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Q1. For cell reaction, Zn + Cu^(2+) ⟶ Zn^(2+)+ Cu Cell representation is
•  Cu^(2+) | Zn || Zn^(2+) | Cu
•  Cu | Zn^(2+) || Zn | Cu^(2+)
•  Cu | Cu^(2+) || Zn^(2+) | Zn
•  Zn | Zn^(2+) || Cu^(2+) | Cu
Solution

Q2.The time required to coat a metal surface of 80 cm^2 with 5 ×〖10〗^(-3) cm thick layer of silver (density 1.05 g cm^(-3) with the passage of 3A current through a silver nitrate solution is :
•  115 sec
•  125 sec
•  135 sec
•  145 sec
Solution
Q3.  Given l/a=0.5cm^(-1),R = 50 ohm, N = 1.0. The equivalent conductance of the electrolytic cell is
•   100Ω^(-1) cm^2 g equiv^(-1)
•  300Ω^(-1) cm^2 g equiv^(-1)
•  20Ω^(-1) cm^2 g equiv^(-1)
•  10Ω^(-1) cm^2 g equiv^(-1)

Q4. The factors which influence the conductance of solution.
•  Solute-solute interaction
•  Solute-solvent interaction
•  Temperature
•  All of the above
Solution
More is solute-solute interaction, lesser is conductance. More is solute-solvent interaction, more is conductance. An increase in temperature also increases conductance due to increase in ionic mobility.

Q5.On passing 1 F of electricity through the electrolytic cells containing Ag^+,Ni^(2+) and Cr^(3+) ions solution, the deposited Ag (at.wt.=108),Ni (at.wt.=59) and Cr (at.wt.=52) is Ag Ni Cr
•  108 g 29.5 g 17.3 g
•  108 g 59.5 g 52.0 g
•  108 g 108 g 108 g
•  108 g 117.5 g 166 g
Solution
Wt. of Ag deposited = eq. wt. of Ag = 108 g
Wt. of Ni deposited = eQ. wt. of Ni = 29.5 g
Wt. of Cr deposited = eq. wt. of Cr = 17.3 g

Q6. When 9.65 C of electricity is passed through a solution of silver nitrate (atomic weight of Ag = 107.87 taking as 108), the amount of silver deposited is
•  5.8 mg
•  10.8 mg
• 15.8 mg
•  20.8 mg
Solution
w_Ag=(E_Ag× Q)/96500=(108×9.65)/96500=1.08×〖10〗^(-2) g=10.8 mg

Q7.The conductivity of N/10 KCl solution at 200C is 0.0212 ohm^(-1) cm^(-1) and the resistance of cell containing this solution at 200 C is 55 ohm. The cell constant is:
•  2.173 cm^(-1)
•  3.324 cm^(-1)
•  1.166 cm^(-1)
•  4.616 cm^(-1)
Solution
Cell constant=k/C=0.0212 × 55 =1.166 cm^(-1)

Q8.Which one will liberate Br_2 from KBr?
•  SO_2
•  Cl_2
•  I_2
•  HI
Solution
Cl is placed below Br in electrochemical series; the non-metal placed below in series, replaces other from its solution.

Q9.Resistance of 0.2 M solution of an electrolyte is 50 Ω . The specific conductance of the solution is 1.3 S m^(-1). If resistance of the 0.4M solution of the same electrolyte is 260Ω , its molar conductivity is
•  6250 Sm^2 mol^(-1)
•  62.5 Sm^2 mol^(-1)
•  625 × 〖10〗^(-4) Sm^2 mol^(-1)
•  6.25 × 〖10〗^(-4) Sm^2 mol^(-1)
Solution

Q10. The molar conductivities of KCl, NaCl and KNO_3 are 152, 128 and 111 S cm^2 mol^(-1) respectively. What is the molar conductivity of NaNO_3 ?
•  101 S cm^2 mol^(-1)
•  87 S cm^2 mol^(-1)
•  -101 S cm^2 mol^(-1)
• -391 S cm^2 mol^(-1) #### Written by: AUTHORNAME

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ELECTROCHEMISTRY QUIZ-9
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