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Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - NEET MCQ


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Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 1

 Cell reaction is spontaneous when

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 2

 Gibbs free energy change for a cell reaction is positive what does it indicates?

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 2

The correct answer is Option C.

Reactions with a negative ∆G release energy, which means that they can proceed without an energy input (are spontaneous). In contrast, reactions with a positive ∆G need an input of energy in order to take place (are non-spontaneous). Reactions with a positive ∆H and negative ∆S are non-spontaneous at all temperatures.

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Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 3

In the equation, ΔG° = – nF E° cell ; F is:

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 3

The correct answer is Option B.
The relationship between ΔGo and Eo is given by the following equation: ΔGo=−nFEo. Here, n is the number of moles of electrons and F is the Faraday constant.

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 4

 Consider the cell reaction:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 4

The correct answer is Option D.

Redox reaction:
Cd(s)→Cd2++2e
Cu2++2e→Cu(s)
Ecell = E°cell − (0.059/2) log ([Cd2+]/ [Cu2+])
Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 5

Conductivity (K) of 0.01 M NaCI solution is 0.00145 Scm-1. What happens to the conductivity if extra 100 mL of H2O be added to the above solution?

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 5


On dilution, molarity decreases and molar conductance increases, Hence, specific conductance decreases.

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 6

Specific conductance of 0.01 N KCI solution is x Scm-1 having conductance y S. Thus, specific conductance of 0.01 N NaCI having conductance zS is (in S cm-1)

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 6

Specific conductance = Conductance x Cell constant
∴ x = k (0.01 N KCI) = y x cell constant
∴ k (0.01N NaCI) = z x cell constant 

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 7

500 mL of an aqueous solution contains 0.1 mole of KCl. If its specific conductance is x Scm-1, its molar conductance will be (in Scm2 mol-1)

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 7

Molar conductance 

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 8

Which quantity is temperature independent?

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 8

When temperature increases, conductance increases. Thus, resistivity and conductivity both are temperature dependent. By Nernst equation,

Thus, Emf of the cell is dependent on temperature.


Thus, cell constant is temperature independent.

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 9

Given, (Scm2 mol-1)for different electrolytes

Thus, of CH3COOH is

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 9

(CH3COOH) = (CH3COONa) + (HCI) - (NaCI)
= 91.0+ 426.2 -126.5 
= 517.2 - 126.5 
= 390.7 Scm2mol-1

Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 10

At 298 K, given specific conductance of saturated
AgCl solution=3.41 x 10-6 Ω-1 cm-1 and that of water used =1.60 x 10-6 Ω-1 cm-1.
Equivalent conductance of saturated AgCl solution= 138.3 Ω-1 cmequiv-1

Thus, solubility product (KSp) of AgCl is

Detailed Solution for Test: Nernst Equation Conductance of Electrolytic Solutions (November 11) - Question 10

Water has also conductance, hence to obtain pure value of conductance of AgCI, conductance of water should be substracted from the conductance of AgCI given.

Pure specific conductance of AgCI = (3.41 x 10-6 - 1.60 x 10-6)

= 1.81 x 10-6Ω -1cm-1

AgCI is sparingly soluble, hence its solubility may be taken its concentration.
Also, in saturated solution sparingly soluble salt

∴   

∴   AgCl (s)  Ag+ + Cl-

∴ Ksp = [Ag+] [Cl-] = S2
= (1.31 x 10-5)2
= 1.72 x 10-10 mol2L-2

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