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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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Showing 241-270 of 381 questions

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241 Medium
241 of 381
Understanding +4 | -1
In the electrochemical reaction,

2Fe3++Zn→Zn2++2Fe2+2\mathrm{F}{\mathrm{e}^{3 + }} + \mathrm{Zn} \to \mathrm{Z}{\mathrm{n}^{2 + }} + 2\mathrm{F}{\mathrm{e}^{2 + }}

increasing the concentration of Fe2+\mathrm{F}{\mathrm{e}^{2 + }}
View answer and explanation
Correct option: C
Decreases the cell emf
Option C is correct. According to the Nernst equation, increasing the concentration of the product (Fe²⁺) will decrease the cell emf.
242 Easy
242 of 381
Remembering +4 | -1
Electrolytic conduction is due to migration of :
View answer and explanation
Correct option: C
Ions
Option C is correct because electrolytic conduction occurs due to the movement of ions in the solution, carrying the electric charge.
243 Hard
243 of 381
Analyzing +4 | -1
The number of correct statements from the
following is ________

(A) Conductivity always decreases with decrease
in concentration for both strong and weak
electrolytes.

(B) The number of ions per unit volume that carry
current in a solution increases on dilution.

(C) Molar conductivity increases with decrease in
concentration.

(D) The variation in molar conductivity is different
for strong and weak electrolytes.

(E) For weak electrolytes, the change in molar
conductivity with dilution is due to decrease in
degree of dissociation.
View answer and explanation
Correct option: A
3
The correct answer is 3 because statements A, B and E are incorrect, while C and D are correct.
244 Medium
244 of 381
Understanding +4 | -1
The E0 values at 250C{25^0}C for A+/A,B2+/B,D3+/D{A^ + }/A,\mathrm{ }{B^{2 + }}/B,\mathrm{ }{D^{3 + }}/D and E+/E{E^ + }/E are -2.15, -1.75, +0. 15 and +0.21 volt respectively. Which of the following is the strongest oxidising agent ?
View answer and explanation
Correct option: A
E+{E^ + }
Option A is correct because the strongest oxidizing agent is the one with the highest (most positive) standard reduction potential. Among the given options, E+ has the highest value (+0.21 V), indicating it has the greatest tendency to be reduced and thus is the strongest oxidizing agent.
245 Hard
245 of 381
Understanding +4 | -1
Which of the following is/are correct statement(s) for the addition of Li,K,Rb{Li}, {K}, {Rb} to the aqueous solution of Na+{Na}^{+}.

(i) The correct order of metals in which they reduce the Na+{Na}^{+}ion is Rb<K<Li{Rb} < {K} < {Li}.

