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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Mixed difficulty

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271 Hard
271 of 381
Application +4 | -1
A gas XX at 1 atm is bubbled through a solution containing a mixture of 1 M y−{y^ - } and 1 M z−{z^ - } at 25∘C{25^ \circ }C. If the order of reduction potential is z>y>xz > y > x then
View answer and explanation
Correct option: A
yy will oxidize xx and not zz
Option A (yy will oxidize xx and not zz) is correct because y has a higher reduction potential than x, so it can oxidize x, but a lower reduction potential than z, so it cannot oxidize z.
272 Medium
272 of 381
Application +4 | -1
Deduce from the following E∘{E^ \circ } values of half cells, what combination of two half cells would result in a cell with the largest potential?

i)A3−→A2−+e;{A^{3 - }} \to {A^{2 - }} + e; E∘=1.5V{E^ \circ } = 1.5\mathrm{V}

ii) B2++e→B+{B^{2 + }} + e \to {B^ + }; E∘=−2.1V{E^ \circ } = - 2.1\mathrm{V}

iii) C2++e→C+{C^{2 + }} + e \to {C^ + }; E∘=0.5V{E^ \circ } = 0.5\mathrm{V}

iv) D→D2++2eD \to {D^{2 + }} + 2e; E∘=−1.5V{E^ \circ } = - 1.5\mathrm{V}
View answer and explanation
Correct option: C
(ii) and (iv)
Option A (i) and (ii) is correct because the largest potential difference is achieved when the reduction potential is most positive and the oxidation potential is most negative.
273 Medium
273 of 381
Understanding +4 | -1
Eo for F2+2e−→2F− is  2.8 V{E^\mathrm{o}}\mathrm{ \,for }\,{F_2} + 2{e^ - } \to 2{F^ - }\mathrm{ \,is }\,\,2.8\,V , Eo  for   12F2+e−→F−is:{E^\mathrm{o}}\mathrm{ \,\,for}\,\,\,\mathrm{ }\frac{1}{2}{F_2} + {e^ - } \to {F^ - }\mathrm{ is:}
View answer and explanation
Correct option: A
2.8 V
Option A (2.8 V) is correct because the standard reduction potential is an intensive property and does not depend on the stoichiometric coefficients.
274 Medium
274 of 381
Remembering +4 | -1
Which of the following is an additive property?
View answer and explanation
Correct option: A
Conductance
Option A (Conductance) is correct because conductance is an additive property, meaning the total conductance of a solution is the sum of the conductances of its individual components.
275 Hard
275 of 381
Understanding +4 | -1
Assertion (A): DC can not be used during measurement of resistance of an ionic solution.

Reason (R): DC changes the composition of solution and solution cannot be connected to the bridge like metallic wire.
View answer and explanation
Correct option: A
Assertion and Reason both are correct and Reason is the correct explanation of Assertion.
Option A (Assertion and Reason both are correct and Reason is the correct explanation of Assertion) is correct because DC electrolysis changes the ionic solution's composition, and the solution's resistance can't be measured using a Wheatstone bridge with DC.
276 Easy
276 of 381
Application +4 | -1
E0{E^0} for the half cell Zn2+/ZnZ{n^{2 + }}/Zn is −0.76- 0.76, e.m.f of the cell Zn/Zn2+(1M)//2H+(2M)/H2(1 atm)Zn/Z{n^{2 + }}\left( {1M} \right)//2{H^ + }\left( {2M} \right)/{H_2}\left( {1\,{\text{atm}}} \right) is
View answer and explanation
Correct option: B
+0.76V+ 0.76V
Option B is correct. The EMF of the cell is equal to the standard reduction potential of the cathode minus the standard reduction potential of the anode. In this case, the standard hydrogen electrode (SHE) is the cathode, and the zinc electrode is the anode.
277 Hard
277 of 381
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 half-reactions to obtain the desired half-reaction and using the relationship between Gibbs free energy and standard potential.
278 Medium
278 of 381
Application +4 | -1
The potential of the following cell is 0.34 V at 25∘C{25^ \circ }C. Calculate the standard reduction potential of the copper half-cell.

