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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 271-300 of 396 questions

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271 Medium
271 of 396
Application +4 | -1
In following cell reaction

Mg(s)+2Ag+(0.001M)⟶Mg2+(0.20M)+2Ag(s){Mg}({s})+2 {Ag}^{+}(0.001 {M}) \longrightarrow {Mg}^{2+}(0.20 {M})+2 {Ag}({s})

Calculate Ecell E_{\text {cell }} for the reaction [E0=3.17 V]\left[E^0=3.17 {~V}\right]
View answer and explanation
Correct option: B
3.04 V3.04 {~V}
Option B (3.04 V3.04 {~V}) is correct because it is the result of applying the Nernst equation with the given concentrations and standard potential.
272 Hard
272 of 396
Application +4 | -1
Sn4++2e−⟶Sn2+E∘=0.13 V\mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} E^{\circ}=0.13 \mathrm{~V}

Br2+2e−⟶2Br−E∘=1.08 V\mathrm{Br}_2+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Br}^{-} \quad E^{\circ}=1.08 \mathrm{~V}

Calculate KeqK_{\mathrm{eq}} for the cell formed by two electrodes
View answer and explanation
Correct option: B
103210^{32}
Option B (103210^{32}) is correct because the equilibrium constant is calculated from the standard cell potential using the Nernst equation.
273 Medium
273 of 396
Analyzing +4 | -1
Four successive members of the first row transition elements listed below with atomic numbers. Which one of them is expected to have the highest EM3+/M2+∘E_{{M^{3 + }}/{M^{2 + }}}^ \circ value is?
View answer and explanation
Correct option: D
Co (Z=27)Co\,(Z = 27)
Option D is correct because Co (Z = 27) has the highest EM3+/M2+∘E_{M^{3+}/M^{2+}}^\circ value due to the stability of Co²⁺.
274 Medium
274 of 396
Application +4 | -1
If the molar conductivity of 10−3 M HA10^{-3} \mathrm{~M} \mathrm{~HA} is 39 S cm2 mol−1,39 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, then its dissociation constant will be

[Given Λm0(HA)=390 S cm2 mol−1]\Lambda_{\mathrm{m}}^0(\mathrm{HA})=390 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}]
View answer and explanation
Correct option: D
10−5M10^{-5} \mathrm{M}
The correct answer is D (10-5 M) because it is calculated using the formula Ka = Cα^2, where α is the degree of dissociation and C is the concentration.
275 Medium
275 of 396
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.
276 Easy
276 of 396
Analyzing +4 | -1
E∘E^\circ values of some redox couples are given below on the basis of these values. Choose the correct option
E∘E^\circ values:
Br2/Br−=+1.90Ag+/Ag(s)=+0.80Cu2+/Cu(s)=+0.34I2(s)/I−=+0.54\begin{array}{l} B{r_2}\mathrm{/}B{r^ - } = + 1.90\\ A{g^ + }\mathrm{/}Ag(s) = + 0.80\\ C{u^{2 + }}\mathrm{/}Cu(s) = + 0.34\\ {I_2}(s)\mathrm{/}{I^ - } = + 0.54 \end{array}
View answer and explanation
Correct option: D
CuCu will reduce Br2B{r_2}
Cu will reduce Br2 because Cu has a lower reduction potential (+0.34) than Br2 (+1.90), making it more likely to be oxidized and thus reduce Br2.
277 Medium
277 of 396
Application +4 | -1
During the electrolysis of a solution of AgNO3,9650C\mathrm{AgN}{\mathrm{O}_3},9650\mathrm{C} of charge is passed through the electroplating bath.
The mass of silver deposited at the cathode will be
View answer and explanation
Correct option: B
10.8 g
Option B (10.8 g) is correct because it calculates the mass of silver deposited using Faraday's laws of electrolysis.
278 Medium
278 of 396
Application +4 | -1
The equivalent conductance of an aqueous solution of 1.0283×10−3 g1.0283 \times 10^{-3} \mathrm{~g} equivalent acetic acid per litre is 48.15π−1 cm248.15 \pi^{-1} \mathrm{~cm}^2 equiv −1^{-1} at 25∘C25^{\circ} \mathrm{C}. At infinite dilution value is 390.7π−1 cm2390.7 \pi^{-1} \mathrm{~cm}^2 equiv −1{ }^{-1}. Calculate the degree of ionisation and ionisation constant of acetic acid.
View answer and explanation
Correct option: A
0.1232,1.78×10−50.1232,1.78 \times 10^{-5}
Option A (0.1232, 1.78×10−51.78 \times 10^{-5}) is correct because it accurately calculates the degree of ionization and ionization constant using the given equivalent conductance values.
279 Easy
279 of 396
Understanding +4 | -1
The oxidation potential of Mg and Al are + 2.37 and + 1.66 volt respectively. The Mg in chemical reactions :
View answer and explanation
Correct option: B
Will replace Al
Option B (Will replace Al) is correct because Mg has a higher oxidation potential than Al, meaning it is more easily oxidized and can displace Al from its compounds.
280 Medium
280 of 396
Remembering +4 | -1
Which of the following electrolytes is used in lead storage battery?
View answer and explanation
Correct option: C
38 % solution of H2SO4{H_2}S{O_4}
Option C (38 % solution of H2SO4{H_2}S{O_4}) is correct because a 38% solution of sulfuric acid is used as the electrolyte in lead storage batteries.
281 Medium
281 of 396
Application +4 | -1
On the basis of the information available from the 43Al+O2⟶23Al2O3,ΔG=−827 kJ mol−1\frac{4}{3} \mathrm{Al}+\mathrm{O}_2 \longrightarrow \frac{2}{3} \mathrm{Al}_2 \mathrm{O}_3, \Delta G=-827 \mathrm{~kJ} \mathrm{~mol}^{-1} of O2\mathrm{O}_2, the minimum emf required to carry out an electrolysis of Al2O3\mathrm{Al}_2 \mathrm{O}_3 is : (F=96500 C mol−1\left(\mathrm{F}=96500 \mathrm{~C} \mathrm{~mol}^{-1}\right. )
View answer and explanation
Correct option: B
2.14 V2.14 \mathrm{~V}
Option B (2.14 V2.14 \mathrm{~V}) is correct because the minimum EMF required is calculated from the Gibbs free energy change and the number of electrons transferred in the reaction.
282 Medium
282 of 396
Application +4 | -1
In a fuel cell hydrogen and oxygen react to produce electricity. In the process hydrogen gas is oxidised at the anode and oxygen at the cathode. If 67.2 L67.2 \mathrm{~L} of H2\mathrm{H}_2 at STP react in 15 minutes, what is the average current produced ? If the entire current is used for electrodeposition of copper from copper (II) solution, how many grams of copper will be deposited ?

