Electrochemistry JEE Main previous year questions and solutions are given in this article. The branch of chemistry that deals with the chemical changes produced by electric current and the electricity generation by chemical changes is known as electrochemistry. An electrochemical cell consists of two metallic conductors known as electrodes in contact with an ionic conductor i.e. an electrolyte. The electrolyte and the electrode comprise an electrode compartment. Electrolytic cells are those in which a non-spontaneous reaction is controlled by an external source of electric current. This article helps students to revise the previous year questions and score higher rank in the JEE exam.
As far as the JEE exam is concerned, electrochemistry is an important topic. The important topics include electrochemical cell, electrolytic cell, Faraday’s laws of electrolysis, Kohlrausch’s Law, electrode potential, etc. The questions given here give you an idea about what type of questions to expect from this topic. BYJU’S provides accurate solutions created by our subject experts. Students can easily download these solutions in PDF format for free.
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JEE Main Previous Year Solved Questions on Electrochemistry
1. In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to
(a) generate heat
(b) create a potential difference between the two electrodes
(c) produce high purity water
(d) remove adsorbed oxygen from the electrode surface.
Solution:
In a hydrogen-oxygen fuel cell, the combustion of hydrogen occurs because of the difference in potential between two electrodes.
Hence option (b) is the answer.
2. Conductivity (unit Siemen’s S) is directly proportional to the area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel then the unit of the constant of proportionality is
(a) Sm mol-1
(b) Sm2 mol-1
(c) S-2m2 mol1
(d) S2m2 mol1
Solution:
S = Km2mol/ m ×m3
K = Sm2 mol-1
Hence option (b) is the answer.
3. Galvanization is applying a coating of
(a) Pb
(b) Cr
(c) Cu
(d) Zn
Solution:
Galvanization is a method of preventing rust by applying zinc coating. Zinc acts as a sacrificial metal.
Hence option (d) is the answer.
4. Identify the correct statement.
(a) Corrosion of iron can be minimized by forming contact with another metal with a higher reduction potential.
(b) Iron corrodes in oxygen-free water.
(c) Corrosion of iron can be minimized by forming an impermeable barrier at its surface.
(d) Iron corrodes more rapidly in saltwater because its electrochemical potential is higher.
Solution:
Corrosion of iron can be minimized by forming an impermeable barrier at its surface.
Hence option (c) is the answer.
5. A solution of Ni(NO3 )2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many moles of Ni will be deposited at the cathode?
(a) 0.10
(b) 0.15
(c) 0.20
(d) 0.05
Solution:
Ni2+ + 2e– → Ni
2 F deposits 1 mole of Ni.
So 0.1 F will deposit (0.1/2) moles of Ni.
(0.1/2) = 0.05 mole.
Hence option (d) is the answer.
6. EMF of a cell in terms of reduction potential of its left and right electrodes is
(a) E = Eleft – Eright
(b) E = Eleft + Eright
(c) E = Eright – Eleft
(d) E = –(Eright + Eleft)
Solution:
Ecell = Reduction potential of the cathode – Reduction potential of the anode
= Eright – Eleft
Hence option (c) is the answer.
7. For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be
(a) 1 × 10–10
(b) 29.5 × 10–2
(c) 10
(d) 1 × 1010
Solution:
Given E0cell = 0.295 V
E0cell = (0.0591/n) log Kc
n = 2
0.295 = (0.0591/2)log Kc
0.295 = 0.0295 log Kc
log Kc = 0.295/0.0295 = 10
Kc = Antilog 10 = 1010
Hence option (d) is the answer.
8. If φ denotes reduction potential, then which is true?
(a) E°cell = φright – φleft
(b) E°cell = φleft + φright
(c) E°cell = φleft – φright
(d) E°cell = –(φleft +φright)
Solution:
E°cell = φright – φleft
Hence option (a) is the answer.
9. The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is [Molar mass of PbSO4 = 303 g mol–1]
(a) 15.2
(b) 11.4
(c) 7.6
(d) 22.8
Solution:
PbSO4(s) + 2OH– → PbO2 + H2SO4 + 2e–
PbSO4 + 2e– + 2H+ → Pb(s) + H2SO4
Total number of moles of PbSO4 = 0.05/2 = 0.025
Mass of PbSO4 = 0.025 × 303 = 7.575g
Hence option (c) is the answer.
10. Calculate the standard cell potential (in V) of the cell in which the following reaction takes place. Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag(s)
Given that E°Ag+/Ag = x V; E°Fe2+/Fe = y V; E°Fe3+/Fe = z V
(a) x – z
(b) x – y
(c) x + 2y – 3z
(d) x + y – z
Solution:
Fe2+(aq) + Ag+(aq) → Fe3+(aq)+ Ag(s)
Fe2+ + 2e – → Fe
E°Fe2+/Fe = y V
∆G°1 = –2Fy …..(i)
Fe3+ + 3e – → Fe
E°Fe3+/Fe = z V
∆G°2 = –3Fz …(ii)
Ag+ + e– → Ag
E° = +x V
∆G°3 = –Fx …(iii)
Adding equations (i) and (iii) and subtracting (ii) from it, we get
Fe2+(aq) + Ag+(aq) → Ag(s) + Fe3+(aq)
∆G = ∆G°1 + ∆G° 3 – ∆G°2
–FE°cell = –2Fy – Fx – (–3Fz)
–E°cell = –2y – x + 3z
E°cell = x + 2y – 3z
Hence option (c) is the answer.
11. The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s), E ° = 2 V at 298 K (Faraday’s constant, F = 96000 C mol–1) (a) –192
(b) 192
(c) –384
(d) 384
Solution:
Given E0 = 2
F = 96000
n = 2
∆G° = –nFE° = –2 × 96000 × 2
= –384000 J/mol
= –384 kJ/mol
Hence option (c) is the answer.
12. Consider the statements S1 and S2:
S1: Conductivity always increases with decrease in the concentration of electrolyte.
S2: Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is
(a) both S1 and S2 are wrong
(b) both S1 and S2 are correct
(c) S1 is wrong and S2 is correct
(d) S1 is correct and S2 is wrong.
Solution:
Conductivity increases with an increase in concentration.
Molar Conductivity increases with decrease in concentration.
Hence option (c) is the answer.
13. If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction,
Zn(s) + Cu2+ (aq) Zn2+ (aq) + Cu(s)
at 300 K is approximately (R = 8 J K–1 mol–1, F = 96500 C mol–1)
(a) e–80
(b) e 160
(c) e –160
(d) e 320
Solution:
Given T = 300 K
∆G° = – nFE°cell
= -2 × 96000 × 2
= -384000 C V mol–1
∆G° = – RT ln K
-384000 = -8 ×300 ln K
ln K = 384000/2400 = 160
So K = e160
Hence option (b) is the answer.
14. When during electrolysis of a solution of AgN03 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be
(a) 10.8 g
(b) 21.6 g
(c) 108 g
(d) 1.08 g
Solution:
No. of moles of Ag = 9650/96500 = 0.1
Mass of Ag deposited = 0.1× 108 = 10.8
Hence option (a) is the answer.
15. The correct order of E0M2+/M values with negative sign for the four successive elements Cr, Mn, Fe and Co is
(a) Mn > Cr > Fe > Co
(b) Cr > Fe > Mn> Co
(c) Fe > Mn > Cr > Co
(d) Cr > Mn > Fe > Co
Solution:
The negative values for standard electrode potential decrease except for Mn because of the stable d5 configuration, across the first transition series.
Mn > Cr > Fe > Co
Hence option (a) is the answer.
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