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Q.1. Given that the standard potentials (E°) and Cu^{2+}/Cu and Cu^{+}/Cu are 0.34 V and 0.522 V respectively, the E° of Cu^{2+}/Cu^{+ }is (2020)
(1) 0.182 V
(2) +0.158 V
(3) −0.182 V
(4) −0.158 V
Ans. (2)
Given,
Cu^{2+} + 2^{e }→ Cu; E° 0.34 V (1)
Cu^{+} + e^{ }→ Cu; E° = 0.522 V (2)
For Cu^{2+} + e^{} → Cu^{+} ; E°_{Cu}^{2+}/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°_{3 }= ΔG°_{1} + ΔG°_{2}
⇒ 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)
(2)
(3)
(4)
Ans. (4)
According to Kohlrausch law of independent migration of ions, the incorrect equation is:
From L.H.S
From R.H.S
Hence, L.H.S is not equal to R.H.S.
Q.3. What would be the electrode potential for the given halfcell reaction at pH = 5?
(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 halfcell reaction,
(1)
Substituting the values in Eq. (1), we get
⇒
⇒ E_{cell} = + 1.23 + 0.059 x 5 [Given: pH = 5]
⇒ E_{cell} = 1.525 V
Q.4. For an electrochemical cell
the ratiowhen this cell attains equilibrium is _______.
(2020)
Ans. (2.15)
For the given electrochemical cell
= 0.13  (0.14) = 0.01 V
From the Nernst equation for given electrochemical cell
(1)
Substituting the given values in Eq. (1), we get
[At Equilibrium, E_{cell} = 0]
(2)
Taking antilog of Eq. (2), we get
Q.5. 108 g of silver (molar mass 108 g mol^{1}) is deposited at cathode from AgNO_{3}(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
At anode:
At cathode:
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
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 halfcell of leadacid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO_{4} electrolyzed (in gm) during the process is: (Molar mass of PbSO_{4} = 303 g mol^{1}) (2019)
(1) 22.8
(2) 15.2
(3) 7.6
(4) 11.4
Ans. (3)
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) + Cu^{2+} (aq) ⇌ Zn^{2+} (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) e^{320}
(4) e^{160}
Ans. (4)
We know that,
After putting the given values, we get
∴ K = e^{160}
Q.9. In the cell
Pt(s)H_{2}(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)
(1) 0.94 V
(2) 0.76 V
(3) 0.40 V
(4) 0.20 V
Ans. (4)
Given that:
Pt(s)H_{2}(g, lbar) / HCl(aq)AgCl(s)/Ag(s)Pt(s)
E_{cell} = 0.92 V
Net cell reaction:
Q.10. For the cell Zn(s)  Zn^{2+}(aq)  M^{x+} (aq)  M(s), different half cells and their standard electrode potentials are given below:
If, which cathode will give a maximum value of E°_{cell} per electron transferred? (2019)
(1) Ag^{+}/Ag
(2) Fe^{3+}/Fe^{2+}
(3) Au^{3+}/Au
(4) Fe^{2+}/Fe
Ans. (1)
Q.11. Given the equilibrium constant:
K_{C} of the reaction:
Cu(s) + 2Ag^{+} (aq) → Cu^{2+} (aq) + 2Ag(s) is
10 x 10^{15}. Calculate the E°_{cell} of this reaction at 298 K. (2019)
(1) 0.04736 mV
(2) 0.4736 mV
(3) 0.4736 V
(4) 0.04736 V
Ans. (3)
Q.12. The standard electrode potential E° and its temperature coefficient for a cell are 2 V and 5 x 10^{4} VK^{1} at 300 K respectively. The cell reaction is:
Zn(s) + Cu^{2+}(aq) → Zn^{2+} (aq) + Cu
The standard reaction enthalpy (Δ_{r}H°) 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)
Q.13. Λ^{0}_{m} for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 cm^{2} 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)
Q.14. Given that
The strongest oxidising agent is: (2019)
(1) Au^{3+}
(2) O_{2}
(3) S_{2}O_{8}^{2}
(4) Br_{2}
Ans. (3)
More positive is the reduction potential stronger is the oxidising agent. Reduction potential is maximum for S_{2}O_{8}^{2}. 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:
Fe^{2+} (aq) + Ag^{+} (aq) → Fe^{3+} (aq) + Ag(s)
Given that (2019)
(1) x  z
(2) x  y
(3) x + 2y  3z
(4) x + y  z
Ans. (3)
(given)
Q.16. The standard Gibbs energy for the given cell reaction in kJ mol^{1} at 298 K is:
Zn(s) + Cu^{2+} (aq) → Zn^{2+} (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)
= 2 (96000) 2 V =  384000 J/mol
= 384 kJ/mol
Q.17. A solution of Ni (NO_{3})_{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,
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 cm^{3} 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 (λ_{m}) 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, (S_{1}) is wrong and (S_{2}) is correct.
