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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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Showing 181-210 of 381 questions

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181 Easy
181 of 381
Understanding +4 | -1
Which of the following is not correct?
View answer and explanation
Correct option: B
The units of electrochemical equivalent are g-coulomb.
Option B is not correct because the units of electrochemical equivalent are g/Coulomb (grams per Coulomb), not g-coulomb.
182 Medium
182 of 381
Remembering +4 | -1
Which of the following chemical reactions represents Hall-Heroult Process?
View answer and explanation
Correct option: B
2Al2O3+3C→4Al+3CO22A{l_2}{O_3} + 3C \to 4Al + 3C{O_2}
Option B (2Al2O3+3C→4Al+3CO22Al_2O_3 + 3C \to 4Al + 3CO_2) is the correct representation of the Hall-Heroult process, which is used for aluminum extraction.
183 Hard
183 of 381
Analyzing +4 | -1
For which cell e.m.f. is independent of the concentrations of electrolytes used?
View answer and explanation
Correct option: A
Fe∣FeO(s)∣KOH(aq)∣NiO(s)∣Ni2O3(s)∣NiFe|FeO(s)|KOH(aq)|NiO(s)|N{i_2}{O_3}(s)|Ni
Option A (Fe∣FeO(s)∣KOH(aq)∣NiO(s)∣Ni2O3(s)∣NiFe|FeO(s)|KOH(aq)|NiO(s)|N{i_2}{O_3}(s)|Ni) is correct because the cell reaction involves only solids and KOH(aq), so the EMF is independent of the concentration of KOH.
184 Easy
184 of 381
Application +4 | -1
Limiting molar ionic conductivities of a uni-univalent electrolyte are 57 and 73. The limiting molar conductivity of the solution will be :
View answer and explanation
Correct option: A
130Scm2mol−1130{\kern 1pt} {\kern 1pt} {\kern 1pt} S{\kern 1pt} {\kern 1pt} \mathrm{c}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{mo}{\mathrm{l}^{ - 1}}
Option A is correct because the limiting molar conductivity of a uni-univalent electrolyte is the sum of the limiting molar ionic conductivities of its cation and anion.
185 Medium
185 of 381
Remembering +4 | -1
The standard electrode potential is measured by
View answer and explanation
Correct option: B
Voltmeter
Option B (Voltmeter) is correct because a voltmeter is the instrument used to measure the potential difference, which includes standard electrode potential.
186 Medium
186 of 381
Analyzing +4 | -1
Assertion (A) Ecell E_{\text {cell }} increases with increase in concentration of Ag+\mathrm{Ag}^{+}ions.

