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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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Showing 151-180 of 381 questions

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151 Easy
151 of 381
Remembering +4 | -1
Which metal is most readily corroded in moist air?
View answer and explanation
Correct option: B
Iron
Option B is correct because iron is more readily corroded in moist air compared to copper, silver, and nickel due to its higher reactivity and tendency to form rust.
152 Easy
152 of 381
Understanding +4 | -1
Four alkali metalsA,B,CA,B,C and DD are having respectively standard electrode potential as −3.05,−1.66,−0.40and0.80- 3.05, - 1.66, - 0.40{\kern 1pt} {\kern 1pt} \mathrm{and}{\kern 1pt} {\kern 1pt} 0.80. Which one will be the most reactive?
View answer and explanation
Correct option: A
AA
Option A is correct because the most reactive alkali metal is the one with the most negative standard electrode potential. A more negative value indicates a greater tendency to lose electrons and undergo oxidation. Among the given options, A has the most negative value (-3.05), making it the most reactive.
153 Easy
153 of 381
Understanding +4 | -1
In the electrodeposition of Ag, the silver ions are:
View answer and explanation
Correct option: B
Reduced at cathode
Option B (Reduced at cathode) is correct because reduction, which is the gain of electrons, always occurs at the cathode in an electrochemical cell.
154 Hard
154 of 381
Understanding +4 | -1
STATEMENT-1 : If (dEcell dT)p>0\left(\frac{d E_{\text {cell }}}{d T}\right)_p>0 for a cell reaction then ΔS\Delta S is positive.

STATEMENT-2 : ΔS=nFT(dEdT)p\Delta S=n F T\left(\frac{d E}{d T}\right)_p
View answer and explanation
Correct option: C
If STATEMENT-1 is TRUE and STATEMENT-2 is FALSE
Statement-1 is TRUE: If (dEcell/dT)_p > 0, then ΔS is positive because ΔG = -nFEcell and ΔG = ΔH - TΔS. Statement-2 is FALSE: The correct relation is ΔS = nF(dE/dT)_p.
155 Hard
155 of 381
Remembering +4 | -1
In a gas lighter, mechanical energy is converted into electrical energy by using crystals of barium titanate. Barium titanate is
View answer and explanation
Correct option: B
both piezoelectric as well as ferroelectric
Option B is correct because barium titanate exhibits both piezoelectricity (generating electricity under mechanical stress) and ferroelectricity (having spontaneous electric polarization that can be reversed by an electric field).
156 Easy
156 of 381
Understanding +4 | -1
Which represents a concentration cell?
View answer and explanation
Correct option: A
PtH2∣HCl∣c1∣HCl∣c2PtH2\mathrm{Pt}{\mathrm{H}_2}\mathop {\left| {\mathrm{HCl}} \right|}\limits_{{c_1}} \mathop {\left| {\mathrm{HCl}} \right|}\limits_{{c_2}} \mathrm{Pt}{\mathrm{H}_2}
Option A (PtH2∣HCl∣c1∣HCl∣c2PtH2\mathrm{Pt}{\mathrm{H}_2}\mathop {\left| {\mathrm{HCl}} \right|}\limits_{{c_1}} \mathop {\left| {\mathrm{HCl}} \right|}\limits_{{c_2}} \mathrm{Pt}{\mathrm{H}_2}) correctly represents a concentration cell because it has the same electrodes and electrolyte but different concentrations.
157 Easy
157 of 381
Application +4 | -1
The atomic weight of Al is 27. When a current of 5 faraday is passed through a solution of Al3+\mathrm{A}{\mathrm{l}^{3 + }} ions, the wt.of Al deposited is :
View answer and explanation
Correct option: D
45 g
Option D (45 g) is correct because passing 5 Faradays of charge through a solution of Al3+Al^{3+} ions will deposit 5/3 moles of Al, which corresponds to 45 g.
158 Medium
158 of 381
Remembering +4 | -1
The value of electronic charge is equal to :
View answer and explanation
Correct option: A
FaradayAv.number\frac{{\mathrm{Faraday}}}{{\mathrm{Av}\mathrm{.}{\kern 1pt} {\kern 1pt} \mathrm{number}}}
Option A is correct because the electronic charge is the total charge of one mole of electrons (Faraday constant) divided by the number of electrons in a mole (Avogadro's number).
