The voltage of the cell given below increases with: 

Cell: Sn(s) + 2Ag⁺(aq) → Sn²⁺(aq) + 2Ag(s)

1. Increase in size of the silver rod.

2. Increase in the concentration of Sn²⁺ ions.

3. Increase in the concentration of Ag⁺ ions.

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: 

If an iron rod is dipped in CuSO₄ solution, then:

1. Blue colour of the solution turns red.

2. Brown layer is deposited on iron rod.

3. No change occurs in the colour of the solution.

4. None of the above.

Without losing it's concentration, ZnCl₂ solution cannot be kept in contact with :

1. Au

2. Al

3. Pb

4. Ag

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