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NEET • CHEMISTRY • ELECTRO CHEMISTRY

Electro chemistry practice · Medium

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91 Medium
91 of 145
Application +4 | -1
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 4 and passing hydrogen gas around the platinum wire at one atm pressure and at 298 K. The reduction potential of the hydrogen electrode would be
View answer and explanation
Correct option: D
– 0.236 V
Option D (– 0.236 V) is correct because it is calculated using the Nernst equation, considering the pH and the standard hydrogen electrode potential.
92 Medium
92 of 145
Understanding +4 | -1
A cell with two electrodes, one of grey tin and the other white tin, both dipping in solution of (NH4)2SnCl6{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{SnC}{\mathrm{l}_6} showed zero e.m.f. at 18∘C{18^ \circ }C. What conclusion may be drawn from this?
View answer and explanation
Correct option: D
The standard Gibbs energy change of the cell becomes zero
Option D is correct because at zero e.m.f., the standard Gibbs energy change, which is directly related to the cell potential, must also be zero.
93 Medium
93 of 145
Understanding +4 | -1
For a given cell reaction; Cr+3H2O+OCl−→Cr+3+3Cl−+6OH−,Cr + 3{H_2}O + OC{l^ - } \to C{r^{+3}}+3C{l^ - } + 6O{H^ - }, the species undergoing reduction is :
View answer and explanation
Correct option: C
OCl−OC{l^ - }
Option C (OCl−OC{l^ - }) is correct because in the given reaction, the oxidation state of chlorine in OCl−OC{l^ - } decreases, indicating reduction.
94 Medium
94 of 145
Analyzing +4 | -1
A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the following solutions should the other electrode be immersed to get maximum e. m. f.?
View answer and explanation
Correct option: D
0.1MH2SO40.1M{\kern 1pt} {H_2}S{O_4}
Option D (0.1MH2SO40.1M{\kern 1pt} {H_2}S{O_4}) is correct because it will produce the highest concentration of H+ ions, leading to the largest potential difference compared to the standard hydrogen electrode.
95 Medium
95 of 145
Remembering +4 | -1
Which equation gives relation between concentration of ions is solution, electrode potential (EE) and standard electrode potential (E0)\left(E^0\right)
View answer and explanation
Correct option: B
Nernst's equation
Option B (Nernst's equation) is the correct answer because it relates the concentration of ions in solution to the electrode potential and standard electrode potential.
96 Medium
96 of 145
Application +4 | -1
A silver cup is plated with silver by passing 965 C of electricity. The amount of Ag deposited is
View answer and explanation
Correct option: D
1.08 g
Option D (1.08 g) is correct because it is calculated using Faraday's law of electrolysis: mass = (equivalent weight * charge) / Faraday's constant.
97 Medium
97 of 145
Application +4 | -1
When during electrolysis of a solution of a AgNO3,9650C\mathrm{AgN}{\mathrm{O}_3},9650\mathrm{C} of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be
View answer and explanation
Correct option: B
10.8 g
Option B (10.8 g) is correct because the mass of silver deposited is calculated using Faraday's laws and the given charge.
98 Medium
98 of 145
Application +4 | -1
The standard electrode potential for the half – cell reactions are
Zn2++2e−→Zn;E∘\mathrm{Z}{\mathrm{n}^{2 + }} + 2{e^ - } \to \mathrm{Zn};{E^ \circ } = -0.76 V
Fe2++2e−→Fe;E∘\mathrm{F}{\mathrm{e}^{2 + }} + 2{e^ - } \to \mathrm{Fe};{E^ \circ } = -0.44 V
The emf of the cell reaction,
Fe2++Zn→Zn2++Fe\mathrm{F}{\mathrm{e}^{2 + }} + \mathrm{Zn} \to \mathrm{Z}{\mathrm{n}^{2 + }} + \mathrm{Fe} is
View answer and explanation
Correct option: D
+0.32 V
Option D (+0.32 V) is correct because it is the calculated EMF of the cell using the standard reduction potentials.
99 Medium
99 of 145
Application +4 | -1
On passing electricity through dil. H2SO4{H_2}S{O_4} solution, the amount of substance liberated at the cathode and anode are in the ratio
View answer and explanation
Correct option: A
0.047222222
Option A (0.047222222) is correct because it represents the ratio of the amount of hydrogen liberated at the cathode to the amount of oxygen liberated at the anode during the electrolysis of dilute H2SO4.
100 Medium
100 of 145
Application +4 | -1
The emf of the cell, (EZn2+/Zn=−0.76V)\left( {{E_{\mathrm{Z}{\mathrm{n}^{2 + }}/\mathrm{Zn}}} = - 0.76V} \right)

