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Q1. If ‘F’ is faraday and ‘N’ is Avogadro number, then charge of electron can be expressed as
•  F × N
•  F^2 N
•  N/F
•  F/N
Solution
If F is Faraday and N is Avogadro number, charge of electron = F/N

Q2.For a given cell reaction; Cr+3H_2 O+OCl^- ⟶Cr^(3+) 3Cl^- +6OH^-, the species undergoing reduction is :
•  Cl^-
•  OCl^-
•  Cr^(6+)
•  Cr
Solution
Cl in OCl^-has oxidation number as +1. Thus, Cl^+ +2e ⟶Cl^- (i.e.,reduction of OCl^-)

Q3.  E° for F_(2+)+2e=2F^- is 2.8 V, E° for 1/2〖 F〗_2+e=F^- is:
•   - 1.4 V
•  - 2.8 V
•  1.4 V
•  2.8 V
Solution
E° does not depend on stoichiometry of change

Q4. Which is correct representation for a cell at equilibrium?
•  ∆G°= -2.303 RT log⁡〖K_(eq.) 〗
•  E°=2.303RT/nF log⁡〖K_(eq.) 〗
•  -∆G°=RT ln⁡〖K_(eq^.) 〗
•  All of the above.
Solution
At equilibrium E_cell=0.

Q5.Cu(II) sulphate solution is treated separately with KCl and KI. In which case, Cu^(2+) be reduced to Cu^+?
•  With KI
•  With KCl
•  With both (a) and (b)
•  None of these
Solution

Q6. For gold plating, the electrolyte used is
•  AuCl_3
•  K[Au(CN)_2]
• HAuCl_4
•  None of these
Solution
For gold plating the electrolyte K [Au(CN)_2] is used.

Q7.The specific conductivity of 0.1 N KCl solution is 0.0129 â„¦^(-1) cm^(-1). The resistance of the solution in the cell 100â„¦. The cell constant of the cell will be
•  1.10
•  2.80
•  1.29
•  0.56
Solution
Specific conductivity (Îº) = 1/R ×cell constant Cell constant=Îº ×R = 0.0129 × 100 = 1.29

Q8.Passage of 96500 coulomb of electricity liberates ….litre of O_2at NTP during electrolysis.J
•  6.5
•  5.6
•  22.2
•  11.2
• Solution
96500 C or 1F will liberate 1 eq. of O_2 or 1/4 mole O_2 or 5.6 litre O_2 at NTP.

Q9.The cell reaction is spontaneous, when
•  E_red^° is negative
•  〖Î”G〗^° is positive
•  E_red^° is positive
•  〖Î”G〗^° is negative
Solution
△G = △H - T△S For a spontaneous cell reaction, △H should be negative and △S should be positive. Hence, △G should be negative.

Q10. Which of the following statements is not applicable to electrolytic conductors?
•  Show a positive temperature coefficient for conductance
•  A single stream of electrons flows from cathode to anode
•  New products show up at the electrodes
• Ions are responsible for carrying the current
Solution
B

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