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JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

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Q.1. Given that the standard potentials (E°) and Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the E° of Cu2+/Cuis    (2020)
(1) 0.182 V
(2) +0.158 V
(3) −0.182 V
(4) −0.158 V

Ans. (2) 

Given,
Cu2+ + 2e- → Cu; E° 0.34 V       (1)
Cu+ + e→ Cu; E° = 0.522 V      (2)
For Cu2+ + e- → Cu+ ; E°Cu2+/Cu+ x V
On reversing Eq. (2), we get
Cu → Cu+ + e- ; E° = - 0.522 V      (3)
From Eq. (1) and Eq. (3), we get
ΔG°= ΔG°1 + ΔG°2
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
⇒ E°3 = 2 x 0.34 + (- 0.522)
⇒ E°3 = 0.68 - 0.522 = 0.158 V

Q.2. The equation that is incorrect is:    (2020)
(1)JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(2)JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(3)JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(4)JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Ans. 
(4) 

According to Kohlrausch law of independent migration of ions, the incorrect equation is:
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
From L.H.S
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
From R.H.S
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Hence, L.H.S is not equal to R.H.S.

Q.3. What would be the electrode potential for the given half-cell reaction at pH = 5?
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(R = 8.314 J and mol−1 K−1; T = 298 K; oxygen under standard atm pressure of 1 bar)    (2020)
Ans. (1.525)

 For the half-cell reaction,
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE    (1)
Substituting the values in Eq. (1), we get
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
⇒ Ecell = + 1.23 + 0.059 x 5    [Given: pH = 5]
⇒ Ecell = 1.525 V

Q.4. For an electrochemical cell
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEEthe ratioJEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEEwhen this cell attains equilibrium is _______.
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE    (2020)

Ans. (2.15)

For the given electrochemical cell
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
= 0.13 - (-0.14) = 0.01 V
From the Nernst equation for given electrochemical cell
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE    (1)
Substituting the given values in Eq. (1), we get
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE [At Equilibrium, Ecell = 0]
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE    (2)
Taking antilog of Eq. (2), we get
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.5. 108 g of silver (molar mass 108 g mol-1) is deposited at cathode from AgNO3(aq) solution by a certain quantity of electricity. The volume (in L) of oxygen gas produced at 273 K and 1 bar pressure from water by the same quantity of electricity is ________ .    (2020)
Ans. (5.67) 

The moles of AgNO3 deposited at cathode
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
At anode: JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
At cathode:JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Thus, 1 F required to deposit 1 mol of Ag and 2 F required to produce 0.5 mol of oxygen gas.
⇒ 1 F will produce 0.25 mol of oxygen gas.
Thus, the volume of oxygen gas produced at 273 K and 1 bar pressure can be calculated using ideal gas equation.
⇒ pV = nRT
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.6. Amongst the following, the form of water with the lowest ionic conductance at 298 K is    (2020)
(1) distilled water
(2) saline water used for intravenous injection
(3) water from a well
(4) sea water

Ans. (1) 

Due to the absence of the mineral in the distilled water, the ionic conductance is minimum among the all liquid forms of water.

Q.7. The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed (in gm) during the process is: (Molar mass of PbSO4 = 303 g mol-1)    (2019)
(1) 22.8
(2) 15.2
(3) 7.6
(4) 11.4
Ans.
(3)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.8. 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= 8JK-1 mol-1, F= 96000 C mol-1)    (2019)
(1) e-80
(2) e-160
(3) e320
(4) e160
Ans.
(4)
We know that, JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
After putting the given values, we get
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
∴ K = e160