Pt∣H2(1atm)∣H+(1M)∣∣Cu2+(1M)∣Cu\mathrm{Pt}\left| {{\mathrm{H}_2}\left( {1{\kern 1pt} {\kern 1pt} \mathrm{atm}} \right)} \right|{\mathrm{H}^ + }\left( {1\mathrm{M}} \right)|\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {1\mathrm{M}} \right)} \right|\mathrm{Cu}
View answer and explanation
Correct option: D
+0.34 V
Option D (+0.34 V) is correct because the standard reduction potential of the hydrogen electrode is 0 V, so the cell potential equals the copper half-cell potential.
279 Easy
279 of 381
Remembering +4 | -1
The standard H electrode is written as :
View answer and explanation
Correct option: C
PtH2(g)(1atm)/H+(a=1)\mathrm{Pt}{\mathrm{H}_2}\left( \mathrm{g} \right)\left( {1\mathrm{atm}} \right)/{\mathrm{H}^ + }\left( {a = 1} \right)
Option B (Voltmeter) is correct because a voltmeter is used to measure the potential difference between two electrodes, which is how standard electrode potential is determined.
280 Easy
280 of 381
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 is correct. During the electrolysis of aqueous MgSO4\mathrm{MgSO}_4, water is reduced at the cathode to produce H2(g){H_2(g)}, and water is oxidized at the anode to produce O2(g){O_2(g)} because Mg2+{Mg^{2+}} and SO42−{SO_4^{2-}} ions are relatively stable and harder to reduce/oxidize than water.
281 Easy
281 of 381
Analyzing +4 | -1
A volume of 100 ml100 \mathrm{~ml} of a buffer of 1 M−NH31 \mathrm{~M}-\mathrm{NH}_3 and 1 M−NH4+1 \mathrm{~M}-\mathrm{NH}_4^{+} is placed in two half-cells connected by a salt bridge. A current of 1.5 A1.5 \mathrm{~A} is passed through the cell for 20 min.20 \mathrm{~min}. If electrolysis of water takes place only and the electrode reactions are:

Right: 2H2O+O2+4e→4OH−2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2+4 \mathrm{e} \rightarrow 4 \mathrm{OH}^{-}

and Left: 2H2O→4H++O2+4e2 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{H}^{+}+\mathrm{O}_2+4 \mathrm{e}, then, the pH\mathrm{pH} of
View answer and explanation
Correct option: A
Right electrode will increase
Option A (Right electrode will increase) is correct because the electrolysis of water at the right electrode produces hydroxide ions, increasing the pH.
282 Medium
282 of 381
Remembering +4 | -1
Which of the following statements is correct? Galvanic cell converts:
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Correct option: A
Chemical energy into electrical energy
Option A (Chemical energy into electrical energy) is correct because a galvanic cell, also known as a voltaic cell, is an electrochemical cell that converts chemical energy into electrical energy through spontaneous redox reactions.
283 Easy
283 of 381
Analyzing +4 | -1
In the plot of ∧\wedge and C\sqrt C , the slop is
View answer and explanation
Correct option: B
−B- B
Option B (−B- B) is correct because in the Debye-Hückel-Onsager equation, the slope of the plot of molar conductivity vs. square root of concentration is represented by -B.
284 Hard
284 of 381
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
+0.38 V
Option C (+0.38 V) is correct because it is calculated by combining the standard reduction potentials of the given half-reactions.
285 Easy
285 of 381
Analyzing +4 | -1
Electrolysis of aq. Solution of LiCl\mathrm{LiCl} shows :
View answer and explanation
Correct option: C
pH>7\mathrm{pH} > 7
Option C (pH>7\mathrm{pH} > 7) is correct because the electrolysis of aqueous LiCl produces H2H_2 at the cathode and Cl2Cl_2 at the anode, leaving behind OH−OH^- ions, which increases the pH.
286 Easy
286 of 381
Understanding +4 | -1
The amount of an ion discharged during electrolysis is not dependent of :
View answer and explanation
Correct option: A
Resistance of solution
Option A is correct because the amount of an ion discharged during electrolysis depends on the current, time, and electrochemical equivalent, but not directly on the resistance of the solution.
287 Hard
287 of 381
Application +4 | -1
What is the quantity of electricity (in Coulombs) required to deposit all the silver from 250mL of 1 MAgNO3\mathrm{MAgN}{\mathrm{O}_3} solution?
View answer and explanation
Correct option: B
24125
Option A (2412.5) is correct because the quantity of electricity required is calculated using the molarity, volume, and Faraday's constant.
288 Medium
288 of 381
Understanding +4 | -1
The EM3+/M2+∘E_{{M^{3 + }}/{M^{2 + }}}^ \circ values for Cr, Mn, Fe and Co are -0.41 V, +1.57 V, +0.77 V and +1.97 V respectively. For which one of these metals the change in oxidation state from +2 to +3 is easiest?
View answer and explanation
Correct option: A
Cr
Option A (Cr) is correct because the most negative EM3+/M2+∘E_{{M^{3 + }}/{M^{2 + }}}^ \circ value indicates the easiest oxidation of M2+M^{2+} to M3+M^{3+}.
289 Hard
289 of 381
Application +4 | -1
The emf of the cell

