An electrochemical cell is shown below Pt, H2(1 atm)| HCI (0.1 M)CH3COOH (0.1 M)|

H2(1 atm), Pt The EMF of the cell will not be zero, because

1. EMF depends on molarities of acids used

2. pH of 0.1 M HCl and 0.1 M CH3COOH is not same

3. the temperature is constant

4. acids used in two compartments are different

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Given below are two half-cell reactions:

Mn2+ + 2e → Mn;          E0 = -1.18V

2Mn3+ + 2e → 2Mn2+;  E0 = +1.51V

The E0 for 3Mn2+ → 2Mn+3  +  Mn  will be:

1. -2.69V; the reaction will not occur

2. -2.69V; the reaction will occur

3. -0.33V; the reaction will not occur

4. -0.33V; the reaction will occur

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E0 for Fe2+ + 2e → Fe is –0.44 volt and E0 for Zn2+ + 2e→ Zn is –0.76 volt, thus:

1. Zn is more electropositive than Fe.

2. Fe is more electropositive than Zn.

3. Zn is more electronegative.

4. None of the above.

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EFe2+Fe=-0.441VandEFe3+Fe2+=0.771V, the standard emf of the reaction

Fe+2Fe3+3Fe2+will be

1. 0.111V

2. 0.330V

3. 1.653V

4. 1.212V

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The standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for the Cr3+/Cr couple is -0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be: 

1. +0.89 V 2. +0.18 V
3. +1.83 V 4. +1.199 V
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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

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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

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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

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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.

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If mercury is used as cathode in the electrolysis of aqueous NaCl solution, the ions discharged at cathode are:

1. H+

2. Na+

3. OH-

4. Cl-

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