Anode reaction : H2+2OH−⟶2H2O+2e−\mathrm{H}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{e}^{-}

Cathode reaction: O2+2H2+2e−⟶4OH−\mathrm{O}_2+2 \mathrm{H}_2+2 \mathrm{e}^{-} \longrightarrow 4 \mathrm{OH}^{-}
View answer and explanation
Correct option: A
190
Option A (190) is correct because it is the approximate mass of copper deposited, calculated based on the stoichiometry of the reaction and Faraday's laws of electrolysis.
283 Medium
283 of 396
Analyzing +4 | -1
Select the correct statement(s) for the given reaction.

Cu(s)+2Ag+(aq)⟶Cu2+(aq)+2Ag(s)\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the provided statements regarding the electrochemical cell representation, cell potential, and electrode reactions are all incorrect for the given reaction.
284 Medium
284 of 396
Analyzing +4 | -1
The electrical conductivity of the flowing aqueous solutions is highest for:
View answer and explanation
Correct option: C
0.1 M CHF2COOH0.1 \mathrm{~M} \mathrm{~CHF}_2 \mathrm{COOH}
Option C (0.1 M CHF2COOH0.1 \mathrm{~M} \mathrm{~CHF}_2 \mathrm{COOH}) is correct because difluoroacetic acid is the strongest acid among the options due to the electron-withdrawing effect of the two fluorine atoms, leading to the highest electrical conductivity.
285 Medium
285 of 396
Understanding +4 | -1
KCl(aq)\mathrm{KCl}\left( {aq} \right) cannot be used as a salt bridge for the cell Cu(s)∣CuSO4(aq)∣∣AgNO3(aq)∣Ag(s)\mathrm{Cu}\left( s \right)\left| {\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right)} \right|\left| {\mathrm{AgN}{\mathrm{O}_3}\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right) because :
View answer and explanation
Correct option: C
AgCl\mathrm{AgCl} is precipitated
Option C (AgCl\mathrm{AgCl} is precipitated) is correct because AgCl\mathrm{AgCl} precipitates when KCl\mathrm{KCl} reacts with AgNO3\mathrm{AgN}{\mathrm{O}_3} in the cell.
286 Easy
286 of 396
Application +4 | -1
A steady current of 30.0 A for 70.2 min corresponds to a
passage of
View answer and explanation
Correct option: D
1.31 Faradays
Option D (1.31 Faradays) is correct because it is calculated by converting the current and time to charge in Coulombs and then to Faradays.
287 Medium
287 of 396
Application +4 | -1
A galvanic cell : Cu∣Cu2+∣∣Ag+∣Ag\mathrm{Cu}\left|\mathrm{Cu}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag} initially contains 1 M Ag+1 \mathrm{~M}\mathrm{~Ag}^{+}and 1 M Cu2+1 \mathrm{~M} \mathrm{~Cu}^{2+} ions. The cell potential after the passage of 9.65 A9.65 \mathrm{~A} of current for 1 h1 \mathrm{~h} will be ___ V.
View answer and explanation
Correct option: A
0.010.01
Option A (0.01) is correct because the cell potential is calculated using the Nernst equation after determining the change in concentrations due to the passage of current.
288 Medium
288 of 396
Understanding +4 | -1
Assertion (A): Λm\Lambda_m for weak electrolytes shows a sharp increase, when the electrolytic solution is diluted.

