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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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Showing 331-360 of 381 questions
331 Hard
331 of 381
Understanding +4 | -1
The equivalent conductance of NaClNaCl at concentration CC and at infinite dilution are λC{\lambda _C} and λ∞{\lambda _\infty }, respectively. The correct relationship between λC{\lambda _C} and λ∞{\lambda _\infty } is given as (where, the constant BB is positive)
View answer and explanation
Correct option: C
λC=λ∞−(B)C{\lambda _C} = {\lambda _\infty } - (B)\sqrt C
Option C is correct because the Debye-Hückel-Onsager equation describes the relationship between molar conductivity at a given concentration and at infinite dilution, and it includes a term with the square root of the concentration.
332 Medium
332 of 381
Understanding +4 | -1
When an electric current is passed through an aqueous solution of sodium chloride :
View answer and explanation
Correct option: D
Its pH progressively increases
Option D (Its pH progressively increases) is correct because during the electrolysis of aqueous NaCl, hydrogen gas is produced at the cathode, leading to an increase in hydroxide ion concentration and thus an increase in pH.
333 Medium
333 of 381
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, making it a stronger reducing agent.
334 Easy
334 of 381
Application +4 | -1
What would be the ratio of moles each of Ag+, Cu+2, Fe+3A{g^ + },\,C{u^{ + 2}},\,F{e^{ + 3}} ions would be deposited by passage of same quantity of electricity through solutions of their salts
View answer and explanation
Correct option: B
1:12:131:\frac{1}{2}:\frac{1}{3}
Option B (1:12:131:\frac{1}{2}:\frac{1}{3}) is correct because the ratio of moles deposited is inversely proportional to the charge on the ion.
335 Medium
335 of 381
Understanding +4 | -1
The cell reaction of the galvanic cell

Cu(s)∣Cu2+(aq)∣∣Hg2+(aq)∣Hg(l)\mathrm{Cu}\left( s \right)\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {aq} \right)} \right|\left| {\mathrm{H}{\mathrm{g}^{2 + }}\left( {aq} \right)} \right|\mathrm{Hg}\left( l \right) is
View answer and explanation
Correct option: D
Cu+Hg2+→Cu2++Hg\mathrm{Cu} + \mathrm{H}{\mathrm{g}^{2 + }} \to \mathrm{C}{\mathrm{u}^{2 + }} + \mathrm{Hg}
Option D (Cu+Hg2+→Cu2++Hg\mathrm{Cu} + \mathrm{H}{\mathrm{g}^{2 + }} \to \mathrm{C}{\mathrm{u}^{2 + }} + \mathrm{Hg}) is correct because copper is oxidized to Cu2+Cu^{2+} and Hg2+Hg^{2+} is reduced to Hg.
336 Easy
336 of 381
Remembering +4 | -1
In an electrolytic cell, the anode and cathode are respectively represented as :
View answer and explanation
Correct option: A
Positive electrode, negative electrode
Option A is correct because, by convention, the anode is the positive electrode where oxidation occurs, and the cathode is the negative electrode where reduction occurs in an electrolytic cell.
337 Medium
337 of 381
Application +4 | -1
Assume that during electrolysis of AgNO3AgN{O_3}, only H2O{H_2}O is electrolysed and O2{O_2} is formed as: 2H2O→4H++O2+4e−2{H_2}O \to 4{H^ + } + {O_2} + 4{e^ - } . O2{O_2} formed at NTP due to passage of 2A of current for 965 s is
View answer and explanation
Correct option: A
0.112 L
Option A is correct. The volume of oxygen formed is calculated using Faraday's laws, considering the current, time, and the stoichiometry of the electrolysis reaction of water.
338 Hard
338 of 381
Application +4 | -1
The reduction electrode potential, E of 0.1 M solution of M+{M^ + } ions

