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JEE • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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211 Hard
211 of 396
Analyzing +4 | -1
Assertion: Copper sulphate solution cannot be stored in iron-vessel.

Reason: Copper atoms are oxidized by iron ions.
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 copper ions are reduced by iron atoms, not the other way around, due to iron being more reactive than copper.
212 Easy
212 of 396
Application +4 | -1
A solution of CuSO4\mathrm{CuSO}_4 is electrolyzed for 7 minutes with a current of 0.6 A0.6 \mathrm{~A}. The amount of electricity passed is equal to:
View answer and explanation
Correct option: D
2.6×10−3 F2.6 \times 10^{-3} \mathrm{~F}
Option D (2.6×10−3 F2.6 \times 10^{-3} \mathrm{~F}) is correct because it accurately calculates the amount of electricity passed in Faradays using the given current and time.
213 Easy
213 of 396
Application +4 | -1
The equivalent conductivity of a solution containing 2.54 g of CuSO4perLis91.0Ω−1cm2eq−1\mathrm{CuS}{\mathrm{O}_4}{\kern 1pt} {\kern 1pt} \mathrm{per L is}{\kern 1pt} {\kern 1pt} {\kern 1pt} 91.0{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}\mathrm{e}{\mathrm{q}^{ - 1}}. Its conductivity would be
View answer and explanation
Correct option: A
2.9×10−3Ω−1cm22.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}
Option A (2.9×10−3Ω−1cm22.9 \times {10^{ - 3}}{\mathrm{\Omega }^{ - 1}}\mathrm{c}{\mathrm{m}^2}) is correct because conductivity is calculated from equivalent conductivity and concentration.
214 Medium
214 of 396
Analyzing +4 | -1
Given, ECl2/Cl−o=1.36 V, ECr3+/Cro=−0.74 VE_{C{l_2}/C{l^ - }}^\mathrm{o} = 1.36\,V,\,E_{C{r^{3 + }}/Cr}^\mathrm{o} = - 0.74\,V
ECr2O72−/Cr3+o=1.33 V, EMnO4−/Mn2+o=1.51 VE_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^\mathrm{o} = 1.33\,V,\,E_{MnO_4^ - /M{n^{2 + }}}^\mathrm{o} = 1.51\,V
Among the following, the strongest reducing agent is
View answer and explanation
Correct option: A
CrCr
Option A (CrCr) is correct because the strongest reducing agent has the lowest (most negative) reduction potential. Among the given options, Cr has the lowest reduction potential (-0.74 V).
215 Medium
215 of 396
Understanding +4 | -1
Select the correct statement(s) for the weak electrolyte.
View answer and explanation
Correct option: D
At infinite dilution, electrolyte dissociate completely but at such low concentration the conductivity of solution is low and cannot be measured accurately.
Option D is correct because weak electrolytes dissociate completely only at infinite dilution, but the conductivity is too low to measure accurately.
216 Medium
216 of 396
Application +4 | -1
An object whose surface area is 80 cm280 \mathrm{~cm}^2 is to be plated with an even layer of gold 8.0×10−4 cm8.0 \times 10^{-4} \mathrm{~cm} thick. The density of gold is 19.7 g/cm3.19.7 \mathrm{~g} / \mathrm{cm}^3. The object is placed in a solution of Au(NO3)3\mathrm{Au}\left(\mathrm{NO}_3\right)_3 and a current of 2.4 A\text{2.4 A} is applied. The time required for the electroplating to be completed, assuming that the layer of gold builds up evenly, is (Au=197)(\mathrm{Au}=197)
View answer and explanation
Correct option: A
772 s772 \mathrm{~s}
Option A (772 s772 \mathrm{~s}) is correct because it is calculated using Faraday's laws of electrolysis and the given parameters.
217 Hard
217 of 396
Remembering +4 | -1
The sum of the two transport number of ions for an electrolyte is always equal to :
View answer and explanation
Correct option: A
1
Option A (1) is correct because the sum of the transport numbers of all ions in an electrolyte solution must equal unity to maintain charge neutrality.
218 Medium
218 of 396
Analyzing +4 | -1
E∘{E^ \circ } values of Mg(2+)M{g^{(2 + )}} / Mg is -2.37 V, of Zn2+\mathrm{Z}{\mathrm{n}^{2 + }} / Zn is -0.76 V and Fe2+\mathrm{F}{\mathrm{e}^{2 + }} / Fe is -0.44 V.
Which of the statements is correct?
View answer and explanation
Correct option: A
Zn will reduce Fe2+\mathrm{F}{\mathrm{e}^{2 + }}
Option A (Zn will reduce Fe²⁺) is correct because Zn has a more negative reduction potential than Fe, meaning it is a stronger reducing agent and can reduce Fe²⁺ to Fe.
219 Medium
219 of 396
Understanding +4 | -1
Hydrogen gas is not liberated when the following metal is added to dil. HCl
View answer and explanation
Correct option: A
Ag
Option A (Ag) is correct because silver is below hydrogen in the electrochemical series and therefore cannot displace hydrogen from dilute HCl.
220 Hard
220 of 396
Analyzing +4 | -1
Following cell has EMF 0.7995 V0.7995 \mathrm{~V}.