Reason (R): For weak electrolytes degree of dissociation increases with dilution of solution.
View answer and explanation
Correct option: A
Assertion and Reason both are correct and Reason is the correct explanation of Assertion.
Option A is correct because as a weak electrolyte is diluted, its degree of dissociation increases, leading to a sharp increase in molar conductivity.
289 Hard
289 of 396
Analyzing +4 | -1
Consider the following reaction :

2Ag+C6H12O6+H2O→2Ag(s)+C6H12O7+2H+2\mathrm{A}{\mathrm{g}^ + }{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + {\mathrm{H}_2}\mathrm{O} \to 2\mathrm{Ag}\left( s \right) + {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_7} + 2{\mathrm{H}^ + }

When ammonia is added to the solution, pH is raised to 11. Which half-cell reaction is affected by pH and by how much?

Ag++e−→Ag;Ered∘=0.8V\mathrm{A}{\mathrm{g}^ + } + {e^ - } \to \mathrm{Ag};E_{\mathrm{red}}^ \circ = 0.8\mathrm{V}

C6H12O6+H2O→C6H12O7+2H++2e−;Ered∘=−0.05V{\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_7} + 2{\mathrm{H}^ + } + 2{e^ - };E_{\mathrm{red}}^ \circ = - 0.05\mathrm{V}

Ag(NH3)2++e−→Ag(s)+NH3;Ered∘=0.337V\mathrm{Ag}\left( {\mathrm{N}{\mathrm{H}_3}} \right)_2^ + + {e^ - } \to \mathrm{Ag}\left( s \right) + \mathrm{N}{\mathrm{H}_3};E_{\mathrm{red}}^ \circ = 0.337\mathrm{V}
View answer and explanation
Correct option: A
Eoxid.∘E_{\mathrm{oxid}\mathrm{.}}^ \circ will increase by a factor of 0.65 from Eoxid.∘E_{\mathrm{oxid}\mathrm{.}}^ \circ
The correct answer is A because the oxidation potential of the glucose/gluconic acid half-cell is pH-dependent, and increasing the pH will increase the oxidation potential.
290 Medium
290 of 396
Application +4 | -1
2MnO4−+aH++bH2O2⟶2Mn+2+cH2O+dO22 \mathrm{MnO}_4^{-}+\mathrm{aH}^{+}+\mathrm{bH}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{Mn}^{+2}+\mathrm{cH}_2 \mathrm{O}+\mathrm{dO}_2. What are the values of a,b,c\mathrm{a}, \mathrm{b}, \mathrm{c} and d\mathrm{d}?
View answer and explanation
Correct option: B
6,5,8,56,5,8,5
Option B (6,5,8,56,5,8,5) is correct because it represents the coefficients that balance the given redox reaction.
291 Medium
291 of 396
Understanding +4 | -1
The correct order of molar conductance at infinite dilution of NaCl,KCl,RbCl{NaCl}, {KCl}, {RbCl} is
View answer and explanation
Correct option: B
RbCl>KCl>NaCl{RbCl}>{KCl}>{NaCl}
Option B is correct because ionic conductance increases with ionic mobility, which is inversely proportional to hydrated size.
292 Medium
292 of 396
Understanding +4 | -1
KCl(aq)\mathrm{KCl}\left( {aq} \right) cannot be used as a salt bridge for the cell Cu(s)∣CuSO4(aq)∣∣AgNO3(aq)∣Ag(s)\mathrm{Cu}\left( s \right)\left| {\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right)} \right|\left| {\mathrm{AgN}{\mathrm{O}_3}\left( {aq} \right)} \right|\mathrm{Ag}\left( s \right) because :
View answer and explanation
Correct option: C
AgCl\mathrm{AgCl} is precipitated
Option C (AgCl is precipitated) is correct because Ag+ ions from AgNO3 react with Cl- ions from KCl to form insoluble AgCl, disrupting the salt bridge function.
293 Hard
293 of 396
Analyzing +4 | -1
Given, E∘=−0.27 V\mathrm{E}^{\circ}=-0.27 \mathrm{~V} for the Cl−∣PbCl2∣Pb\mathrm{Cl}^{-}\left|\mathrm{PbCl}_2\right| \mathrm{Pb} couple and 0.12 V0.12 \mathrm{~V} for the Pb2+∣Pb\mathrm{Pb}^{2+} \mid \mathrm{Pb} couple, determine Ksp\mathrm{K}_{\mathrm{sp}} for PbCl2\mathrm{PbCl}_2 at 25∘C25^{\circ} \mathrm{C} ? (Take 2.303RTF=0.06\frac{2.303 R T}{F}=0.06 )
View answer and explanation
Correct option: A
10−510^{-5}
The correct answer is A (10−510^{-5}) because it is calculated using the Nernst equation and the given standard electrode potentials.
294 Medium
294 of 396
Application +4 | -1
Following is the concentration cell in which electrode is
reversible with respect to anion
Pt(Cl2)1  bar∣Cl−(C1)∣∣Cl−(C2)∣Pt(Cl2)1  bar\mathop {Pt\left( {C{l_2}} \right)}\limits_{1\,\,{\text{bar}}} \left| {C{l^ - }\left( {{C_1}} \right)} \right|\left| {C{l^ - }\left( {{C_2}} \right)} \right|\mathop {Pt\left( {C{l_2}} \right)}\limits_{1\,\,{\text{bar}}}
The cell reaction is spontaneous, if
View answer and explanation
Correct option: A
C1>C2{C_1} > {C_2}
Option A (C1>C2{C_1} > {C_2}) is correct because the cell reaction is spontaneous when the concentration of the anion at the left electrode is greater than the concentration at the right electrode.
295 Easy
295 of 396
Understanding +4 | -1
Assertion (A): KMnO4{KMnO}_4 act as only oxidizing agent.