(ERP=−2.36V)\left( {{E_{RP}} = - 2.36V} \right) is
View answer and explanation
Correct option: B
-2.41 V
Option B (-2.41 V) is correct because the reduction potential is calculated using the Nernst equation with the given standard reduction potential and concentration.
339 Easy
339 of 381
Application +4 | -1
What weight of copper will be deposited by passing 2 faraday of electricity through a solution of Cu(II) salt?
View answer and explanation
Correct option: B
63.5 g
Option B (63.5 g) is correct because 2 Faradays of electricity will deposit 1 mole of Cu from a Cu(II) salt solution, which is equal to the molar mass of copper (63.5 g).
340 Easy
340 of 381
Application +4 | -1
In the electrochemical cell
Zn∣ZnSO4(0.01M)∣∣CuSO4(1.0M)∣Cu,\mathrm{Zn|ZnS}{\mathrm{O}_4}(0.01\mathrm{M})||\mathrm{CuS}{\mathrm{O}_\mathrm{4}}(1.0\mathrm{M})\mathrm{|Cu,} the emf of this Daniel cell is E1.{\mathrm{E}_1}. When the concentration of ZnSO4\mathrm{ZnS}{\mathrm{O}_4} is changed to 1.0 M and that of CuSO4\mathrm{CuS}{\mathrm{O}_\mathrm{4}} changed to 0.01 M, the emf changes to E2.{\mathrm{E}_2}. From the following , which one is the relationship between E1  and   E2 ?{\mathrm{E}_\mathrm{1}}\mathrm{ \;and\;}\,{\mathrm{E}_2}\,? (Given, RTF=0.059\frac{{\mathrm{RT}}}{\mathrm{F}} = 0.059)
View answer and explanation
Correct option: D
E1>E2{\mathrm{E}_1} > {\mathrm{E}_2}
Option D (E1>E2{\mathrm{E}_1} > {\mathrm{E}_2}) is correct because, according to the Nernst equation, increasing the concentration of the reactant and decreasing the concentration of the product will increase the cell potential.
341 Hard
341 of 381
Remembering +4 | -1
The electrochemical equivalent of a substance is the mass of the substance liberated on passing a current of
View answer and explanation
Correct option: A
One coulomb
Option A (One coulomb) is correct because the electrochemical equivalent is defined as the mass of a substance liberated by one coulomb of charge.
342 Hard
342 of 381
Application +4 | -1
The ionic conductance of Ba2+\mathrm{B}{\mathrm{a}^{2 + }} and Cl−\mathrm{C}{\mathrm{l}^ - } are respectively 127 and 76 Ω−1cm2{\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2} at infinite dilution. The equivalent conductance (in Ω−1cm2{\Omega ^{ - 1}}\mathrm{c}{\mathrm{m}^2}) of BaCl2\mathrm{BaC}{\mathrm{l}_2} at infinite dilution will be
View answer and explanation
Correct option: A
139.5
Option A (139.5) is correct because it correctly calculates the equivalent conductance of BaCl₂ using the ionic conductances of Ba²⁺ and Cl⁻.
343 Hard
343 of 381
Analyzing +4 | -1
At 298 K, the standard electrode potentials of Cu2+/Cu,Zn2+/Zn,Fe2+/Fe  and  Ag+/AgC{u^{2 + }}/Cu,Z{n^{2 + }}/Zn,F{e^{2 + }}/Fe\,\,\mathrm{and}\,\,A{g^ + }/Ag are 0.34 V, - 0.76 V, - 0.44 V and 0.80 V, respectively.
On the basis of standard electrode potential, predict which of the following reaction cannot occur?
View answer and explanation
Correct option: A
2CuSO4(aq)+2Ag(s)→2Cu(s)+Ag2SO4(aq)2CuS{O_4}(aq) + 2Ag(s) \to 2Cu(s) + A{g_2}S{O_4}(aq)
Option A (2CuSO4(aq)+2Ag(s)→2Cu(s)+Ag2SO4(aq)2CuS{O_4}(aq) + 2Ag(s) \to 2Cu(s) + A{g_2}S{O_4}(aq)) is correct because the cell potential for this reaction is negative, indicating it is non-spontaneous.
344 Medium
344 of 381
Understanding +4 | -1
A cell necessarily does not contain :
View answer and explanation
Correct option: B
A cathode
Option D (A porous diaphragm) is correct because while anodes, cathodes, and electrolytes/fuels are essential components of a cell, a porous diaphragm is not always necessary; some cells function without it.
345 Easy
345 of 381
Analyzing +4 | -1
Select the correct statements among the following.

(i) Copper sulphate solution can't be stored in a zinc pot.

(ii) The potential of individual half cell can be measured at 298 K298 \mathrm{~K} and at 1 bar pressure.

(iii) Copper can dissolve in dil. HCl\mathrm{HCl} but it cannot dissolve in nitric acid due to passive nature of nitrate ion towards copper.