Reason (R) Ecell E_{\text {cell }} has positive value.
View answer and explanation
Correct option: B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
Option B is correct because both the assertion and reason are true, but the reason does not explain the assertion. The Nernst equation explains the effect of concentration on cell potential.
187 Easy
187 of 381
Application +4 | -1
The equivalent conductivity of two strong electrolytes at infinite dilution are :
ΛCH3COONa(aq.)∘=91.0Scm2eq.−1{\mathop \mathrm{\Lambda }\limits^ \circ _{\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}\left( {\mathrm{aq}\mathrm{.}} \right)}} = 91.0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}
ΛHCl(aq.)∘=426.2Scm2eq.−1{\mathop \mathrm{\Lambda }\limits^ \circ _{\mathrm{HCl}\left( {\mathrm{aq}\mathrm{.}} \right)}} = 426.2{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\rm{S {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt} \mathrm{eq}{\mathrm{.}^{ - 1}}
What additional information one needs to calculate Λ∘\mathop \mathrm{\Lambda }\limits^ \circ ̊ of an aqueous CH3COOH?\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}?
View answer and explanation
Correct option: C
Λ∘ofNaCl\mathop \mathrm{\Lambda }\limits^ \circ \mathrm{of}{\kern 1pt} \mathrm{NaCl}
Option C is correct because, according to Kohlrausch's law, to calculate the limiting molar conductivity of CH3COOH, you need the limiting molar conductivities of CH3COONa, HCl, and NaCl.
188 Medium
188 of 381
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 (0.1 HCl) is correct because HCl is a strong acid and provides the highest concentration of highly mobile H+ ions, leading to maximum specific conductance.
189 Medium
189 of 381
Application +4 | -1
How much charge is present on 3 grams of CO32−CO_3^{2 - } ions?
View answer and explanation
Correct option: B
9650 C
Option B is correct. The charge is calculated by finding the number of moles of carbonate ions and then multiplying by the charge per mole of ions (2F).
190 Easy
190 of 381
Remembering +4 | -1
The standard e.m.f. of a galvanic cell can be calculated from :
View answer and explanation
Correct option: D
The E∘{E^ \circ } values of the two half cells
Option D (The E∘{E^ \circ } values of the two half cells) is correct because the standard EMF of a galvanic cell is calculated from the difference in standard reduction potentials of the two half-cells.
191 Medium
191 of 381
Remembering +4 | -1
Which of the following is the use of electrolysis?
View answer and explanation
Correct option: C
Both (1) &\& (2)
Option C (Both (1) & (2)) is correct because electrolysis is used in both electrorefining and electroplating.
192 Hard
192 of 381
Analyzing +4 | -1
Assertion : E∘E^{\circ} for Mn3+/Mn2+{Mn}^{3+} / {Mn}^{2+} is more positive than Cr3/Cr2+{Cr}^3 / {Cr}^{2+}.

Reason : The third ionization energy of Mn{Mn} is larger than that of Cr{Cr}.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.
Option A (If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.) is correct because the higher third ionization energy of Mn makes Mn2+ more stable, leading to a more positive reduction potential for Mn3+/Mn2+.
193 Hard
193 of 381
Application +4 | -1
E.m.f. of cell Ni∣Ni2+(0.1M)∣∣Au3+(1.0M)∣Auis…..,\mathrm{Ni}\left| {\mathrm{N}{\mathrm{i}^{2 + }}\left( {0.1M} \right)} \right|\left| {\mathrm{A}{\mathrm{u}^{3 + }}\left( {1.0M} \right)} \right|\mathrm{Au}{\kern 1pt} {\kern 1pt} is \ldots .., if E∘{E^ \circ } for Ni2+∣Ni\mathrm{N}{\mathrm{i}^{2 + }}|\mathrm{Ni} is −0.25V,E∘- 0.25\mathrm{V},{E^ \circ } for Au3+∣Au\mathrm{A}{\mathrm{u}^{3 + }}|\mathrm{Au} is 1.50 V.
View answer and explanation
Correct option: C
+ 1.75 V
Option C (+ 1.75 V) is correct because the EMF of the cell is calculated as the difference between the standard reduction potentials of the cathode and anode, with a correction for non-standard conditions using the Nernst equation.
194 Hard
194 of 381
Understanding +4 | -1
Assertion: On electrolysis of aq. NaCl,Na\mathrm{NaCl}, \mathrm{Na} is obtained.
Reason: Na+\mathrm{Na}^{+}is reduced at cathode.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because on electrolysis of aqueous NaCl, H2H_2 is obtained instead of Na, and H2OH_2O is reduced at the cathode, not Na+Na^+.
195 Medium
195 of 381
Understanding +4 | -1
The products formed when an aqueous solution of NaBr{NaBr} is electrolysed in a cell having inert electrodes are:
View answer and explanation
Correct option: C
H2,Br2{H}_2, {Br}_2 and NaOH{NaOH}
Option C (H2,Br2{H}_2, {Br}_2 and NaOH{NaOH}) is correct because during the electrolysis of aqueous NaBr, water is reduced at the cathode to produce hydrogen gas and hydroxide ions, while bromide ions are oxidized at the anode to produce bromine gas.
196 Easy
196 of 381
Understanding +4 | -1
The best way to prevent rusting of iron is
View answer and explanation
Correct option: A
Making it cathode
Option A (Making it cathode) is correct because making iron the cathode in an electrochemical cell provides cathodic protection, preventing oxidation (rusting).
197 Medium
197 of 381
Understanding +4 | -1
Which process involves corrosion?
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the listed processes (brown deposits on iron, green deposits on battery terminals, and black deposits on silver) are examples of corrosion.
198 Hard
198 of 381
Understanding +4 | -1
O2\mathrm{O}_2 gas will be evolved as a product of electrolysis of:

(A) an aqueous solution of AgNO3\mathrm{AgNO}_3 using silver electrodes.

(B) an aqueous solution of AgNO3\mathrm{AgNO}_3 using platinum electrodes.

(C) a dilute solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 using platinum electrodes.

(D) a high concentration solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 using platinum electrodes.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(B) and (C) only
Option A ((B) and (C) only) is correct because O2O_2 is evolved from the electrolysis of aqueous AgNO3AgNO_3 with platinum electrodes and dilute H2SO4H_2SO_4 with platinum electrodes.
199 Hard
199 of 381
Understanding +4 | -1
It is not advisable to
View answer and explanation
Correct option: C
Both (1) and (2)
Option C is correct. It is not advisable to stir either copper sulphate solution with a zinc spoon or silver nitrate solution with a copper spoon because zinc is more reactive than copper and copper is more reactive than silver, leading to displacement reactions.
200 Medium
200 of 381
Remembering +4 | -1
Which is the correct relationship between equilibrium constant and standard reduction potential of cell?
View answer and explanation
Correct option: B
Ecell0=0.059nlog⁡KcE_{\text {cell}}^0=\frac{0.059}{n} \log K_c
Option B (Ecell0=0.059nlog⁡KcE_{\text {cell}}^0=\frac{0.059}{n} \log K_c) is the correct relationship between the standard cell potential and the equilibrium constant at 298 K.
201 Easy
201 of 381
Remembering +4 | -1
For gold plating, the electrolyte used is
View answer and explanation
Correct option: C
K[Au(CN)2]\mathrm{K}\left[ {\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}} \right]
Option C (K[Au(CN)2]\mathrm{K}\left[ {\mathrm{Au}{{\left( {\mathrm{CN}} \right)}_2}} \right]) is correct because it is the electrolyte commonly used for gold plating due to its ability to form stable gold complexes.
202 Easy
202 of 381
Analyzing +4 | -1
Which of the following reactions cannot be A base for electrochemical cell?
View answer and explanation
Correct option: D
KMnO4+FeSO4+H2SO4→K2SO4+Fe2(SO4)3+MnSO4+H2O\mathrm{KMn}{\mathrm{O}_4} + \mathrm{FeS}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4} + \mathrm{F}{\mathrm{e}_2}{(\mathrm{S}{\mathrm{O}_4})_3} + \mathrm{MnS}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{O}
Option C is correct because AgNO3+NaCl→AgCl↓+NaNO2\mathrm{AgN}{\mathrm{O}_3} + \mathrm{NaCl} \to \mathrm{AgCl} \downarrow + \mathrm{NaN}{\mathrm{O}_2} is a precipitation reaction, not a redox reaction, and therefore cannot be a basis for an electrochemical cell.
203 Medium
203 of 381
Remembering +4 | -1
Galvanization is applying a coating of:
View answer and explanation
Correct option: C
ZnZn
Option C (ZnZn) is correct because galvanization is the process of coating iron or steel with zinc to protect it from corrosion.
204 Medium
204 of 381
Application +4 | -1
One mole of electrons passes through each of the solution of AgNO3, CuSO4AgN{O_3},\,CuS{O_4} and AlCl3AlC{l_3} when Ag,CuAg,Cu and AlAl are deposited at cathode. The molar ratio of Ag,Cu  and AlAg,Cu\,\,\mathrm{and}\,Al deposited are
View answer and explanation
Correct option: B
6:3:26:3:2