(ii) Reduction of metal ions would not take place.
View answer and explanation
Correct option: C
Statement (ii) is correct only.
Option C is correct. Statement (ii) is correct because Na+{Na}^{+} is already a stable ion and further reduction by Li,K,Rb{Li}, {K}, {Rb} is not favorable due to their higher reduction potentials compared to sodium.
246 Easy
246 of 381
Remembering +4 | -1
What flows in the internal circuit of A galvanic cell?
View answer and explanation
Correct option: A
Ions
Option A (Ions) is correct because ions are the charge carriers within the electrolyte solution in the internal circuit of a galvanic cell.
247 Hard
247 of 381
Application +4 | -1
The standard reduction potential&nbsp;E∘{E^ \circ } for the half reactions are as
Zn→Zn2++2e−,E∘=0.76V\mathrm{Zn} \to \mathrm{Z}{\mathrm{n}^{2 + }} + 2{e^ - },{E^ \circ } = 0.76V
Cu→Cu2++2e−,E∘=0.34V\mathrm{Cu} \to \mathrm{C}{\mathrm{u}^{2 + }} + 2{e^ - },{E^ \circ } = 0.34V
The emf for the cell reaction,
Zn+Cu2+→Zn2++Cu\mathrm{Zn} + \mathrm{C}{\mathrm{u}^{2 + }} \to \mathrm{Z}{\mathrm{n}^{2 + }} + \mathrm{Cu}
View answer and explanation
Correct option: A
0.42 V
Option A (0.42 V) is correct because it is the cell potential calculated by subtracting the oxidation potential of copper from the oxidation potential of zinc.
248 Easy
248 of 381
Remembering +4 | -1
Faraday’s laws hold good at :
View answer and explanation
Correct option: A
All pressures
Option D is correct because Faraday's laws are fundamental and hold true under all normal conditions, including varying pressures and solvents.
249 Medium
249 of 381
Understanding +4 | -1
When Zn piece is kept in CuSO4\mathrm{CuS}{\mathrm{O}_4} solution, copper gets precipitated because:
View answer and explanation
Correct option: B
Standard reduction potential of zinc is less than copper
Option B is correct. Copper precipitates because zinc has a lower standard reduction potential than copper, meaning zinc is more easily oxidized and can displace copper from the solution.
250 Easy
250 of 381
Application +4 | -1
Molar conductivities (Λm∘{\mathop \mathrm{\Lambda }\limits^ \circ _m}) at infinite dilution of NaCl,HCl\mathrm{NaCl},\mathrm{HCl} and CH3COONa\mathrm{C}{\mathrm{H}_3}\mathrm{COONa} are 126.4, 425.9 and 91.0 Scm2mol−1{\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}} respectively.Λm∘{\mathop \mathrm{\Lambda }\limits^ \circ _m} for CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} will be :
View answer and explanation
Correct option: D
390.5 Scm2mol−1{\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}
Option D is correct because, according to Kohlrausch's law, the molar conductivity of CH3COOH at infinite dilution is the sum of the molar conductivities of CH3COONa and HCl, minus the molar conductivity of NaCl, resulting in 390.5 S cm² mol⁻¹.
251 Easy
251 of 381
Understanding +4 | -1
During electrolysis of fused CaH2,H2\mathrm{Ca}{\mathrm{H}_2},{\mathrm{H}_2} is liberated at :
View answer and explanation
Correct option: A
Anode
Option A (Anode) is correct because in the electrolysis of fused CaH2, hydride ions (H-) migrate to the anode and are oxidized to form hydrogen gas.
252 Hard
252 of 381
Understanding +4 | -1
Which colourless gas evolves when NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} reacts with zinc in A dry cell battery?
View answer and explanation
Correct option: C
H2{\mathrm{H}_2}
Option C (H2{\mathrm{H}_2}) is correct because zinc reacts with ammonium chloride to produce zinc chloride, ammonia, and hydrogen gas.
253 Medium
253 of 381
Analyzing +4 | -1
Statement I: When iron rod is immersed in NaCl\mathrm{NaCl} solution such that half of its length is corroded and other half part when exposed to air is not corroded.

Statement II: Metal portion with less access to O2\mathrm{O}_2 corrodes.
View answer and explanation
Correct option: A
If both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
Option A is correct because Statement I and II are both true, and Statement II correctly explains why the iron rod corrodes more where it has less access to oxygen.
254 Medium
254 of 381
Application +4 | -1
The equilibrium constant for the reaction at 298 K is
View answer and explanation
Correct option: D
1037.22{10^{37.22}}
Without the reaction, it is impossible to determine the equilibrium constant. This question is incomplete and cannot be answered.
255 Hard
255 of 381
Understanding +4 | -1
Select the correct statement for Λm=Λm∘−AC1/2\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-A C^{1 / 2}.
View answer and explanation
Correct option: D
Value of AA depends on the charges of cation and anion
Option D is correct because the constant 'A' in the equation Λm=Λm∘−AC1/2\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-A C^{1 / 2} depends on the charges of the cation and anion of the electrolyte.
256 Medium
256 of 381
Analyzing +4 | -1
Statement I: During electrolysis of aqueous
sodium acetate solution, the molar ratio of gases
formed at cathode and anode is 1:3.

