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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

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 half-cell 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)**

For the half-cell 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)**

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)**

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**

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 PbSO _{4} electrolyzed (in gm) during the process is: (Molar mass of PbSO_{4} = 303 g mol^{-1}) (2019)** (3)

(1) 22.8

(2) 15.2

(3) 7.6

(4) 11.4

Ans.

Zn(s) + Cu

at 300 K is approximately

(R= 8JK

(1) e

(2) e

(3) e

(4) e

Ans.

We know that,

After putting the given values, we get

∴ K = e

Pt(s)|H

(1) 0.94 V

(2) 0.76 V

(3) 0.40 V

(4) 0.20 V

Ans.

Given that:

Pt(s)|H

E

Net cell reaction:

If, which cathode will give a maximum value of E°

(1) Ag

(2) Fe

(3) Au

(4) Fe

Ans.

K

Cu(s) + 2Ag

10 x 10

(1) 0.04736 mV

(2) 0.4736 mV

(3) 0.4736 V

(4) 0.04736 V

Ans.

Zn(s) + Cu

The standard reaction enthalpy (Δ

[Use R= 8 JK

(1) -412.8

(2) -384.0

(3) 192.0

(4) 206.4

Ans.

(1) 0.50

(2) 0.25

(3) 0.125

(4) 0.75

Ans.

The strongest oxidising agent is: (2019)

(1) Au

(2) O

(3) S

(4) Br

Ans.

More positive is the reduction potential stronger is the oxidising agent. Reduction potential is maximum for S

Fe

Given that (2019)

(1) x - z

(2) x - y

(3) x + 2y - 3z

(4) x + y - z

Ans.

(given)

Zn(s) + Cu

E° = 2 V at 298 K

(Faraday's constant, F = 96000 C mol

(1) - 384

(2) 384

(3) 192

(4) - 192

Ans.

= -2 (96000) 2 V = - 384000 J/mol

= -384 kJ/mol

(1) 0.05

(2) 0.20

(3) 0.15

(4) 0.10

Ans.

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

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.

Conductivity of an electrolyte is the conductance of 1 cm

(1)

Both NaCl and KCl are strong electrolytes and as Na

Pb

Ce

Bi

oxidizing power of the species will increase in the order: (2019)

(1) Ce

(2) Bi

(3) Co

(4) Co

Ans.

Higher the reduction potential, higher will be oxidising power. So, Bi

(Atomic weight of B = 10.8 u)

(1) 6.4 hours

(2) 0.8 hours

(3) 3.2 hours

(4) 1.6 hours

Ans.

B

According to balanced equation:

27.66 g B

1 mole O

∴ 3 mole O

∴ Now applying

Q = It

12 x 96500 C = 100 x t(s)

t = 3.2 hours

(1) 0.1

(2) 0.5

(3) 1.0

(4) 2.0

Ans.

**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 gAns.** (3)

(v.f.) = 4

W = 9.81 g

**Q.24. GivenAmong the following, the strongest reducing agent is (2017)(1) Cr(2) Mn** (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**(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 VAns. **(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** (3)

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.

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

Fact

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