Option B (6:3:2) is correct because it reflects the molar ratio of Ag, Cu, and Al deposited based on their respective valencies and the passage of one mole of electrons.
205 Easy
205 of 381
Application +4 | -1
If ECu2+/Cu∘=+0.34  V{E}_{{Cu}^{2+} / {Cu}}^{\circ}=+0.34 \,{~V} then reduction potential of Cu2+/Cu{Cu}^{2+} / {Cu} electrode at pH=14{pH}=14 is (Ksp\left({K}_{{sp}}\right. of Cu(OH)2=1{Cu}({OH})_2=1 ×10−19\times 10^{-19} )
View answer and explanation
Correct option: B
−0.22  V-0.22 \,{~V}
Option B is correct because it uses the Nernst equation and the Ksp of Cu(OH)2 to calculate the reduction potential at pH=14.
206 Medium
206 of 381
Application +4 | -1
ECu∘=0.34V,EZn∘=0.76VE_{Cu}^ \circ = 0.34V,E_{Zn}^ \circ = 0.76V . A Daniel cell contains 0.1 M ZnSO4\mathrm{ZnS}{\mathrm{O}_4} solution and 0.01 M CuSO4\mathrm{CuS}{\mathrm{O}_4} solution at its electrodes. EMF of the cell is
View answer and explanation
Correct option: A
1.10 V
Option A (1.10 V) is correct because, under standard conditions (and close enough to it), the EMF of the Daniel cell is approximately the difference between the standard reduction potentials of copper and zinc.
207 Medium
207 of 381
Application +4 | -1
The equivalent conductivity of KCl\mathrm{KCl} at infinite dilution is 130 mhocm2eq−1{\rm{mho {\kern 1pt} {\kern 1pt}c}}{\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{e}{\mathrm{q}^{ - 1}}. The transport number of Cl−\mathrm{C}{\mathrm{l}^ - } ion in KCl\mathrm{KCl} at the same temperature is 0.505. The transport number of K+{\mathrm{K}^ + } ion is :
View answer and explanation
Correct option: A
0.495
Option A (0.495) is correct because the sum of the transport numbers of the cation and anion in an electrolyte solution must equal 1.
208 Medium
208 of 381
Analyzing +4 | -1
The number of electrons transferred (lost or gained) during the reactions Fe+H2O→Fe3O4+H2Fe + {H_2}O \to F{e_3}{O_4} + {H_2}
View answer and explanation
Correct option: A
8
Option A (8) is correct because the oxidation state of Fe changes from 0 to +8/3, and the total number of electrons transferred is 8.
209 Hard
209 of 381
Understanding +4 | -1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: In equation ΔrG=−nFEcell\Delta_r G=-n F E_{\text{cell}} value of ΔrG\Delta_r G depends on nn.

Reasons R: Ecell E_{\text {cell }} is an intensive property and ΔrG\Delta_r G is an extensive property.

In the light of the above statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: C
Both A and R are true and R is the correct explanation of A
Option D is correct because the Gibbs free energy change depends on the amount of substance (extensive property), while the cell potential is independent of the amount (intensive property), and this difference explains why ΔrG\Delta_r G depends on n.
210 Hard
210 of 381
Application +4 | -1
The cost of 2 Rs/ kWh2 \mathrm{~Rs} / \mathrm{~kWh} of operating an electric motor for 10 hours takes 10 amp10 \mathrm{~amp} at 110 V110 \mathrm{~V} is :
View answer and explanation
Correct option: D
22 Rs22 \mathrm{~Rs}
Option D (22 Rs22 \mathrm{~Rs}) is correct because the cost is calculated by multiplying the power (in kW) by the time (in hours) and the cost per kWh.
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