Q.19. Which one of the following graphs between molar conductivity (Λ_{m}) versus √C is correct? (2019)
(1)
(2)
(3)
(4)
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 :
Co^{3+} + e^{} → Co^{2+}; E° = + 1.81 V
Pb^{4+} + 2e^{} → Pb^{2+}; E° = + 1.67 V
Ce^{4+} + e^{} → Ce^{3+}; E° = + 1.61 V
Bi^{3+} + 3e^{} → Bi; E° = + 0.20 V
oxidizing power of the species will increase in the order: (2019)
(1) Ce^{4+} < Pb^{4+} < Bi^{3+} < Co^{3+}
(2) Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}
(3) Co^{3+} < Ce^{4+} < Bi^{3+} < Pb^{4+}
(4) Co^{3+} < Pb^{4+} < Ce^{4+} < Bi^{3+}
Ans. (2)
Higher the reduction potential, higher will be oxidising power. So, Bi^{3+} < Ce^{4+} < Pb^{4+} < Co^{3+}
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)
B_{2}H_{6} + 3O_{2} → B_{2}O_{3} + 3H_{2}O
According to balanced equation:
27.66 g B_{2}H_{6} i.e. 1 mole B_{2}H_{6} requires 3 mole of O_{2}. Now this oxygen is produced by electrolysis of water.
1 mole O_{2} is produced by 4 F charge
∴ 3 mole O_{2} will be produced by 12 F charge
∴ Now applying
Q = It
12 x 96500 C = 100 x t(s)
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)
Q.23. When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of paminophenol produced is: (2018)
(1) 109.0 g
(2) 98.1 g
(3) 9.81 g
(4) 10.9 g
Ans. (3)
(v.f.) = 4
W = 9.81 g
Q.24. Given
Among the following, the strongest reducing agent is (2017)
(1) Cr
(2) Mn^{2+}
(3) Cr^{3+}
(4) Cl^{}
Ans. (1)
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:
Based on these data, which of the following statements is correct? (2017)
(1) Tl^{3+} is more stable than Al^{3+}
(2) Al^{+} is more stable than Al^{3+}
(3) Tl^{3+} 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 Fe^{3+} → Fe?
Given that: (2017)
(1) –0.057 V
(2) + 0.30 V
(3) – 0.30 V
(4) + 0.057 V
Ans. (1)
(i) Fe^{2+} + 2e^{} → Fe; E° = 0.47V;
(ii) Fe^{3+} + e^{} → Fe^{2+}; E° = +0.77V;
(iii) Fe^{3+} + 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 (Fe^{3+} → Fe)
=
Q.27. To find the standard potential of M^{3+}/M electrode, the following cell is constituted: Pt/M/M^{3+} (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 M^{3+} + 3e^{} → M at 298 K will be: (Given E^{Θ}_{Ag}+_{/ 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)
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.
Q.29. What will occur if a block of copper metal is dropped into a beaker containing a solution of 1M ZnSO_{4}? (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 ZnSO_{4}, no reaction will occur because
Hence Cu can't displace Zn from ZnSO_{4} 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 oxygenfree 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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