159 Hard
159 of 381
Analyzing +4 | -1
For the cell reaction Fe+2Fe3+=3Fe2+\mathrm{Fe} + 2\mathrm{F}{\mathrm{e}^{3 + }} = 3\mathrm{F}{\mathrm{e}^{2 + }}, which is not possible?
View answer and explanation
Correct option: A
One cell can be constructed
Option A (One cell can be constructed) is correct because only one cell reaction is given, and therefore only one cell can be constructed based on that specific reaction.
160 Easy
160 of 381
Application +4 | -1
TheE∘{E^ \circ } for half-cell Fe/Fe2+\mathrm{Fe}/\mathrm{F}{\mathrm{e}^{2 + }} and Cu/Cu2+\mathrm{Cu}/\mathrm{C}{\mathrm{u}^{2 + }} are−0.44V- 0.44\mathrm{V} and +0.32V+ 0.32\mathrm{V} respectively, then
View answer and explanation
Correct option: A
Cu2+oxidisesFe\mathrm{C}{\mathrm{u}^{2 + }}{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{oxidises}{\kern 1pt} {\kern 1pt} \mathrm{ Fe}
Option A (Cu2+oxidisesFe\mathrm{C}{\mathrm{u}^{2 + }}{\kern 1pt} {\kern 1pt} {\kern 1pt} \mathrm{oxidises}{\kern 1pt} {\kern 1pt} \mathrm{ Fe}) is correct because Cu2+\mathrm{C}{\mathrm{u}^{2 + }} has a higher reduction potential than Fe2+\mathrm{F}{\mathrm{e}^{2 + }}, making it a stronger oxidizing agent.
161 Medium
161 of 381
Application +4 | -1
E1,E2{E_1},{E_2} , and E3{E_3} are the emfs of the following three galvanic cells respectively
I. Zn (s) |Zn2+Z{n^{2 + }} (1 M) || Cu2+\mathrm{C}{\mathrm{u}^{2 + }} (0.1 M) | Cu (s)
II. Zn (s) |Zn2+Z{n^{2 + }} (1 M) || Cu2+\mathrm{C}{\mathrm{u}^{2 + }} (1 M) | Cu (s)
III. Zn (s) | Zn2+Z{n^{2 + }} (0.1 M) || Cu2+\mathrm{C}{\mathrm{u}^{2 + }} (1 M) | Cu (s)
Which one of the following is true?
View answer and explanation
Correct option: D
E3>E2>E1{E_3} > {E_2} > {E_1}
Option D (E3>E2>E1{E_3} > {E_2} > {E_1}) is correct because, based on the Nernst equation, increasing the concentration of reactants and decreasing the concentration of products increases the cell emf.
162 Medium
162 of 381
Understanding +4 | -1
Which of the following reactions is correct for a given electrochemical cell at 25∘C{25^ \circ }C?

Pt∣Br2(g)∣Br−(g)∣∣Cl−(aq)∣Cl2(g)∣Pt\mathrm{Pt}\left| {\mathrm{B}{\mathrm{r}_2}\left( \mathrm{g} \right)} \right|\mathrm{B}{\mathrm{r}^ - }\left( \mathrm{g} \right)||\mathrm{C}{\mathrm{l}^ - }\left( {aq} \right)\left| {\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)} \right|\mathrm{Pt}
View answer and explanation
Correct option: A
2Br−(aq)+Cl2(g)→2Cl−(aq)+Br2(g)2\mathrm{B}{\mathrm{r}^ - }\left( {\mathrm{aq}} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) \to 2\mathrm{C}{\mathrm{l}^ - }\left( {\mathrm{aq}} \right) + \mathrm{B}{\mathrm{r}_2}\left( \mathrm{g} \right)
Option A is correct because it represents the spontaneous reaction occurring in the electrochemical cell, where chlorine oxidizes bromide ions.
163 Medium
163 of 381
Application +4 | -1
The hydrogen electrode is dipped in a solution of pH = 3 at 25∘C{25^ \circ }C. The reduction potential of the cell would be :
View answer and explanation
Correct option: B
⎯ 0.177 V
Option B (⎯ 0.177 V) is correct because the reduction potential is calculated using the Nernst equation for the hydrogen electrode at the given pH.