Statement II: Acetate ion discharges at anode and
H+\text{H}^+ ion, at cathode.
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 both statements are correct, but Statement II does not explain Statement I. The gas ratio is due to the products of acetate electrolysis, not simply H+ discharge.
257 Hard
257 of 381
Application +4 | -1
A current of 12 A is passed through an electrolytic cell containing aqueous NiSO4\mathrm{NiS}{\mathrm{O}_4} solution. Both Ni and H2{\mathrm{H}_2} gas are formed at the cathode. The current efficiency is 60%. What is the mass of nickel deposited on the cathode per hour?
View answer and explanation
Correct option: A
7.883 g
Option A (7.883 g) is correct because the mass of nickel deposited is calculated considering the current efficiency and Faraday's laws.
258 Hard
258 of 381
Analyzing +4 | -1
100 mL100 {~mL} of 0.05MCuSO4(0.05 {MCuSO}_4( aq) solution was electrolyzed using inert electrodes by passing current till the pH{pH} of the resulting solution was 2 . The solution after electrolysis was neutralized and then treated with excess KI{KI} and formed I2{I}_2 titrated with 0.04MNa2 S2O30.04 M {Na}_2{~S}_2 {O}_3. Calculate the required volume (in mL{mL} ) of Na2 S2O3{Na}_2 {~S}_2 {O}_3 :
View answer and explanation
Correct option: A
112.5 mL112.5 {~mL}
Option A is correct. The required volume of Na2S2O3\mathrm{Na}_2 \mathrm{S}_2 \mathrm{O}_3 is 112.5 mL.
259 Hard
259 of 381
Application +4 | -1
The equilibrium constant for the reaction,&nbsp;Cu(s)+2Ag+(aq)⇌Cu(2+)(aq)+2Ag(s),Cu(s) + 2A{g^ + }(aq) \rightleftharpoons C{u^{(2 + )}}(aq) + 2Ag(s), E∘=0.46 V{E^ \circ } = 0.46\text{ V} at 298 K is:
View answer and explanation
Correct option: C
4.0×10154.0 \times {10^{15}}
Option C (4.0×10154.0 \times {10^{15}}) is correct because it is calculated using the relationship between standard EMF and the equilibrium constant, K=10nE00.0591K = 10^{\frac{{nE^0}}{{0.0591}}}.
260 Hard
260 of 381
Application +4 | -1
The hydrogen electrode is dipped in a solution of pH=3at25∘C\mathrm{pH} = 3{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{at}{\kern 1pt} {\kern 1pt} {\kern 1pt} {25^ \circ }C. The potential of the cell would be (thevalueof2.303RTTis0.059V\mathrm{the value of}\frac{{2.303RT}}{T}\mathrm{is}{\kern 1pt} {\kern 1pt} {\kern 1pt} 0.059\mathrm{V})
View answer and explanation
Correct option: C
-0.177 V
Option C (-0.177 V) is correct because it is the calculated potential of the hydrogen electrode using the Nernst equation with the given pH.
261 Easy
261 of 381
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}
Option D (CuCu will reduce Br2B{r_2}) is correct because copper has a lower reduction potential than bromine, meaning it can be oxidized (act as a reducing agent) to reduce bromine.
262 Hard
262 of 381
Understanding +4 | -1
Which one of the following statements is correct for electrolysis of brine solution?
View answer and explanation
Correct option: D
OH−\mathrm{OH}^{-}is formed at cathode
Option D (OH−OH^- is formed at cathode) is correct because during the electrolysis of brine solution, H2OH_2O is reduced to H2H_2 and OH−OH^- at the cathode.
263 Hard
263 of 381
Analyzing +4 | -1
A student made the following observations in the laboratory,
i) Clean copper metal did not react with 1 molar Pb(NO3)2\mathrm{Pb}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2} solution.
ii) Clean lead metal dissolved in a 1 molar AgNO3\mathrm{AgN}{\mathrm{O}_3} solution and crystals of Ag\mathrm{Ag} metal appeared.
iii) Clean silver metal did not react with 1 molar Cu(NO3)2\mathrm{Cu}{\left( {\mathrm{N}{\mathrm{O}_3}} \right)_2} solution.
The order of decreasing reducing character of the three metals is :
View answer and explanation
Correct option: C
Pb,Cu,Ag\mathrm{Pb},\mathrm{Cu},\mathrm{Ag}
Option C (Pb,Cu,Ag\mathrm{Pb},\mathrm{Cu},\mathrm{Ag}) is correct because Pb dissolved in AgNO3, Cu did not react with Pb(NO3)2, and Ag did not react with Cu(NO3)2, indicating the order of reducing character.
264 Easy
264 of 381
Understanding +4 | -1
In which cell, electrical energy is converted into chemical energy?
View answer and explanation
Correct option: D
Either of these
Option D (Either of these) is correct because both water voltameter, silver voltameter and coulometer convert electrical energy into chemical energy.
265 Medium
265 of 381
Application +4 | -1
On passing 0.5 F electricity through molten sodium chloride, sodium deposited at cathode will be
View answer and explanation
Correct option: B
11.50 g
Option B (11.50 g) is correct because the mass of sodium deposited is calculated using Faraday's laws and the given quantity of electricity.
266 Medium
266 of 381
Application +4 | -1
1.8 g of metal were deposited by a current of 3 amperes for 50 minute. The equivalent wt. of metal is :
View answer and explanation
Correct option: C
19.3
Option C (19.3) is correct because it is the calculated equivalent weight of the metal using Faraday's laws of electrolysis.
267 Hard
267 of 381
Analyzing +4 | -1
Cu+2HCl→CuCl2+H2(g)\mathrm{Cu} + 2\mathrm{HCl} \to \mathrm{CuC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