(iv) Platinum and gold electrodes can be used both as anode and cathode.
View answer and explanation
Correct option: A
Only (i) and (iv)
Option A (Only (i) and (iv)) is correct because copper sulphate cannot be stored in a zinc pot due to a redox reaction, and platinum/gold can be used as both anode and cathode due to their inertness.
346 Medium
346 of 381
Remembering +4 | -1
Which of the following metal has the highest conductivity at room temperature ?
View answer and explanation
Correct option: C
Ag
Option C (Ag) is correct because silver (Ag) has the highest electrical conductivity at room temperature among the given options.
347 Easy
347 of 381
Understanding +4 | -1
Stainless steel does not rust because
View answer and explanation
Correct option: B
Chromium forms an oxide layer and protects iron from rusting
Option B (Chromium forms an oxide layer and protects iron from rusting) is correct because chromium forms a passive layer of chromium oxide (Cr2O3Cr_2O_3) that prevents further oxidation of the iron.
348 Easy
348 of 381
Remembering +4 | -1
Rust is a mixture of :
View answer and explanation
Correct option: C
Fe2O3andFe(OH)3\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}{\kern 1pt} {\kern 1pt} {\rm{and{\kern 1pt} {\kern 1pt} Fe}}{\left( {\mathrm{OH}} \right)_3}
Option C (Fe2O3andFe(OH)3\mathrm{F}{\mathrm{e}_2}{\mathrm{O}_3}{\kern 1pt} {\kern 1pt} {\rm{and{\kern 1pt} {\kern 1pt} Fe}}{\left( {\mathrm{OH}} \right)_3}) is correct because rust is a hydrated form of iron(III) oxide and iron(III) hydroxide.
349 Hard
349 of 381
Analyzing +4 | -1
Consider the reaction of extraction of gold from its ore:

Au(s)+2CN−(aq)+14O2(g)+12H2O(l)→Au(CN)2−(aq)+OH−(aq)\begin{aligned} & \mathrm{Au}(\mathrm{s})+2 \mathrm{CN}^{-}(\mathrm{aq})+\frac{1}{4} \mathrm{O}_2(\mathrm{g})+\frac{1}{2} \mathrm{H}_2 \mathrm{O}({l}) \rightarrow \mathrm{Au}(\mathrm{CN})_2{ }^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}

Use the following data to calculate ΔG∘\Delta G^{\circ} for the above reaction.