Pt∣H2(1 atm)∣HNO3(1M)∥AgNO3(1M)∣Ag\mathrm{Pt}\left|\mathrm{H}_2(1 \mathrm{~atm})\right| \mathrm{HNO}_3(1 \mathrm{M}) \| \mathrm{AgNO}_3(1 \mathrm{M}) \mid \mathrm{Ag}

If we add enough KCl\mathrm{KCl} to the Ag\mathrm{Ag} cell so that the
final Cl−\mathrm{Cl}^{-}is 1M1 \mathrm{M}. Now the measured emf of the cell is 0.222 V0.222 \mathrm{~V}. The Ksp\mathrm{K}_{s p} of AgCl\mathrm{AgCl} would be-
View answer and explanation
Correct option: A
1×10−9.81 \times 10^{-9.8}
Option A (1×10−9.81 \times 10^{-9.8}) is correct because it represents the KspK_{sp} calculated from the Nernst equation and the given EMF values.
221 Easy
221 of 396
Application +4 | -1
The quantity of charge required to obtain one mole of aluminium from Al2O3A{l_2}{O_3} is
View answer and explanation
Correct option: C
3 F
Option C (3 F) is correct because 3 moles of electrons are required to obtain one mole of aluminum from Al2O3.
222 Medium
222 of 396
Application +4 | -1
For given cell, Mg/Mg2+(0.01M)∥Ag+(0.01M)/Ag\mathrm{Mg} / \mathrm{Mg}^{2+}(0.01 \mathrm{M}) \| \mathrm{Ag}^{+}(0.01 \mathrm{M}) / \mathrm{Ag}, if Ecel ∘=3.17 V\mathrm{E}_{\text {cel }}^{\circ}=3.17 \mathrm{~V}, then Ecell \mathrm{E}_{\text {cell }} will be
View answer and explanation
Correct option: C
3.11 volt
Option C (3.11 volt) is correct because the cell potential is calculated using the Nernst equation, considering the concentrations of the ions.
223 Hard
223 of 396
Application +4 | -1
Two electrolytic cells one containing acidified ferrous chloride and another acidified ferric chloride are connected in series. The ratio of iron deposited at cathodes in the two cells when electricity is passed through the cells will be
View answer and explanation
Correct option: D
0.126388889
Option D (03:02) is correct because the ratio of iron deposited is determined by the ratio of their equivalent weights.
224 Medium
224 of 396
Application +4 | -1
A certain current liberates 0.504 g of hydrogen in 2 hr. How many gram of copper can be liberated by the same current flowing for the same time in CuSO4\mathrm{CuS}{\mathrm{O}_4} solution?
View answer and explanation
Correct option: B
16
Option B (16) is correct because the mass of copper liberated is calculated using the electrochemical equivalence and Faraday's laws.
225 Medium
225 of 396
Understanding +4 | -1
Which process occurs in the electrolysis of aqueous solution of nickel chloride at nickel anode?
View answer and explanation
Correct option: D
Ni→Ni2++2e\mathrm{Ni} \to \mathrm{N}{\mathrm{i}^{2 + }} + 2e
Option D (Ni→Ni2++2e\mathrm{Ni} \to \mathrm{N}{\mathrm{i}^{2 + }} + 2e) is correct because at the nickel anode, nickel metal is oxidized to nickel ions.
226 Hard
226 of 396
Application +4 | -1
I2(s)∣I−(0.1M){\mathrm{I}_2}\left( s \right)|{\mathrm{I}^ - }\left( {0.1\mathrm{M}} \right) half-cell is connected to a H+(aq)∣H2(1bar)∣{\mathrm{H}^ + }\left( {aq} \right)\left| {{\mathrm{H}_2}\left( {1\mathrm{bar}} \right)} \right| Pt half-cell and emf is found to be 0.7714 V. If EI2/I−∘=0.535VE_{{I^2}/{I^ - }}^ \circ = 0.535\mathrm{V}, find the pH\mathrm{pH} of H+/H2{\mathrm{H}^ + }/{\mathrm{H}_2} half-cell
View answer and explanation
Correct option: C
3
Option C (3) is correct because the pH is calculated using the Nernst equation and the given cell emf and standard electrode potential.
227 Medium
227 of 396
Analyzing +4 | -1
Statement I: on dilution of the solution the electrode potential increases