164 Easy
164 of 381
Application +4 | -1
The conductance of a solution of an electrolyte is same as that of its conductivity. The cell used can be said to have cell constant equal to :
View answer and explanation
Correct option: A
1
Option A is correct because conductance equals conductivity when the cell constant (length/area) is equal to 1.
165 Easy
165 of 381
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 less electropositive than aluminium and cannot displace it from its salt solution.
166 Easy
166 of 381
Application +4 | -1
The reaction (1/2) H2(g)+AgCl(s)=H+(aq)+Cl−(aq)+Ag(s)(1/2)\,{H_2}(g) + AgCl(s) = {H^ + }(aq) + C{l^ - }(aq) + Ag(s) occurs in the galvanic cell:
View answer and explanation
Correct option: C
Pt∣H2(g)∣HCl(soln.)∣AgCl(s) ∣AgPt|{H_2}(g)|HCl({\text{soln}}{\text{.)}}|AgCl(s)\,|Ag
Option C is correct because it correctly represents the galvanic cell where hydrogen gas is oxidized to H+ ions at the platinum electrode, and AgCl is reduced to Ag at the silver electrode.
167 Hard
167 of 381
Remembering +4 | -1
The tendency of an electrode to lose electrons is known as
View answer and explanation
Correct option: C
Oxidation potential
Option C is correct because oxidation potential is the measure of the tendency of a chemical species to lose electrons and become oxidized at an electrode.
168 Medium
168 of 381
Analyzing +4 | -1
Assertion: When aqueous sodium chloride solution is electrolysed, Oxygen gas is produced at the anode.

Reason: It is due to the overpotential for oxidation of water to oxygen.
View answer and explanation
Correct option: A
Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.
Option A is correct because the assertion and reason are both true, and the overpotential for water oxidation explains why chloride ions are oxidized instead of water at the anode during the electrolysis of aqueous NaCl.
169 Easy
169 of 381
Understanding +4 | -1
The passage of electricity in the Daniell cell when Zn and Cu electrodes are connected:
View answer and explanation
Correct option: B
From Cu to Zn outside the cell
Option B (From Cu to Zn outside the cell) is correct because electrons flow from the zinc electrode (anode) to the copper electrode (cathode) through the external circuit.
170 Medium
170 of 381
Application +4 | -1
EMF of hydrogen electrode in term of pH is (at 1 atm pressure)
View answer and explanation
Correct option: D
EH2={E_{{H_2}}} = - 0.0591 pH
Option D (EH2={E_{{H_2}}} = - 0.0591 pH) is correct because it directly relates the EMF of a hydrogen electrode to the pH of the solution at 1 atm pressure.
171 Easy
171 of 381
Remembering +4 | -1
If ϕ denotes standard reduction potential, which is true:
View answer and explanation
Correct option: A
Ecell∘=ϕR−ϕLE_{\mathrm{cell}}^ \circ = {\phi _R} - {\phi _L}
Option A (Ecell∘=ϕR−ϕLE_{\mathrm{cell}}^ \circ = {\phi _R} - {\phi _L}) is correct because the standard cell potential is calculated by subtracting the standard reduction potential of the left-hand electrode (anode) from the standard reduction potential of the right-hand electrode (cathode).
172 Easy
172 of 381
Understanding +4 | -1
The cell reaction is spontaneous, when
View answer and explanation
Correct option: C
ΔG∘\Delta {G^ \circ } is negative
Option C (ΔG∘\Delta {G^ \circ } is negative) is correct because a spontaneous cell reaction requires a negative change in Gibbs free energy.
173 Medium
173 of 381
Remembering +4 | -1
Indicator electrode is :
View answer and explanation
Correct option: D
Quinhydrone electrode
Option D is correct because a quinhydrone electrode is a type of redox electrode used as an indicator electrode to measure hydrogen ion concentration (pH).
174 Hard
174 of 381
Analyzing +4 | -1
Small quantities of solutions of compounds TX,TYTX,TY and TZTZ are put into separate test tubes containing X, Y and Z solutions. TXTX does not react with any of these. TYTY reacts with both X and Z. TZTZ reacts with X. The decreasing order of ease of oxidation of the anions X−,Y−,Z−{X^ - },{Y^ - },{Z^ - } is
View answer and explanation
Correct option: A
Y−,Z−,X−{Y^ - },{Z^ - },{X^ - }
Option A (Y−,Z−,X−{Y^ - },{Z^ - },{X^ - }) is correct because Y- can reduce both X and Z, Z- can reduce X, and X- cannot reduce any, indicating their relative ease of oxidation.
