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

Given below are two half-cell reactions:

Mn^₂₊ + 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

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  

$E_{\raisebox{1ex}{${\mathrm{Fe}}^{2+}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Fe}$}\right.}^{\circ } = -0.441 \mathrm{V} and E_{\raisebox{1ex}{${\mathrm{Fe}}^{3+}$}\!\left/ \!\raisebox{-1ex}{${\mathrm{Fe}}^{2+}$}\right.}^{\circ } =0.771 \mathrm{V}$, the standard emf of the reaction

$\mathrm{Fe} + 2{\mathrm{Fe}}^{3+} \stackrel{ }{\to }3{\mathrm{Fe}}^{2+}$will be

1. 0.111V

2. 0.330V

3. 1.653V

4. 1.212V

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: 

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

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

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

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.

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^-