Reason (R): In KMnO4,Mn{KMnO}_4, {Mn} is in its highest oxidation state +7.
View answer and explanation
Correct option: A
Both (A) and (R) true and (R) is the correct explanation of (A).
Option A (Both (A) and (R) true and (R) is the correct explanation of (A)) is correct because KMnO4 acts as an oxidizing agent due to Mn being in its highest oxidation state (+7), which allows it to accept electrons and get reduced.
296 Medium
296 of 396
Analyzing +4 | -1
Assertion: The cell constant of a cell depends upon the nature of the material of the electrodes.

Reason: The observed conductance depends upon the nature of the electrolyte and the concentration of the solution.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because the cell constant depends on the geometry of the cell, not the electrode material, and conductance depends on electrolyte nature and concentration.
297 Easy
297 of 396
Application +4 | -1
The position of some metals in the electrochemical series in decreasing electropositive character is given as Mg>Al>Zn>Cu>Ag\mathrm{Mg} > Al > Zn > Cu > Ag. What will happen, if a copper spoon is used to stir a solution of aluminium nitrate?
View answer and explanation
Correct option: D
There is no reaction
Option D (There is no reaction) is correct because copper is below aluminum in the electrochemical series, meaning it cannot displace aluminum from its solution.
298 Easy
298 of 396
Remembering +4 | -1
The laws of electrolysis were proposed by
View answer and explanation
Correct option: B
Faraday
Option B (Faraday) is correct because Michael Faraday formulated the laws of electrolysis in the 19th century.
299 Medium
299 of 396
Analyzing +4 | -1
Red hot carbon will remove oxygen from the oxides XO, and YO but not from ZO. Y will remove oxygen from XO. Use this evidence to reduce the order of activity of the three metals X, Y and Z putting the most reactive first:
View answer and explanation
Correct option: B
Z, Y, X
Option B (Z, Y, X) is correct because Z is the most reactive as carbon cannot remove oxygen from ZO, Y is more reactive than X as Y removes oxygen from XO.
300 Easy
300 of 396
Remembering +4 | -1
Rust is
View answer and explanation
Correct option: C
Fe2O3+Fe(OH)2\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3} + \mathrm{Fe}{\left( {\mathrm{OH}} \right)_2}
Option D (Fe2O3andFe(OH)3\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}{\kern 1pt} {\kern 1pt} \mathrm{and }{\kern 1pt} {\kern 1pt} \mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}) is correct because rust is primarily composed of iron(III) oxide (Fe2O3\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}) and iron(III) hydroxide (Fe(OH)3\mathrm{Fe}{\left( {\mathrm{OH}} \right)_3}).
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