175 Easy
175 of 381
Application +4 | -1
Conductance ‘CC’ (inSS) in directly proportional to the area of the electrode and concentration and inversely proportional to length of separation of electrode, the unit of constant of proportionality is:
View answer and explanation
Correct option: B
Sm2mol−1\mathrm{S}{\kern 1pt} {\kern 1pt} {\mathrm{m}^2}{\kern 1pt} {\kern 1pt}\mathrm{mo}{\mathrm{l}^{ - 1}}
Option B is correct because the constant of proportionality is conductivity, and its unit is S m² mol⁻¹ when conductance is directly proportional to area and concentration and inversely proportional to length.
176 Easy
176 of 381
Understanding +4 | -1
The minimum equivalent conductance in fused state is shown by :
View answer and explanation
Correct option: B
BeCl2\mathrm{BeC}{\mathrm{l}_2}
Option B (BeCl2\mathrm{BeC}{\mathrm{l}_2}) is correct because beryllium has the smallest ionic size among the given options, leading to lower mobility and thus lower equivalent conductance in the fused state.
177 Easy
177 of 381
Application +4 | -1
1 liter of 1M CuSO4CuS{O_4} solution is electrolysed. After passing 2 F charge, the molarity of CuSO4CuS{O_4} will be
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because passing 2F of charge through 1 liter of 1M CuSO4CuSO_4 will completely deposit all the Cu2+Cu^{2+} ions, leaving a molarity of zero.
178 Medium
178 of 381
Analyzing +4 | -1
Cu(II)\mathrm{Cu}\left( {\mathrm{II}} \right) sulphate solution is treated separately with KCl\mathrm{KCl} and KI. In which case, Cu2+\mathrm{C}{\mathrm{u}^{2 + }} be reduced to Cu+?\mathrm{C}{\mathrm{u}^ + }?
View answer and explanation
Correct option: B
With KI
Option B (With KI) is correct because I- is a stronger reducing agent than Cl-, and Cu2+ is reduced to Cu+ by I-.
179 Medium
179 of 381
Analyzing +4 | -1
Equal quantities of electricity are passed through three voltameters containing FeSO4,Fe2(SO4)3\mathrm{FeSO}_4, \mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3, and Fe(NO3)3\mathrm{Fe}\left(\mathrm{NO}_3\right)_3. Consider the following statements in this regard:
(1) The amount of iron deposited in FeSO4\mathrm{FeSO}_4 and Fe2(SO4)3\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3 is equal
(2) The amount of iron deposited in Fe(NO3)3\mathrm{Fe}\left(\mathrm{NO}_3\right)_3 is two thirds of the amount of iron deposited in FeSO4\mathrm{FeSO}_4
(3) The amount of iron deposited in Fe2(SO4)3 Fe(NO3)3\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3\,\mathrm{Fe}\left(\mathrm{NO}_3\right)_3 is equal
Of these statements:
View answer and explanation
Correct option: B
2 and 3 are correct
Option B (2 and 3 are correct) is correct because the amount of iron deposited depends on the charge required to deposit one mole of iron, and statements 2 and 3 accurately reflect the ratios of iron deposited in the different compounds.
180 Easy
180 of 381
Remembering +4 | -1
Daniel cell, anode and cathode are respectively
View answer and explanation
Correct option: A
Zn∣Zn2+andCu2+∣Cu\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\mathrm{and}{\kern 1pt} {\kern 1pt} \mathrm{C}{\mathrm{u}^{2 + }}} \right|\mathrm{Cu}
Option A (Zn∣Zn2+andCu2+∣Cu\mathrm{Zn}\left| {\mathrm{Z}{\mathrm{n}^{2 + }}\mathrm{and}{\kern 1pt} {\kern 1pt} \mathrm{C}{\mathrm{u}^{2 + }}} \right|\mathrm{Cu}) is correct because in a Daniel cell, the anode is zinc immersed in a zinc sulfate solution, and the cathode is copper immersed in a copper sulfate solution.
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