The standard reduction potential at 290 K for the following half reactions are,

(i) Zn2+ + 2e— → Zn(s);      E° = -0.762 V

(ii) Cr3+ + 3e → Cr(s);          E° = -0.740 V

(iii) 2H+ + 2e → H2(g); ·      E° = +0.000 V

(iv) Fe3+ + e → Fe2+;         E° = +0.77V

Which is the strongest reducing agent?

1. Zn

2. Cr

3. Fe2+

4. H2

Subtopic:  Electrode & Electrode Potential |
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Which graph correctly correlates Ecell as a function of concentrations for the cell (for different values of M and M') ?

Zn(s) + Cu2+(M)           Zn2+(M') + Cu(s);
                Ecell = 1.10 V
X-axis : log10Zn2+Cu2+, Y-axis : Ecell 

1.                         2. 
3.                         4. 

Subtopic:  Nernst Equation |
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The most convenient method to protect the bottom of the ship made of iron is 

1. coating it with red lead oxide

2. white tin plating 

3. connecting it with Mg block

4. connecting it with Pb block

Subtopic:  Corrosion |
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When 0.1 mole of MnO2-4 is oxidised, the quantity of electricity required to completely oxidise MnO2-4 to MnO-4 is 

1. 96500 C

2. 2*96500 C 

3. 9650 C

4. 96.50 C

Subtopic:  Faraday’s Law of Electrolysis |
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The Zn acts as sacrificial or cathodic protection to prevent rusting of iron because:

1. EOP of Zn < EOP of Fe

2. EOP of Zn > EOP of Fe

3. EOP of Zn = EOP of fe

4. Zn is cheaper than iron

Subtopic:  Corrosion |
 74%
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The standard reduction potential for Fe2+|Fe and Sn2+|Sn electrodes are -0.44 V and -0.14 V respectively. For the cell reaction,

Fe2+ + Sn   Fe + Sn2+, the standard Emf is - 

1. +0.30 V

2. 0.58 V

3. +0.58 V

4. -0.30 V

Subtopic:  Electrode & Electrode Potential |
 60%
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On electrolysing a solution of dilute H2SO4 between platinum electrodes, the gas evolved at the anode and cathode are respectively:

1. SO2 and O2

2. SO3 and H2

3. O2 and H2

4. H2 and O2

Subtopic:  Electrode & Electrode Potential |
 64%
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What mass of copper will be deposited by passing 2 faraday of electricity through a solution of Cu(II) salt?

1. 35.6 g

2. 63.5 g

3. 6.35 g

4. 3.56 g

Subtopic:  Faraday’s Law of Electrolysis |
 85%
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In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam because :

1. Hg is more inert than Pt

2. More voltage is required to reduce H+ at Hg than at Pt

3. Na is dissolved in Hg while it does not dissolve in Pt

4. Concentration of H+ ions is larger when Pt electrode is taken.

Subtopic:  Electrode & Electrode Potential |
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The same amount of electricity was passed through two cells containing molten Al2O3 and molten NaCl. If 1.8 g of Al were liberated in one cell, the amount of Na liberated in the other cell is:

1. 4.6 g

2. 2.3 g

3. 6.4 g

4. 3.2 g

Subtopic:  Faraday’s Law of Electrolysis |
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