Q.9. In the cell
Pt(s)|H2(g, lbar) / HCl(aq)||AgCl(s)/Ag(s)|Pt(s), the cell potential is 0.92 V when a 10-6 molal HCl solution is used. The standard electrode potential of (AgCl/Ag, Cl-) electrode is:    (2019)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(1) 0.94 V
(2) 0.76 V
(3) 0.40 V 
(4) 0.20 V
Ans.
(4)
Given that:
Pt(s)|H2(g, lbar) / HCl(aq)||AgCl(s)/Ag(s)|Pt(s)
Ecell = 0.92 V
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Net cell reaction:
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.10. For the cell Zn(s) | Zn2+(aq) || Mx+ (aq) | M(s), different half cells and their standard electrode potentials are given below:
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
IfJEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE, which cathode will give a maximum value of E°cell per electron transferred?    (2019)
(1) Ag+/Ag
(2) Fe3+/Fe2+
(3) Au3+/Au
(4) Fe2+/Fe
Ans.
(1)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.11. Given the equilibrium constant:
KC of the reaction:
Cu(s) + 2Ag+ (aq) → Cu2+ (aq) + 2Ag(s) is
10 x 1015. Calculate the E°cell of this reaction at 298 K.    (2019)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(1) 0.04736 mV
(2) 0.4736 mV
(3) 0.4736 V
(4) 0.04736 V
Ans.
(3)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.12. The standard electrode potential E° and its temperature coefficient JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE for a cell are 2 V and -5 x 10-4 VK-1 at 300 K respectively. The cell reaction is:
Zn(s) + Cu2+(aq) → Zn2+ (aq) + Cu
The standard reaction enthalpy (ΔrH°) at 300 K in kJ mol-1 is,
[Use R= 8 JK-1 mol-1 and F = 96,000 C mol-1]    (2019)
(1) -412.8
(2) -384.0
(3) 192.0
(4) 206.4
Ans. 
(1)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.13. Λ0m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 cm2 mol-1, respectively. If the conductivity of 0.001 M HA is 5 x 10-5 Scm-1, degree of dissociation of HA is:    (2019)
(1) 0.50
 (2) 0.25
(3) 0.125
(4) 0.75
Ans. 
(3)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.14. Given thatJEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
The strongest oxidising agent is:    (2019)
(1) Au3+ 
(2) O2
(3) S2O82- 
(4) Br2
Ans.
(3)
More positive is the reduction potential stronger is the oxidising agent. Reduction potential is maximum for S2O82-. Therefore, it is the strongest oxidising agent amongst the given species.

Q.15. Calculate the standard cell potential (in V) of the cell in which following reaction takes place:
Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag(s)
Given that    (2019)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE 
(1) x - z
(2) x - y
(3) x + 2y - 3z 
(4) x + y - z
Ans.
(3)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE (given)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.16. 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)    (2019)
(1) - 384
(2) 384
(3) 192
(4) - 192
Ans.
(1)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
= -2 (96000) 2 V = - 384000 J/mol
= -384 kJ/mol

Q.17. 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?    (2019)
(1) 0.05
(2) 0.20
(3) 0.15
(4) 0.10
Ans.
(1)
According to the Faraday’s law of electrolysis, nF of current is required for the deposition of 1 mol According to the reaction,
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
2 F of current deposits = 1 mol
∴ 0.1 F of current deposits = 0.1/2 = 0.05 mol

Q.18. 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:    (2019)
(1) Both S1 and S2 are wrong
(2) S1 is wrong and S2 is correct
(3) Both S1 and S2 are correct
(4) S1 is correct and S2 is wrong
Ans.
(2)
Conductivity of an electrolyte is the conductance of 1 cm3 of the given electrolyte. It increases with the increase in concentration of electrolyte due to increase in the number of ions per unit volume. Molar conductivity (λm) is the conductance of a solution containing 1 mole of the electrolyte. It increases with the decrease of concentration due to increase in the total volume having one mole of electrolyte. Thus, interionic attraction increases and degree of ionisation decreases. Therefore, (S1) is wrong and (S2) is correct.

Q.19. Which one of the following graphs between molar conductivity (Λm) versus √C is correct?    (2019)
(1) JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

(2) JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(3) JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
(4) JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Ans. (2)
Both NaCl and KCl are strong electrolytes and as Na+(aq⋅) has less conductance than K+(aq⋅) due to more hydration. Therefore, the graph of option (2) is correct.

