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

Which graph correctly correlates E_cell as a function of concentrations for the cell (for different values of M and M') ?

$\mathrm{Zn}\left(\mathrm{s}\right) + {\mathrm{Cu}}^{2+}\left(M\right) \stackrel{ }{\to }{\mathrm{Zn}}^{2+}(M\text{'}) + \mathrm{Cu}\left(\mathrm{s}\right);$
$E_{\mathrm{cell}}^{\circ } = 1.10 \mathrm{V}$
$X-\mathrm{axis} : {\log }_{10}\frac{\left[{\mathrm{Zn}}^{2+}\right]}{\left[{\mathrm{Cu}}^{2+}\right]}, Y-\mathrm{axis} : E_{\mathrm{cell}}$

1.                         2. 
3.                         4. 

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

The electrode potentials for

${\mathrm{Cu}}^{2+}\left(\mathrm{aq}\right) + {\mathrm{e}}^{-} \stackrel{ }{\to }{\mathrm{Cu}}^{+}\left(\mathrm{aq}\right)$
$\mathrm{and} {\mathrm{Cu}}^{+}\left(\mathrm{aq}\right) + {\mathrm{e}}^{-} \stackrel{ }{\to \mathrm{Cu}\left(\mathrm{s}\right)}$

are +0.15 V and +0.50 V respectively. The value of $E_{\raisebox{1ex}{${\mathrm{Cu}}^{2+}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{Cu}$}\right.}^{\circ }\mathit{ }$will be

1. 0.325 V

2. 0.650 V

3. 0.150 V

4. 0.500 V

The Zn acts as sacrificial or cathodic protection to prevent rusting of iron because:

1. $E_{OP}^{\circ }$ of Zn < $E_{OP}^{\circ }$ of Fe

2. $E_{OP}^{\circ }$ of Zn > $E_{OP}^{\circ }$ of Fe

3. $E_{OP}^{\circ }$ of Zn = $E_{OP}^{\circ }$ of fe

4. Zn is cheaper than iron

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^-

The standard reduction electrode potentials of three metals X, Y, and Z are given as -1.2 V, +0.5 V, and -3.0 V, respectively. Based on these values, what is the correct order of their reducing power?