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Electrochemistry — Chemistry STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry4 Marks
Question
The electrochemical cell shown below is concentration cell.

M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm-3) |M The emfof the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059V.

The following questions are multiple choice questions. Choose the most appropriate answer:

  1. The solubility product (Ksp' mol3 dm-9) of MX2 at 298 K based on the information available for the given concentration cell is $(\text{take }2.303 \times \text{R}\times \frac{298}{\text{F}} = 0.059)$
  1. 2 × 10-15
  2. 4 × 10-15
  3. 3 × 10-12
  4. 1 × 10-12
  1. 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
  1. 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 × 1012
  2. 6.5 × 1010
  3. 5.2 × 109
  4. 3.4 × 1012
  1. 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:

  1. 3.359 × 10-12 mol3 L-3
  2. 2.287 × 10-12 mol3 L-3
  3. 1.158 × 10-12 mol3 L-3
  4. 4.135 × 10-12 mol3 L-3
  1. To calculate the emf of the cell, which of the foUowing options is correct?
  1. emf = Ecathode - Eanode
  2. emf = Eanode - Ecathode
  3. emf = Eanode + Ecathode
  4. None of these.
✓

Answer

  1. (b) 4 × 10-15

Explanation:

$0.059=\frac{+0.059}{2}\log\frac{0.001}{[\text{M}^{2+}]}$

$\log\frac{0.001}{[\text{M}^{2+}]}=2$ or $[\text{M}^{2+}]=10^{-5}$

Let solubility of sah be S mol/ litre.

Thus, $\text{MX}_2\xrightarrow{\ \ \ \ }\text{M}^{2+}+2\text{X}^-\\\ \ \text{S}\ \ \ \ \ \ \ \ \ \ \ \ \text{S}\ \ \ \ \ \ \ \ \ \ 2\text{S}$

$\therefore\text{K}_\text{sp}=4\text{S}^3=4\times(10^{-5})^3=4\times10^{-15}$

  1. (d) -11.4

Explanation:

$\triangle\text{G}=\text{nFE}=-2\times96500\times0.059$

= -11387 J mol-1 = -11.4 kJ mol-1

  1. (a) 7.6 × 1012

Explanation:

$\text{E}^\circ_\text{cell}=\frac{0.059}{1}\log\text{K}_\text{C}$

$\text{E}^\circ_\text{cell}=\text{E}^\circ_\frac{{\text{Fe}^{2+}}}{{\text{Fe}^{3+}}}+\text{E}^\circ_\frac{\text{Ce}^{4+}}{\text{Ce}^{3+}}$

$= -0.68 + 1.44 = 0.76 \text{V}$

$\log_{10}\text{K}_\text{C}=\frac{0.76}{0.059}=12.88$

$\text{K}_\text{C}=7.6\times10^{12}$

  1. (b) 2.287 × 10-12 mol3 L-3

Explanation:

$\text{E}_\text{cell}=\frac{0.059}{1}\log\frac{[\text{Ag}^+]_\text{RHS}}{[\text{Ag}^+]_\text{LHS}}$

$0.164=\frac{0.059}{1}\log\frac{0.1}{[\text{Ag}^+]_\text{LHS}}$

$[\text{Ag}^+]_\text{LHS}=1.66\times10^{-4}\text{M}$

So, $[\text{CrO}^{2-}_4]=\frac{1.66\times10^{-4}}{2}$

$\text{K}_\text{Sp}(\text{Ag}_2\text{CrO}_4)=[\text{Ag}^+]^2[\text{CrO}_4^{2-}]$

$= (1.66\times10^{-4} )^2\Big(\frac{1.66\times10^{-4}}{2}\Big)$

$=2.287\times10^{-12} \text{mol}^{3} \text{L}^{-3}$

  1. (a) emf = Ecathode - Eanode

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