Q.20. Given :
Co3+ + e- → Co2+; E° = + 1.81 V
Pb4+ + 2e- → Pb2+; E° = + 1.67 V
Ce4+ + e- → Ce3+; E° = + 1.61 V
Bi3+ + 3e- → Bi; E° = + 0.20 V
oxidizing power of the species will increase in the order:    (2019)
(1) Ce4+ < Pb4+ < Bi3+ < Co3+
(2) Bi3+ < Ce4+ < Pb4+ < Co3+
(3) Co3+ < Ce4+ < Bi3+ < Pb4+
(4) Co3+ < Pb4+ < Ce4+ < Bi3+
Ans. 
(2)
Higher the reduction potential, higher will be oxidising power. So, Bi3+ < Ce4+ < Pb4+ < Co3+

Q.21. How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane?    (2018)
(Atomic weight of B = 10.8 u)
(1) 6.4 hours
(2) 0.8 hours
(3) 3.2 hours
(4) 1.6 hours
Ans.
(3)
B2H6 + 3O2 →  B2O3 + 3H2O
According to balanced equation:
27.66 g B2H6 i.e. 1 mole B2H6 requires 3 mole of O2. Now this oxygen is produced by electrolysis of water.
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
1 mole O2 is produced by 4 F charge
∴ 3 mole O2 will be produced by 12 F charge
∴ Now applying
Q = It
12 x 96500 C = 100 x t(s)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
t = 3.2 hours

Q.22. When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is:    (2018)
(1) 0.1
(2) 0.5
(3) 1.0
(4) 2.0
Ans.
(3)

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Q.23. When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is:    (2018)
(1) 109.0 g
(2) 98.1 g
(3) 9.81 g
(4) 10.9 g
Ans.
(3)

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

(v.f.) = 4 JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

W = 9.81 g


Q.24. Given
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Among the following, the strongest reducing agent is    (2017)
(1) Cr
(2) Mn2+
(3) Cr3+
(4) Cl-
Ans.
(1)

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Positive E° is for Cr, hence it is strongest reducing agent.


Q.25. Consider the following standard electrode potentials (E° in volts) in aqueous solution:
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE
Based on these data, which of the following statements is correct?    (2017)
(1) Tl3+ is more stable than Al3+
(2) Al+ is more stable than Al3+
(3) Tl3+ is more stable than Al+
(4) Tl+ is more stable than Al+
Ans. 
(4)

ΔG is -ve


Q.26. What is the standard reduction potential (E°) for Fe3+ → Fe?
Given that:    (2017)
JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE 
(1) –0.057 V
(2) + 0.30 V
(3) – 0.30 V
(4) + 0.057 V
Ans. 
(1)

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

(i) Fe2+ + 2e- → Fe; E° = -0.47V;

(ii) Fe3+ + e- → Fe2+; E° = +0.77V;

(iii) Fe3+ + 3e- → Fe

ΔG° = -nFE° = -2(-0.47)F = 0.94F

ΔG° = -nFE° = -1(-0.77)F = 0.77F

On adding : ΔG° = +0.17F

ΔG° = -nFE°

E° for (Fe3+ → Fe)

= JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE 


Q.27. To find the standard potential of M3+/M electrode, the following cell is constituted: Pt/M/M3+ (0.001 mol L-1)/Ag+ (0.01 mol L-1)/Ag
The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction M3+ + 3e- → M at 298 K will be: (Given EΘAg+/ Ag at 298 K = 0.80 volt)    (2017)
(1) 0.38 volt
(2) 1.28 volt
(3) 0.32 volt
(4) 0.66 volt
Ans.
(3)

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE


Q.28. Galvanization is applying a coating of:    (2016)
(1) Pb
(2) Cr
(3) Cu
(4) Zn
Ans. 
(4)

Galvanization is applying a coating of zinc.

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE


Q.29. What will occur if a block of copper metal is dropped into a beaker containing a solution of 1M ZnSO4?    (2016)
(1) The copper metal will dissolve and zinc metal will be deposited.
(2) The copper metal will dissolve with evolution of oxygen gas.
(3) The copper metal will dissolve with evolution of hydrogen gas.
(4) No reaction will occur.
Ans.
(4)

If a block of copper metal is dropped into a beaker containing solution of 1 M ZnSO4, no reaction will occur because

JEE Main Previous year questions (2016-20): Electrochemistry Notes | Study Chemistry for JEE - JEE

Hence Cu can't displace Zn from ZnSO4 solution.


Q.30. Identify the correct statement:    (2016)
(1) Corrosion of iron can be minimized by forming an impermeable barrier at its surface.
(2) Iron corrodes in oxygen-free water.

(3) Iron corrodes more rapidly in salt water because its electrochemical potential is higher.
(4) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential.
Ans. 
(1)

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