- The solubility product (Ksp' mol3 dm-9) of MX2 at 298 K based on the information available for the given concentration cell is $(\text{take }2.303 \times \text{R}\times \frac{298}{\text{F}} = 0.059)$
- 2 × 10-15
- 4 × 10-15
- 3 × 10-12
- 1 × 10-12
- The value of $\triangle\text{G}$ (in kJ mol-1) for the given cell is (take 1 F = 96500 C mol-1)
- 3.7
- -3.7
- 10.5
- -11.4
- The equilibrium constant for the foUowing reaction is:
$\text{Fe}^{2+}+\text{Ce}^{4+}\rightleftharpoons\text{Ce}^{3+}+\text{Fe}^{3+}$
(Given, $\text{E}^\circ_\frac{\text{Ce}^{4+}}{\text{Ce}^{3+}}=1.44\text{V}$ and $\text{E}^\circ_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}=0.68\text{V}$)
- 7.6 × 1012
- 6.5 × 1010
- 5.2 × 109
- 3.4 × 1012
- The solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the emf of the cell
Ag|Ag+ (satd. Ag2CrO4 soln) || Ag+ (0.1 M) | Ag
is 0.164V at 298 K, is:
- 3.359 × 10-12 mol3 L-3
- 2.287 × 10-12 mol3 L-3
- 1.158 × 10-12 mol3 L-3
- 4.135 × 10-12 mol3 L-3
- To calculate the emf of the cell, which of the foUowing options is correct?
- emf = Ecathode - Eanode
- emf = Eanode - Ecathode
- emf = Eanode + Ecathode
- None of these.