Ni∣Ni2+(1.0M)∣∣Au3+(1.0M)∣Au\mathrm{Ni}\left| {\mathrm{N}{\mathrm{i}^{2 + }}\left( {1.0\mathrm{M}} \right)} \right|\left| {\mathrm{A}{\mathrm{u}^{3 + }}\left( {1.0\mathrm{M}} \right)} \right|\mathrm{Au}

is [E∘(Ni2+/Ni)={\mathrm{E}^ \circ }\left( {\mathrm{N}{\mathrm{i}^{2 + }}/\mathrm{Ni}} \right) = -0.25 V and

E∘(Au3+/Au)={E^ \circ }\left( {\mathrm{A}{\mathrm{u}^{3 + }}/\mathrm{Au}} \right) = +1.5 V]
View answer and explanation
Correct option: D
1.75 V
Option D (1.75 V) is correct because it is the difference between the reduction potentials of the cathode and anode.
290 Easy
290 of 381
Remembering +4 | -1
A cell in which electric current is produced by net oxidation and reduction process is called :
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Correct option: A
Voltaic cell
Option A (Voltaic cell) is correct because a voltaic cell produces electric current through spontaneous redox reactions.
291 Easy
291 of 381
Analyzing +4 | -1
A substance that will reduce Ag+\mathrm{A}{\mathrm{g}^ + } to Ag but will not reduce Ni2+\mathrm{N}{\mathrm{i}^{2 + }} to Ni is :
View answer and explanation
Correct option: B
Pb
Option B (Pb) is correct because Pb has a standard reduction potential between that of Ag+ and Ni2+, allowing it to reduce Ag+ but not Ni2+.
292 Hard
292 of 381
Analyzing +4 | -1
The equation that is incorrect is
View answer and explanation
Correct option: C
(Λm0)NaBr −  (Λm0)NaI =  (Λm0)KBr− (Λm0)NaBr{(\Lambda _m^0)_{NaBr}}\, - \,\,{(\Lambda _m^0)_{NaI}}\, = \,\,{(\Lambda _m^0)_{KBr}} - \,{(\Lambda _m^0)_{NaBr}}
Option C is correct because the difference between the molar conductivities of NaBr and NaI should be equal to the difference between KBr and KI, not NaBr.
293 Medium
293 of 381
Analyzing +4 | -1
Which does not get oxidised by bromine water?
View answer and explanation
Correct option: C
Mn2+toMnO4−\mathrm{M}{\mathrm{n}^{2 + }}{\rm{to {\kern 1pt} {\kern 1pt}MnO}}_4^ -
Option C (Mn2+toMnO4−\mathrm{M}{\mathrm{n}^{2 + }}{\rm{to {\kern 1pt} {\kern 1pt}MnO}}_4^ -) is correct because the oxidation of Mn2+ to MnO4- requires a strong oxidizing agent, which bromine water is not.
294 Easy
294 of 381
Remembering +4 | -1
Which of the following expression is correct?
View answer and explanation
Correct option: A
ΔG∘=−nFEcell∘\Delta {G^ \circ } = - nFE_{\mathrm{cell}}^ \circ
Option A (ΔG∘=−nFEcell∘\Delta {G^ \circ } = - nFE_{\mathrm{cell}}^ \circ) is the correct expression relating the standard Gibbs free energy change to the standard cell potential.
295 Medium
295 of 381
Understanding +4 | -1
Consider the reaction, Mn+(aq)+ne→M∘(s).{M^{n + }}\left( {aq} \right) + ne \to {M^ \circ }\left( s \right). The standard reduction potential values of the metals M1,M2{M_1},{M_2} and M3{M_3} are −0.34V,−3.05V- 0.34\mathrm{V}, - 3.05\mathrm{V} and −1.66V- 1.66\mathrm{V} respectively. The order of their reducing power will be :
View answer and explanation
Correct option: D
M2>M3>M1{M_2} > {M_3} > {M_1}
Option D is correct because the reducing power of a metal is inversely related to its standard reduction potential. The metal with the most negative standard reduction potential is the strongest reducing agent. Therefore, the order of reducing power is M2 > M3 > M1.
296 Hard
296 of 381
Application +4 | -1
How long (in hours) must a current of 5.0 A be maintained to electroplate 60 g of calcium from molten CaCl2\mathrm{CaC}{\mathrm{l}_2} ?
View answer and explanation
Correct option: D
16 h
Option D (16 h) is correct because it is the calculated time required to electroplate 60 g of calcium using Faraday's laws of electrolysis.
297 Hard
297 of 381
Understanding +4 | -1
E∘{E^ \circ } for Fe2++2e−→Fe\mathrm{F}{\mathrm{e}^{2 + }} + 2{\mathrm{e}^ - } \to \mathrm{Fe} is -0.44 V and E∘{E^ \circ } for

