~M})\right| \mathrm{M}(\mathrm{s}) [Given : Kw=10−14,EM+/Mo=0.03 V\mathrm{K}_{\mathrm{w}}=10^{-14}, \mathrm{E}_{\mathrm{M}^{+} / \mathrm{M}}^{\mathrm{o}}=0.03 \mathrm{~V}]. The correct statement for the given Galvanic cell is:
View answer and explanation
Correct option: C
ΔGreaction ∘<0,Ecell <0\Delta \mathrm{G}_{\text {reaction }}^{\circ} < 0, \mathrm{E}_{\text {cell }} < 0
Option C (ΔGreaction ∘<0,Ecell <0\Delta \mathrm{G}_{\text {reaction }}^{\circ} < 0, \mathrm{E}_{\text {cell }} < 0) is correct because the standard Gibbs free energy change is negative, indicating spontaneity, but the cell potential is negative under the given non-standard conditions.
364 Medium
364 of 396
Understanding +4 | -1
Which of the following solutions of KCl{KCl} will have the highest value of specific conductance?
View answer and explanation
Correct option: A
1.0 N1.0 {~N}
Option A (1.0 N1.0 {~N}) is correct because specific conductance increases with concentration, meaning a higher concentration of KCl will have a higher specific conductance.
365 Easy
365 of 396
Application +4 | -1
The molar conductivity at infinite dilution of AgNO3,NaCl\mathrm{AgN}{\mathrm{O}_3},\mathrm{NaCl} and NaNO3\mathrm{NaN}{\mathrm{O}_3} are 116.5, 110.3 and 105.2 mhocm2mol−1{\rm{mho{\kern 1pt} {\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}} respectively. The molar conductivity of AgCl is :
View answer and explanation
Correct option: A
121.6
Option A (121.6) is correct because it is calculated by applying Kohlrausch's law of independent migration of ions.
366 Easy
366 of 396
Understanding +4 | -1
Which of the following statements is correct
View answer and explanation
Correct option: A
Oxidation occur at onode in both galvanic and electrolytic cell
Option A (Oxidation occur at onode in both galvanic and electrolytic cell) is correct because oxidation always occurs at the anode, regardless of whether it's a galvanic or electrolytic cell.
367 Medium
367 of 396
Understanding +4 | -1
Effect of dilution on conductivity of solution:
View answer and explanation
Correct option: A
Increases
Option A (ZnZn) is correct because Zinc (ZnZn) has a completely filled d-orbital (d10\mathrm{d^{10}}) and thus weaker metallic bonding, resulting in the least enthalpy of atomization.
368 Medium
368 of 396
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 electrolytic bath for gold plating contains gold chloride and sodium cyanide to form a complex.
369 Easy
369 of 396
Application +4 | -1
The value of Λeq∞\mathrm{\Lambda }_{eq}^\infty for NH4Cl,NaOHandNaCl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl},{\rm{NaOH {\kern 1pt}{\kern 1pt}and {\kern 1pt}{\kern 1pt}NaCl}} are respectively, 149.74, 248.1 and 126.4 Ω−1cm2equiv−1{\Omega ^{ - 1}}{\kern 1pt}{\kern 1pt}\mathrm{c}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt}\mathrm{equi}{\mathrm{v}^{ - 1}}). The value of Λeq∞\mathrm{\Lambda }_{eq}^\infty of NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH} is
View answer and explanation
Correct option: B
271.44
Option B (271.44) is correct because it is calculated using Kohlrausch's law of independent migration of ions.
370 Easy
370 of 396
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 (Anode) is correct because zinc has a higher oxidation potential than hydrogen, so it will be oxidized at the anode.
371 Easy
371 of 396
Application +4 | -1
Conductance &lsquo;CC&rsquo; (inSS) in directly proportional to the area of the electrode and concentration and inversely proportional to length of separation of electrode, the unit of constant of proportionality is:
View answer and explanation
Correct option: B
Sm2mol−1\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{mo}{\mathrm{l}^{ - 1}}
The correct answer is B (Sm2mol−1\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{mo}{\mathrm{l}^{ - 1}}) because it represents the units of molar conductivity, which is the constant of proportionality in the given relationship.
372 Hard
372 of 396
Application +4 | -1
When electric current is passed through acidified water for 1930 s, 1120mL of H2{\mathrm{H}_2} gas is collected (at STP) at the cathode. What is the current passed in amperes?
View answer and explanation
Correct option: C
5
Option C (5) is correct because the current passed is calculated using Faraday's laws of electrolysis, relating the amount of hydrogen gas collected to the charge passed through the solution.
373 Medium
373 of 396
Analyzing +4 | -1
Consider the following reaction at 1100∘C1100^{\circ} \mathrm{C}.