Zn/Zn2+(1M)∣∣Cu2+(1M)∣Cu\mathrm{Zn}/\mathrm{Z}{\mathrm{n}^{2 + }}\left( {1\mathrm{M}} \right)|\left| {\mathrm{C}{\mathrm{u}^{2 + }}\left( {1\mathrm{M}} \right)} \right|\mathrm{Cu}

(ECu2+/Cu=+0.34V)\left( {{E_{\mathrm{C}{\mathrm{u}^{2 + }}/\mathrm{Cu}}} = + 0.34V} \right) will be
View answer and explanation
Correct option: A
+1.10 V
Option A is correct. The EMF of the cell is calculated by summing the standard oxidation potential of the zinc electrode and the standard reduction potential of the copper electrode.
101 Medium
101 of 145
Analyzing +4 | -1
Statement (S): A graph plotted between Λv vs. c1/2{\Lambda _v}\,vs.\,{c^{1/2}} gives a straight line with slope equal to −A- A and intercept equal to Λ∞{\Lambda _\infty }
Explanation (E): The value of AA for a given solvent and temperature depends upon the nature of electrolyte, i.e., charge on cation or anion furnished by electrolyte on dissolution.
View answer and explanation
Correct option: D
Both S and E are correct and E is not correct explanation of S.
Option D (Both S and E are correct and E is not correct explanation of S) is correct because both the statement about the graph and the explanation about A's dependence are true, but the explanation doesn't directly explain the statement.
102 Medium
102 of 145
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.
103 Medium
103 of 145
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.
104 Medium
104 of 145
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.
105 Medium
105 of 145
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.
106 Medium
106 of 145
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.
107 Medium
107 of 145
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.
108 Medium
108 of 145
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.
109 Medium
109 of 145
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.
110 Medium
110 of 145
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.
111 Medium
111 of 145
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.
112 Medium
112 of 145
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.
113 Medium
113 of 145
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.
114 Medium
114 of 145
Remembering +4 | -1
Which of the following statements is correct? Galvanic cell converts:
View answer and explanation
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.
115 Medium
115 of 145
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+}.
116 Medium
116 of 145
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.
117 Medium
117 of 145
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.
118 Medium
118 of 145
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.
119 Medium
119 of 145
Application +4 | -1
A current of strength 2.5 A was passed through CuSO4\mathrm{CuS}{\mathrm{O}_4} solution for 6 min 26 s. The amount of copper deposited is (At. Wt. of Cu = 63.5, 1F = 96500 C)
View answer and explanation
Correct option: A
0.3175 g
Option A (0.3175 g) is correct because the mass of copper deposited is calculated using Faraday's laws, relating current, time, and the electrochemical equivalent of copper.
120 Medium
120 of 145
Analyzing +4 | -1
A certain metal fails to liberate H2{\mathrm{H}_2} gas from a moderately conc. HCl solution. However, it displaces Ag\mathrm{Ag} from AgNO3\mathrm{AgN}{\mathrm{O}_3} solution. Which among the followings may it be?
View answer and explanation
Correct option: C
Cu\mathrm{Cu}
Option C (Cu) is correct because Cu is less reactive than H and cannot liberate H2 from HCl, but it is more reactive than Ag and can displace it from AgNO3 solution.
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