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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 211-230 of 230 questions
211 Medium
211 of 230
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.
212 Medium
212 of 230
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.
213 Medium
213 of 230
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.
214 Medium
214 of 230
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.
215 Medium
215 of 230
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.
216 Medium
216 of 230
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.
217 Medium
217 of 230
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.
218 Medium
218 of 230
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.
219 Medium
219 of 230
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.
220 Medium
220 of 230
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.
221 Medium
221 of 230
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.
222 Medium
222 of 230
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.
223 Medium
223 of 230
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.
224 Medium
224 of 230
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+.
225 Medium
225 of 230
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.
226 Medium
226 of 230
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).
227 Medium
227 of 230
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.
228 Medium
228 of 230
Understanding +4 | -1
The product, which is not obtained during the electrolysis of brine solution is
View answer and explanation
Correct option: B
HCl\mathrm{HCl}
Option B (HClHCl) is correct because during the electrolysis of brine solution, hydrogen gas, chlorine gas, and sodium hydroxide are produced, but hydrochloric acid is not.
229 Medium
229 of 230
Understanding +4 | -1
Which cell will measure standard electrode potential of copper electrode?
View answer and explanation
Correct option: C
Pt(s)∣H2(g,1\mathrm{Pt}(s) \mid \mathrm{H}_2(g, 1 bar )∣H+(aq,1M)∣∣Cu2+(aq,1M)∣Cu)\left|\mathrm{H}^{+}(a q, 1 \mathrm{M})\right|\left|\mathrm{Cu}^{2+}(a q, 1 \mathrm{M})\right| \mathrm{Cu}
Option C is correct because a standard electrode potential requires all species to be at standard conditions (1 bar for gases, 1 M for solutions).
230 Medium
230 of 230
Application +4 | -1
Resistance of a 0.1 M KC10.1 \mathrm{~M} \mathrm{~KC1} solution in a conductance cell is 300 ohm300 \mathrm{~ohm} and specific conductance of 0.1 M KC10.1 \mathrm{~M} \mathrm{~KC1} is 1.33×10−2 ohm−1.33 \times 10^{-2} \mathrm{~ohm}^{-} 1 cm−1{ }^1 \mathrm{~cm}^{-1}. The resistance of 0.1M NaCl0.1 \mathrm{M} \mathrm{~NaCl} solution in the same cell is 400 ohm400 \mathrm{~ohm}. The equivalent conductance of the 0.1 M NaCl0.1 \mathrm{~M} \mathrm{~NaCl} solution (in ohm −1 cm2/{ }^{-1} \mathrm{~cm}^2 / gmeq.) is
View answer and explanation
Correct option: A
100
Option A (100) is correct because it calculates the equivalent conductance using the given resistance and specific conductance.
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