[ECu2+/Cu∘=+0.34V]\left[ {E_{C{u^{2 + }}/Cu}^ \circ = + 0.34V} \right]

Zn+2HCl→ZnCl2+H2(g)\mathrm{Zn} + 2\mathrm{HCl} \to \mathrm{ZnC}{\mathrm{l}_2} + {\mathrm{H}_2}\left( \mathrm{g} \right)

[EZn2+/Zn∘=−0.76V]\left[ {E_{Z{n^{2 + }}/Zn}^ \circ = - 0.76V} \right]

Ag+2HCl→AgCl+12H2(g)\mathrm{Ag} + 2\mathrm{HCl} \to \mathrm{AgCl} + \frac{1}{2}{\mathrm{H}_2}\left( \mathrm{g} \right)

[EAg+/Ag∘=+0.80V]\left[ {E_{A{g^ + }/Ag}^ \circ = + 0.80V} \right]

Which of the following reaction is feasible ?

View answer and explanation
Correct option: A
(ii)
Option A ((ii)) is correct because the reaction with Zinc (Zn) is feasible as it has a negative standard reduction potential, indicating it can be oxidized by HCl.
268 Easy
268 of 381
Analyzing +4 | -1
What will be pH of aqueous solution of electrolyte in electrolytic cell during electrolysis of CuSO4(aq)\mathrm{CuS}{\mathrm{O}_4}\left( {aq} \right) between graphite electrodes?
View answer and explanation
Correct option: C
pH < 7.0
Option C (pH < 7.0) is correct because the electrolysis of CuSO4(aq) produces H2SO4, which is acidic.
269 Hard
269 of 381
Analyzing +4 | -1
Which one of the following solutions will have highest conductivity?
View answer and explanation
Correct option: D
0.1 M HCl
Option D is correct because HCl is a strong acid that completely dissociates in solution, resulting in the highest concentration of ions and therefore the highest conductivity.
270 Hard
270 of 381
Remembering +4 | -1
Which one is correct relation :
View answer and explanation
Correct option: A
ΔS=(∂E∂T)P×nF\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} \times nF
Option A (ΔS=(∂E∂T)P×nF\Delta S = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_P} \times nF) is the correct relation because it links the change in entropy to the temperature coefficient of the cell potential at constant pressure.
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