Kf[Au(CN)2−]=XO2+2H2O+4e−→4OH−;E∘=+0.41 VAu3++3e−→Au;E∘=+1.50 VAu3++2e−→Au+;E∘=+1.40 V\begin{aligned} & K_{\mathrm{f}}\left[\mathrm{Au}(\mathrm{CN})_2^{-}\right] =X \\ & \mathrm{O}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{e}^{-} \rightarrow 4 \mathrm{OH}^{-} ; E^{\circ}=+0.41 \mathrm{~V} \\ & \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} ; E^{\circ}=+1.50 \mathrm{~V} \\ & \mathrm{Au}^{3+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Au}^{+} ; E^{\circ}=+1.40 \mathrm{~V} \end{aligned}
View answer and explanation
Correct option: A
−RTln⁡X+1.29 F-R T \ln X+1.29 \mathrm{~F}
Option A is correct because the overall reaction involves the formation of a complex ion and a redox reaction, and the Gibbs free energy change can be calculated using the given data.
350 Medium
350 of 381
Understanding +4 | -1
Normal Al−AlCl3\mathrm{Al} - \mathrm{AlC}{\mathrm{l}_3} coupled with normal hydrogen electrode gives an e.m.f. of 1.66 V. The standard oxidation electrode potential of aluminium is :
View answer and explanation
Correct option: B
+1.66V+ 1.66{\kern 1pt} {\kern 1pt} \mathrm{V}
Option B (+1.66V+ 1.66{\kern 1pt} {\kern 1pt} \mathrm{V}) is correct because the standard oxidation potential is the negative of the standard reduction potential, and since the emf is given as 1.66 V, the oxidation potential of aluminum is +1.66 V.
351 Medium
351 of 381
Remembering +4 | -1
According to Faraday’s first law of electrolysis mass of substances liberated is equal to
View answer and explanation
Correct option: B
eQ
Option B is correct because, according to Faraday's first law of electrolysis, the mass of a substance liberated at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte, represented by the equation m = eQ, where e is the electrochemical equivalent and Q is the charge.
352 Easy
352 of 381
Remembering +4 | -1
The electroplating with chromium is undertaken because :
View answer and explanation
Correct option: C
Chromium gives a protective and decorative coating to the base metal
Option C is correct because chromium electroplating provides a protective and decorative coating to the base metal, enhancing its corrosion resistance and appearance.
353 Medium
353 of 381
Application +4 | -1
The resistance of 0.01 N solution of an electrolyte was found to be 210 ohm at 298 K, using a conductivity cell of cell constant 0.66cm−10.66\mathrm{c}{\mathrm{m}^{ - 1}} . The equivalent conductivity of solution is :
View answer and explanation
Correct option: A
314.28mhocm2eq.−1314.28{\kern 1pt}{\kern 1pt}{\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}
Option A (314.28mhocm2eq.−1314.28{\kern 1pt}{\kern 1pt}{\rm{mho {\kern 1pt}{\kern 1pt} c}}{\mathrm{m}^2}{\kern 1pt}{\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}) is correct because it is the result of correctly calculating the equivalent conductivity using the given resistance, cell constant, and normality.
354 Easy
354 of 381
Remembering +4 | -1
Coulomb is the quantity of current defined as :
View answer and explanation
Correct option: D
All of the above
Option D is correct because a coulomb is defined as the quantity of charge transferred by a current of one ampere flowing for one second, and it's also related to the deposition of silver and electrochemical equivalence.
355 Hard
355 of 381
Application +4 | -1
The standard emf of a galvanic cell involving 2 moles of electrons in its redox reaction is 0.59 V. The equilibrium constant for the redox reaction of the cell is
View answer and explanation
Correct option: A
1020{10^{20}}
Option A (1020{10^{20}}) is correct because it accurately calculates the equilibrium constant using the Nernst equation and the given standard emf.
356 Medium
356 of 381
Analyzing +4 | -1
Given that EFe3+/Fe2+∘=−0.36 V\mathrm{E}_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\circ}=-0.36 \mathrm{~V} and EFe2+/Fe∘=−0.439 V\mathrm{E}_{\mathrm{Fe}^{2+}/\mathrm{Fe}}^{\circ}=-0.439 \mathrm{~V} respectively. Therefore value of EF3+,Fe2+/Pto\mathrm{E}_{\mathrm{F}^{3+}, \mathrm{Fe}^{2+} /\mathrm{Pt}}^{\mathrm{o}} is:
View answer and explanation
Correct option: D
[3(−0.036)−2(−[3(-0.036)-2(- 0.439)] V
Option D ([3(−0.036)−2(−[3(-0.036)-2(- 0.439)] V) is correct because it correctly applies the formula for calculating the standard reduction potential of the overall reaction using the given standard reduction potentials and the number of electrons involved.
357 Hard
357 of 381
Understanding +4 | -1
H2{\mathrm{H}_2} cannot be displaced by
View answer and explanation
Correct option: D
Ag+\mathrm{A}{\mathrm{g}^ + }
Option D (Ag+\mathrm{A}{\mathrm{g}^ + }) is correct because silver is a less reactive metal than hydrogen and cannot displace it from acids or water.
358 Hard
358 of 381
Understanding +4 | -1
When silver nitrate solution is added to potassium iodide solution then the sol produced is
View answer and explanation
Correct option: A
Agl/I−Agl/{I^ - }
Option A (Agl/I−Agl/{I^-}) is correct because when silver nitrate is added to potassium iodide, AgI precipitates, and I−I^- ions are adsorbed on the surface, forming a negatively charged sol.
359 Hard
359 of 381
Application +4 | -1
Electrolysis of a solution MnSO4\mathrm{MnSO}_4 in aqueous sulphuric acid is a method for the preparation of MnO2\mathrm{MnO}_2. Passing a current of 27 A for 24 hours gives 1 kg1 \mathrm{~kg} of MnO2\mathrm{MnO}_2. The current efficiency is
View answer and explanation
Correct option: B
95.185%95.185 \%
Option B (95.185%95.185 \%) is correct because it accurately calculates the current efficiency based on the mass of MnO2 produced during electrolysis.
360 Hard
360 of 381
Understanding +4 | -1
Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are
View answer and explanation
Correct option: A
Fe is oxidised to Fe2+\mathrm{F}{\mathrm{e}^{2 + }} and dissolved oxygen in water is reduced to OH−\mathrm{O}{\mathrm{H}^ - }
Option A (Fe is oxidised to Fe2+\mathrm{F}{\mathrm{e}^{2 + }} and dissolved oxygen in water is reduced to OH−\mathrm{O}{\mathrm{H}^ - }) is correct because iron is oxidized at the anode and oxygen is reduced at the cathode in the electrochemical process of corrosion.
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