(1) 2C+O2⟶2CO ΔG∘=−460 kJ mol−12 \mathrm{C}+\mathrm{O}_2 \longrightarrow 2 \mathrm{CO} \,\Delta \mathrm{G}^{\circ}=-460 \mathrm{~kJ} \mathrm{~mol}^{-1}
(2) 2Zn+O2⟶2ZnO ΔG∘=−360 kJ mol−12 \mathrm{Zn}+\mathrm{O}_2 \longrightarrow 2 \mathrm{ZnO} \,\Delta \mathrm{G}^{\circ}=-360 \mathrm{~kJ} \mathrm{~mol}^{-1}
based on these, select correct alternate
View answer and explanation
Correct option: B
Zinc oxide can be reduced by carbon
Option B (Zinc oxide can be reduced by carbon) is correct because the reaction ZnO + C -> Zn + CO has a negative change in Gibbs free energy, making it spontaneous.
374 Easy
374 of 396
Remembering +4 | -1
Anode in the Leclanche cell is
View answer and explanation
Correct option: B
Zinc container
Option B (Zinc container) is correct because in a Leclanche cell (dry cell), the zinc container acts as the anode (negative electrode) where oxidation occurs.
375 Medium
375 of 396
Application +4 | -1
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A0.10 \mathrm{~A} for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced at STP is ____cm3\mathrm{cm}^3. (Nearest integer)
[Given : Faraday constant F=96500Cmol−1\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1} at STP, molar volume of an ideal gas is 22.7 L mol−122.7 \mathrm{~L} \mathrm{~mol}^{-1} ]
View answer and explanation
Correct option: A
127  cm3127\,\,\text{cm}^3
Option A (127 cm³) is correct because the total volume of gases produced is calculated using Faraday's laws of electrolysis and the molar volume of an ideal gas at STP.
376 Medium
376 of 396
Understanding +4 | -1
Which of the following aqueous solution will have the maximum specific conductance?
View answer and explanation
Correct option: A
0.1 HCl
Option A is correct because HCl is a strong electrolyte that dissociates completely in water, resulting in a higher concentration of ions and thus maximum specific conductance compared to the other options.
377 Easy
377 of 396
Understanding +4 | -1
The factors which influence the conductance of solution.
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because solute-solute interactions, solute-solvent interactions, and temperature all influence the conductance of a solution.
378 Medium
378 of 396
Understanding +4 | -1
Two half cells have reduction potentials −0.76 V-0.76 \mathrm{~V} and 0.13 V0.13 \mathrm{~V} respectively. A galvanic cell is made from these two half cells. Which of the following statements is correct?
View answer and explanation
Correct option: B
Electrode of half-cell potential −0.76 V-0.76 \mathrm{~V} serve as anode
Option B (Electrode of half-cell potential −0.76 V-0.76 \mathrm{~V} serve as anode) is correct because the electrode with the more negative reduction potential acts as the anode in a galvanic cell.
379 Easy
379 of 396
Understanding +4 | -1
Which set of relationships could apply to the same electrochemical cell?
View answer and explanation
Correct option: D
ΔGo<0 ; ∈o>0\Delta {G^\mathrm{o}} < 0\,;\,{ \in ^\mathrm{o}} > 0
Option D is correct because a spontaneous electrochemical cell has a negative Gibbs free energy change and a positive standard cell potential.
380 Easy
380 of 396
Remembering +4 | -1
Cell constant has the unit:
View answer and explanation
Correct option: B
cm−1\mathrm{c}{\mathrm{m}^{ - 1}}
Option B (cm−1\mathrm{c}{\mathrm{m}^{ - 1}}) is correct because cell constant is defined as the ratio of length to area, resulting in units of inverse length.
381 Medium
381 of 396
Application +4 | -1
Conductivity (Seimen&#39;s S)S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then constant of proportionality is expressed in:
View answer and explanation
Correct option: C
S m2 mol−1{S~m}^2 {~mol}^{-1}
The correct answer is C (S m2 mol−1{S~m}^2 {~mol}^{-1}). It is derived from the proportionality relation between conductivity, area, concentration, and length.
382 Easy
382 of 396
Remembering +4 | -1
SI unit of conductivity is :
View answer and explanation
Correct option: B
ohm−1cm−1orSm−1\mathrm{oh}{\mathrm{m}^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{or {\kern 1pt} {\kern 1pt}S}}{\kern 1pt} {\kern 1pt}{\mathrm{m}^{ - 1}}
Option B (ohm−1cm−1orSm−1\mathrm{oh}{\mathrm{m}^{ - 1}}{\kern 1pt} {\kern 1pt}\mathrm{c}{\mathrm{m}^{ - 1}}{\kern 1pt} {\kern 1pt}{\rm{or {\kern 1pt} {\kern 1pt}S}}{\kern 1pt} {\kern 1pt}{\mathrm{m}^{ - 1}}) is the correct SI unit of conductivity, which is Siemens per meter (S/m) or ohm−1^{-1}m−1^{-1}.
383 Medium
383 of 396
Remembering +4 | -1
The value of electronic charge is equal to :
View answer and explanation
Correct option: A
FaradayAv.number\frac{{\mathrm{Faraday}}}{{\mathrm{Av}\mathrm{.}{\kern 1pt} {\kern 1pt} \mathrm{number}}}
Option A (FaradayAv.number\frac{{\mathrm{Faraday}}}{{\mathrm{Av}\mathrm{.}{\kern 1pt} {\kern 1pt} \mathrm{number}}}) is correct because electronic charge is equal to Faraday's constant divided by Avogadro's number.