Statement II: Effect of dilution on electrode potential depends on electrode whether it is metal or non-metal
View answer and explanation
Correct option: D
Statement I is false but Statement II is correct
Option D is correct because Statement I is false, and Statement II is correct regarding the effect of dilution on electrode potential depending on the electrode type.
228 Easy
228 of 396
Application +4 | -1
Give the products available on the cathode and the anode respectively during the electrolysis of an aqueous solution of MgSO4\mathrm{MgS}{\mathrm{O}_4} between inert electrodes.
View answer and explanation
Correct option: A
H2(g)andO2{\mathrm{H}_2}\left( \mathrm{g} \right)\mathrm{and}{\kern 1pt} {\kern 1pt} {\mathrm{O}_2} (g)
Option A (H2(g)andO2{\mathrm{H}_2}\left( \mathrm{g} \right)\mathrm{and}{\kern 1pt} {\kern 1pt} {\mathrm{O}_2} (g)) is correct because during electrolysis of aqueous MgSO4, water is reduced at the cathode to produce H2 and oxidized at the anode to produce O2.
229 Hard
229 of 396
Analyzing +4 | -1
MnO4−+8H++5e−→Mn2++4H2O;E∘=1.51V\mathrm{MnO}_4^ - + 8{\mathrm{H}^ + } + 5{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 4{\mathrm{H}_2}\mathrm{O};{E^ \circ } = 1.51\mathrm{V}

MnO4+4H++2e−→Mn2++2H2O;E∘=1.23V\mathrm{Mn}{\mathrm{O}_4} + 4{\mathrm{H}^ + } + 2{e^ - } \to \mathrm{M}{\mathrm{n}^{2 + }} + 2{\mathrm{H}_2}\mathrm{O};{E^ \circ } = 1.23\mathrm{V}

EMnO4−∣MnO2∘E_{\mathrm{MnO}_4^ - |\mathrm{Mn}{\mathrm{O}_2}}^ \circ is
View answer and explanation
Correct option: A
1.70 V
Option A (1.70 V) is correct because it is calculated by combining the given reduction potentials using Gibbs free energy changes to find the reduction potential for the desired reaction.
230 Easy
230 of 396
Remembering +4 | -1
The amount of electricity required to liberate 1 g-equiv of Cu is
View answer and explanation
Correct option: B
1 F
Option B (1 F) is correct because 1 Faraday of electricity is required to liberate 1 gram equivalent of any substance.
231 Hard
231 of 396
Application +4 | -1
Consider the following cell

Pt∣H2(g)∣HCl(aq)∥Cl−(aq)∣AgCl(s)∣Ag(s)\mathrm{Pt}\left|\mathrm{H}_2(\mathrm{g})\right| \mathrm{HCl}(\mathrm{aq}) \| \mathrm{Cl}^{-}(\mathrm{aq})|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}(\mathrm{s})

E.M.F of the above cell is 0.355 V at 25∘C25^{\circ} \mathrm{C} and 0.305 V at 30∘C.30^{\circ} \mathrm{C}. The enthalpy change per mole of the reaction taking place at 25∘C25^{\circ} \mathrm{C} is

[Consider 1 mole of Ag is formed]
View answer and explanation
Correct option: A
-321.83 kJ
Option A (-321.83 kJ) is correct because it is the calculated enthalpy change per mole of the reaction based on the given EMF values at different temperatures.
232 Medium
232 of 396
Analyzing +4 | -1
Assertion (A): In weak electrolytes, the change in Λm\Lambda_{\mathrm{m}} with dilution is mainly due to increase in degree of dissociation
Reason (R): At a given concentration of weak electrolyte the degree of dissociation α=Λm0Λm\alpha=\frac{\Lambda_{\mathrm{m}}^0}{\Lambda_{\mathrm{m}}}
View answer and explanation
Correct option: D
A is correct but R is not correct
Option D is correct because the degree of dissociation is calculated as α=ΛmΛm0\alpha = \frac{{\Lambda_m}}{{\Lambda_m^0}} and not α=Λm0Λm\alpha=\frac{{\Lambda_{\mathrm{m}}^0}}{{\Lambda_{\mathrm{m}}}}.
233 Hard
233 of 396
Analyzing +4 | -1
Assertion : For cell reaction; at 298KZn(s)+Cu+2(1M)⇌Zn+2(1M)+Cu(s)298 \mathrm{KZn}(\mathrm{s})+\mathrm{Cu}^{+2}(1 \mathrm{M}) \rightleftharpoons \mathrm{Zn}^{+2}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s}) Ecell ∘=0\mathrm{E}_{\text {cell }}^{\circ}=0

Reason: At standard condition ; cell potential is always zero.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option C is correct because both Li and Na impart color to the oxidizing flame. Li imparts a crimson red color, and Na imparts a yellow color.
234 Easy
234 of 396
Understanding +4 | -1
At infinite dilution stage, the solution of CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} in water does not contain :
View answer and explanation
Correct option: C
CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} molecule
Option C (CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} molecule) is correct because at infinite dilution, weak electrolytes like acetic acid are considered to be 100% ionized, meaning there are virtually no unionized molecules left.
235 Medium
235 of 396
Analyzing +4 | -1
Assertion: Sodium ions are discharged at the Hg\mathrm{Hg} electrode during electrolysis in preference to H+\mathrm{H}^{+}ions.

