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The electrochemical cell shown below is concentration cell.
$M \mid M ^{2+}$ (saturated solution of a sparingly soluble salt, $\left.MX _2\right)\left|\left| M ^{2+}\left(0.001 mol dm ^{-3}\right)\right| M\right.$ The emfof the cell depends on the difference in concentrations of $M ^{2+}$ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V . The following questions are multiple choice questions. Choose the most appropriate answer:

1. The solubility product ($K_{sp'} mol^3 dm^{-9})$ of $MX_2$ at 298 K based on the information available for the given concentration cell is $(\text{take }2.303 \times \text{R}\times \frac{298}{\text{F}} = 0.059)$

1. $2 \times 10^{-15}$
2. $4 \times 10^{-15}​​​​​​​$
3. $3 \times 10^{-12}​​​​​​​$
4. $1 \times 10^{-12}​​​​​​​$

2. The value of $\triangle\text{G}$ (in kJ $mol^{-1}​​​​​​​$) for the given cell is (take $1 F = 96500 C mol^{-1})$

1. 3.7
2. -3.7
3. 10.5
4. -11.4

3. 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 $\text{E}^\circ_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}=0.68\text{V}$)

1. $7.6 \times 10^{12}$​​​​​​​
2. $6.5 \times 10^{10}$
3. $5.2 \times 10^9$
4. $3.4 \times 10^{12}​​​​​​​$

4. The solubility product of a saturated solution of $Ag_2CrO_4 $in water at 298 K if the emf of the cell

$Ag|Ag^+ (satd. Ag_2CrO_4 soln) || Ag^+ (0.1 M) | Ag$

is 0.164V at 298 K, is:

1. $3.359 \times 10^{-12} mol^3 L^{-3}$​​​​​​​
2. $2.287 \times 10^{-12} mol^3 L^{-3}$
3. $1.158 \times 10^{-12} mol^3 L^{-3}$
4. $4.135 \times 10^{-12} mol^3 L^{-3}$

5. To calculate the emf of the cell, which of the foUowing options is correct?

1. emf $= E_{cathode}- E_{anode}$​​​​​​​
2. emf $= E_{anode}- E_{cathode}$
3. emf $= E_{anode} + E_{cathode}$
4. None of these.