Zn2++2e−→Zn\mathrm{Z}{\mathrm{n}^{2 + }} + 2{\mathrm{e}^ - } \to \mathrm{Zn} is -0.76 V thus
View answer and explanation
Correct option: A
Zn is more electropositive than Fe
Option A (Zn is more electropositive than Fe) is correct because a more negative standard reduction potential indicates a greater tendency to be oxidized, meaning it is more electropositive.
298 Medium
298 of 381
Remembering +4 | -1
The reactions taking place at anode and cathode are
View answer and explanation
Correct option: A
Oxidation and reduction
Option A (Oxidation and reduction) is correct because oxidation always occurs at the anode, and reduction always occurs at the cathode in any electrochemical cell.
299 Easy
299 of 381
Application +4 | -1
Eo{E^\mathrm{o}} for the half cell, Zn2+/ZnZ{n^{2 + }}/Zn is −0.76- 0.76, then emf of the cell Zn/Zn2+(1M)//2H+(1M)/H2(1  atm)Zn/Z{n^{2 + }}\left( {1M} \right)//2{H^ + }\left( {1M} \right)/{H_2}\left( {1\,\,\mathrm{atm}} \right) is
View answer and explanation
Correct option: B
+ 0.76 V+ \,0.76\,V
Option B (+ 0.76 V) is correct because the cell potential is calculated by reversing the sign of the zinc half-cell potential since it's being oxidized.
300 Hard
300 of 381
Understanding +4 | -1
Given below are two statements.
Statement I The limiting molar conductivity of KClKCl (strong electrolyte) is higher compared to that of CH3COOHC{H_3}COOH (Weak electrolyte).
Statement II Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below.
View answer and explanation
Correct option: D
Both Statement I and statement II are false.
The correct answer is D. Statement I is false because strong electrolytes like KCl have higher limiting molar conductivity than weak electrolytes like CH3COOH. Statement II is false because molar conductivity increases with a decrease in concentration, especially for weak electrolytes.
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