Reason: The nature of electrode also affects the order of discharge of cations.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A is correct because both the assertion and reason are true, and the reason correctly explains why sodium ions are discharged at the Hg electrode in preference to H+ ions.
236 Medium
236 of 396
Application +4 | -1
The conductivity of a 0.1NKCl0.1N{\kern 1pt} \mathrm{KCl} solution at 230C is 0.012 ohm−1cm−1\mathrm{oh}{\mathrm{m}^{ - 1}}{\kern 1pt}{\kern 1pt}\mathrm{c}{\mathrm{m}^{ - 1}} .The resistance of the cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be :
View answer and explanation
Correct option: B
0.66 cm−1\mathrm{c}{\mathrm{m}^{ - 1}}
Option B (0.66 cm−1\mathrm{c}{\mathrm{m}^{ - 1}}) is incorrect. The correct calculation using the conductivity and resistance yields 0.66 cm−1\mathrm{c}{\mathrm{m}^{ - 1}}.
237 Hard
237 of 396
Application +4 | -1
Calculate equilibrium constant (approx) at 25∘C25^{\circ} \mathrm{C} for the cell reaction

Fe(s)+CuSO4⇌FeSO4+Cu(s)\mathrm{Fe}_{(\mathrm{s})}+\mathrm{CuSO}_4 \rightleftharpoons \mathrm{FeSO}_4+\mathrm{Cu}_{(\mathrm{s})}
Given EFe/Fc+2o=+0.453 volt ; ECu/Cu+2o=−0.3435 volt \begin{aligned} & \mathrm{E}_{\mathrm{Fe} / \mathrm{F} \mathrm{c}^{+2}}^{\mathrm{o}}=+0.453 \text { volt } ; \ \mathrm{E}_{\mathrm{Cu} / \mathrm{Cu}^{+2}}^{\mathrm{o}}=-0.3435 \text { volt }\end{aligned}
View answer and explanation
Correct option: C
1.0×10271.0 \times 10^{27}
Option C (1.0×10271.0 \times 10^{27}) is correct because it is calculated using the Nernst equation and the relationship between the equilibrium constant and the standard cell potential.
238 Medium
238 of 396
Application +4 | -1
The quantity of charge required to completely reduce 0.5 mol0.5 \mathrm{~mol} of MnO4−\mathrm{MnO}_4^{-}to Mn2O3\mathrm{Mn}_2 \mathrm{O}_3 is
View answer and explanation
Correct option: C
2 F2 \mathrm{~F}
Option C (2 F) is correct because the reduction of 0.5 mol of MnO₄⁻ to Mn₂O₃ requires a change of 4 electrons per Mn atom, resulting in a total charge of 2 Faradays.
239 Easy
239 of 396
Remembering +4 | -1
The amount of substance deposited by the passage of 1 A of current for 1 s is equal to
View answer and explanation
Correct option: C
Electrochemical equivalent
Option C (Electrochemical equivalent) is correct because according to Faraday's first law of electrolysis, the mass of a substance deposited is directly proportional to the quantity of electricity passed, and the electrochemical equivalent (Z) is the mass deposited by 1 coulomb (1 A for 1 s).
240 Hard
240 of 396
Application +4 | -1
Cu+(aq)\mathrm{C}{\mathrm{u}^ + }\left( {aq} \right) is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction
2Cu+(aq)⇋Cu2+(aq)+Cu(s)2\mathrm{C}{\mathrm{u}^ + }\left( {aq} \right)\leftrightharpoons\mathrm{C}{\mathrm{u}^{2 + }}\left( {aq} \right) + \mathrm{Cu}\left( s \right)
choose correct E∘{E^ \circ } for the above reaction if
ECu2+/Cu∘=0.34V,ECu2+/Cu+∘=0.15VE_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{Cu}}^ \circ = 0.34\mathrm{V},E_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{C}{\mathrm{u}^ + }}^ \circ = 0.15\mathrm{V}
View answer and explanation
Correct option: C
#ERROR!
Option C (+0.38 V) is correct because the standard cell potential for the disproportionation reaction is calculated by combining the reduction potentials of the half-reactions.
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