384 Medium
384 of 396
Analyzing +4 | -1
Which does not get oxidised by bromine water?
View answer and explanation
Correct option: C
Mn2+toMnO4−\mathrm{M}{\mathrm{n}^{2 + }}{\rm{to {\kern 1pt} {\kern 1pt}MnO}}_4^ -
Option C (Mn2+toMnO4−\mathrm{M}{\mathrm{n}^{2 + }}{\rm{to {\kern 1pt} {\kern 1pt}MnO}}_4^ -) is correct because MnO4−\mathrm{MnO}_4^ - is a strong oxidizing agent itself, making further oxidation by bromine water unfavorable.
385 Medium
385 of 396
Application +4 | -1
Equivalent conductance of 0.1 M HA0.1 \mathrm{~M} \mathrm{~HA} (weak acid) solution is 10 S cm2 equivalent−110 \mathrm{~S~cm}^2 \mathrm{~equivalent}^{-1} and that at infinite dilution is 200 S cm2 equivalent−1.200 \mathrm{~S~cm}^2 \mathrm{~equivalent}^{-1}. Hence pH\mathrm{pH} of HA\mathrm{HA} solution is
View answer and explanation
Correct option: C
2.32.3
Option C (2.3) is correct because the pH is calculated using the degree of dissociation and the concentration of H+ ions.
386 Easy
386 of 396
Analyzing +4 | -1
A substance that will reduce Ag+\mathrm{A}{\mathrm{g}^ + } to Ag but will not reduce Ni2+\mathrm{N}{\mathrm{i}^{2 + }} to Ni is :
View answer and explanation
Correct option: B
Pb
Option B is correct because Pb is above Ag but below Ni in the electrochemical series, meaning it can reduce Ag+\mathrm{A}{\mathrm{g}^ + } but not Ni2+\mathrm{N}{\mathrm{i}^{2 + }}.
387 Medium
387 of 396
Application +4 | -1
Consider the following Eo{E^\mathrm{o}} Values.
EFe3+/Fe2+o=0.77 V ; ESn2+/Sno=−0.14 VE_{F{e^{3 + }}/F{e^{2 + }}}^\mathrm{o} = 0.77\,V\,;\,E_{S{n^{2 + }}/Sn}^\mathrm{o} = - 0.14\,V
Under standard conditions, the potential for the reaction:
Sn(s)+2Fe3+(aq)→2Fe2+(aq)+Sn2+(aq)Sn\left( s \right) + 2F{e^{3 + }}\left( {aq} \right) \to 2F{e^{2 + }}\left( {aq} \right) + S{n^{2 + }}\left( {aq} \right)
View answer and explanation
Correct option: C
0.91 V
Option C (0.91 V) is correct because the cell potential is calculated by adding the reduction potential of Fe3+/Fe2+ and the oxidation potential of Sn/Sn2+.
388 Medium
388 of 396
Application +4 | -1
Calculate KaK_{\mathrm{a}} of acetic acid if its 0.05 M0.05 \mathrm{~M} solution has molar conductivity of 7.814×10−4 Ω−1 m2 mol−17.814 \times 10^{-4} \, \Omega^{-1} \mathrm{~m}^2 \mathrm{~mol}^{-1} at 25∘C.25^{\circ} \mathrm{C}. Given: Λm∘\Lambda_{\mathrm{m}}^{\circ} for CH3COOH=3.907×10−2 Ω−1 m2 mol−1.\mathrm{CH}_3 \mathrm{COOH}=3.907 \times 10^{-2} \, \Omega^{-1} \mathrm{~m}^2 \mathrm{~mol}^{-1}.
View answer and explanation
Correct option: A
2×10−52 \times 10^{-5}
Option A (2×10−52 \times 10^{-5}) is correct because it is calculated using the formula for Ka based on the degree of dissociation and concentration.
389 Medium
389 of 396
Remembering +4 | -1
Which of the following has negative value?
View answer and explanation
Correct option: C
ECu/Cu2+∘\mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{\mathrm{\circ}}
Option C (ECu/Cu2+∘\mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{2+}}^{\mathrm{\circ}}) is correct because the standard reduction potential for Cu/Cu2+ is negative (-0.34 V).
390 Medium
390 of 396
Application +4 | -1
When 1 F of electricity is passed through acidulated water, O2{\mathrm{O}_2} evolved is
View answer and explanation
Correct option: B
5.6dm35.6{\kern 1pt} {\kern 1pt} \mathrm{d}{\kern 1pt} {\mathrm{m}^3}
Option B (5.6 dm³) is correct because 1 Faraday of electricity liberates 1/4 mole of O₂ during the electrolysis of water, which corresponds to 5.6 dm³ at STP.
Revision notes and key takeaways

Review ELECTRO CHEMISTRY in your textbook or class notes, then use this set to check your understanding.

Showing 361-390 of 396 questions

Your practice

0 / 396 answered · 396 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.