The solubility product of silver iodide is 8.3×10-17and the standard potential (reduction) of Ag,Ag+ electrode is + 0.800 volts at 25°C.The standard potential of Ag, AgI/I- electrode (reduction) from these data is-

(1) – 0.30 V                            

(2) + 0.15 V

(3) + 0.10 V                            

(4) – 0.15 V

Subtopic:  Relation between Emf, G, Kc & pH |
Level 3: 35%-60%
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How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-

1. 26.805 h                                   

2. 53.61 h

3. 107.22 h                                   

4. None of these

Subtopic:  Faraday’s Law of Electrolysis |
 62%
Level 2: 60%+
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The standard redox potentials Eof the following systems are-
 

System Eo(Volts)
\(MnO_{4}^{-}+8 H^{+}+5 e \rightarrow M n^{2+}+4 H_{2}O\) 1.51
Sn4+ +   2e-   →   S n2+ 0.15
Cr2O72- + 14H+ +  6e- →  2Cr3+ +  7H2O 1.33
C e4+  +  e- →   Ce3+   1.61

The oxidizing power of the various species decreases in the order-

1. Ce4+>Cr2O72->Sn4+>MnO4-

2. Ce4+>MnO4->Cr2O72->Sn4+

3. Cr2O72->Sn4+>Ce4++MnO4-

4. MnO4->Ce4+>Sn4+>Cr2O72-

Subtopic:  Electrode & Electrode Potential |
 86%
Level 1: 80%+
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Calculate the maximum work that can be obtained from the Daniell cell given below -

Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s).

Given that EZn2+/Zn°=-0.76VandECu2+/Cu°=+0.34V

(1) – 212300 J                                        

(2) – 202100 J

(3) – 513100 J                                        

(4) – 232120 J

 

Subtopic:  Relation between Emf, G, Kc & pH |
 82%
Level 1: 80%+
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The metal that cannot be produced on reduction of its oxide by aluminium is :

(1) K                                                        

(2) Mn

(3) Cr                                                      

(4) Fe

Subtopic:  Electrode & Electrode Potential |
 76%
Level 2: 60%+
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We have taken a saturated solution of AgBr, Ksp of AgBr is 12 × 10-14.   If 10-7 mole of AgNO3  is added to 1 litre of this solution then the conductivity of this solution in terms of  10-10   Sm-1 units will be:

[λ°Ag+=6×10-3Sm2mol-1;
λ°Br-=8×10-3Sm2mol-1,
λ°NO3-=7×10-3Sm2mol-1]

1. 39 
2. 55
3. 15 
4. 41

Subtopic:  Electrode & Electrode Potential |
 65%
Level 2: 60%+
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Calculate the e.m.f. for the following cell reaction:
\(F e^{2+}+Z n → Z n^{2+}+F e\)
The standard potentials \(E^{\circ} \text {,for the half reactions are as}\)

\(Z n →Z n^{2+}+2 e ; E^{\circ}=+0.76 \mathrm{~V}\)
\(F e → F e^{2+}+2 e ; E^{\circ}=+0.41 \mathrm{~V}\)

1. – 0.35 V                                            

2. + 0.35 V

3. + 1.17 V                                           

4. – 1.17 V

Subtopic:  Electrode & Electrode Potential |
 70%
Level 2: 60%+
Hints

Which of following cell can produce more electrical work.

1. pt,H2|NH4Cl||0.1MCH3COOH|H2,pt

2. pt,H2|0.1MHCl||0.1MNaOH|H2,pt

3. pt,H2|0.1MHCl||0.1MCH3COOK|H2,pt

4. pt,H2|0.1MCH3COOK||0.1MHCl|H2,pt

Subtopic:  Nernst Equation |
 50%
Level 3: 35%-60%
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The calomel electrode is reversible with respect to-

1. Mercury                                       

2. H+

3. Hg2+                                          

4. Cl-

Subtopic:  Electrode & Electrode Potential |
Level 4: Below 35%
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Zn amalgam is prepared by electrolysis of aqueous ZnCl2 using Hg cathode (9 gm). How much current is to be passed through ZnCl2 solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by wt. (Zn = 65.4)

(1) 5.6 amp 

(2) 7.2 amp

(3) 8.85 amp 

(4) 11.2 amp

Subtopic:  Faraday’s Law of Electrolysis |
 56%
Level 3: 35%-60%
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