Acetic acid has Ka = 1.8 × 10-5 while formic acid has Ka = 2.1 × 10-4. What would be the magnitude of the emf of the cell 

                  PtH20.1 M acetic acid + 0.1 M sodium acetate 0.1 M formic acid  0.1 M sodium formatePtH2 at 25°C ?

(1) 0.0315 volt                            

(2) 0.0629 volt

(3) 0.0455 volt                            

(4) 0.0545 volt

Subtopic:  Nernst Equation |
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Following cell has EMF 0.7995 V 

Pt | H2 (1 atm) | HNO3 (1M) || AgNO3 (1M) | Ag

If we add enough KCl to the Ag cell so that the final Cl-is 1M. Now the measured emf of the cell is 0.222 V. The KSP of AgCl would be :

(1) 1 × 109.8 

(2) 1 × 1019.6

(3) 2 × 1010  

(4) 2.64 × 1014

Subtopic:  Nernst Equation |
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A flashlight cell has the cathodic reaction 2MnO2 (s) + Zn+2+ 2'e- Zn Mn2O4 (S)

If the flashlight cell is to give out 4.825 mA, how long could it run if initially 8.7 g of the

limiting reagent MnO2 is present? [Mn = 55, O = 16]

(1) 2 × 106sec                                         

(2) 4 × 106sec

(3) 6× 106sec                                          

(4) 8 × 106sec

Subtopic:  Faraday’s Law of Electrolysis |
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An excess of granular zinc was added to 500 mL of 1M nickel nitrate till equilibrium was established. Find out the concentration of nickel at the equilibrium, if the standard electrode potential of Zn2+ / Zn and Ni2+/Ni  are – 0.75 and – 0.24 V, respectively.
(1) 5.56 x 10-18 M
(2) 0.56 x 10-18 M
(3) 2.16 x 10-18 M
(4) 4.12 x 10-18 M

Subtopic:  Nernst Equation |
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Calculate the cell EMF in mV for Pt | H2 (1atm) | HCl (0.01 M) | AgCl(s) | Ag(s) at 298 K 

If Gr° values are at 25°C -109.56kJmol for AgCl(s) & 130.79kJmol for (H++ Cl-) (aq)

(1) 456 mV                         

(2) 654 mV

(3) 546 mV                         

(4) None of these

Subtopic:  Faraday’s Law of Electrolysis |
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A current of 0.1A was passed for 2hr through a solution of cuprocyanide and 0.3745 g of

copper was deposited on the cathode. Calculate the current efficiency for the copper

deposition. (Cu – 63.5)

1. 79%                                 

2. 39.5%

3. 63.25%                             

4. 63.5%

Subtopic:  Faraday’s Law of Electrolysis |
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In acidic medium MnO4- is an oxidising agent. MnO4- + 8H+ + 5e-  Mn2+ + 4H2O. If H+ ion concentration is doubled, electrode potential of the half cell will : 

(1) Increase by 28.46 mV                             

(2) Decrease by 28.46 mV

(3) Increase by 14.23 mV                             

(4) Decrease by 142.30 mV

Subtopic:  Electrode & Electrode Potential |
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Calculate the EMF of the cell at 298 K Pt | H2 (1atm) | NaOH (xM), NaCl (xM) | AgCl (s) | Ag E°Cl-/AgCl/Ag=+0.222 V

(1) – 0.604 V                                                  

(2) – 0.04 V

(3) 1.048 V                                  

(4) emf depends on x and cannot be determined unless value of x is given

 

Subtopic:  Nernst Equation |
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Assertion : Fluorine cannot be prepared from fluorides by chemical oxidation.

Reason : Fluorine is the strongest oxidizing agent due to its highly positive standard potential.

  1. If both the assertion and the reason are true and the reason is a correct explanation of the assertion
  2. If both the assertion and reason are true but the reason is not a correct explanation of the assertion
  3. If the assertion is true but the reason is false
  4. If both the assertion and reason are false
Subtopic:  Electrode & Electrode Potential |
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Assertion : : Sodium reacts with water at room temperature, whereas magnesium decomposes only hot water.

Reason : Sodium is more electropositive than magnesium.

  1. If both the assertion and the reason are true and the reason is a correct explanation of the assertion
  2. If both the assertion and reason are true but the reason is not a correct explanation of the assertion
  3. If the assertion is true but the reason is false
  4. If both the assertion and reason are false
Subtopic:  Electrode & Electrode Potential |
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