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STD 12 - 2. Electrochemistry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electrochemistry4 Marks
MCQ
The electrochemical cell shown below is a concentration cell.

$M \mid M ^{2+}$ (saturated solution of a sparingly soluble salt, $\left.MX _2\right) \| M ^{2+}\left(0.001\right.$ mol dm $\left.d ^{-3}\right) \| M$ The emf of the cell depends on the difference in concetration of $M ^{2+}$ ions at the two electrodes. The emf of the cell at $298$ is $0.059 \ V$

$1.$ The solubility product $\left( K _{ sp } ; mol ^3 dm ^{-9}\right)$ of $MX _2$ at $298$ based on the information available the given concentration cell is (take $2.303 \times R \times 298 / F =0.059 \ V$ )

$(A)$ $1 \times 10^{-15}$ $(B)$ $4 \times 10^{-15}$

$(C)$ $1 \times 10^{-12}$ $(D)$ $4 \times 10^{-12}$

$2.$ The value of $\Delta G \left( kJ \ mol ^{-1}\right)$ for the given cell is (take $1 F =96500 \ C \ mol ^{-1}$ )

$(A)$ $-5.7$ $(B)$ $5.7$ $(C)$ $11.4$ $(D)$ $-11.4$

Give the answer question $1$ and $2.$

  • ✓
    $(B,D)$
  • B
    $(B,C)$
  • C
    $(A,D)$
  • D
    $(C,D)$
✓

Answer

Correct option: A.
$(B,D)$
a
$1.$ $M \mid M ^{2+}( aq )\left\| M ^{2+}( aq )\right\| M $

$0.001 M $

$\text { Anode : } \quad M \longrightarrow M ^{2+}( aq )+2 e ^{-} $

$\text {Cathode: } \quad M ^{2+}( aq )+2 e ^{-} \longrightarrow M $

$M ^{2+}( aq )_{ c } \rightleftharpoons M ^{2+}( aq )_{ a } $

$E _{\text {cell }}=0-\frac{0.059}{2} \log \left\{\frac{ M ^{2+}( aq )_{ a }}{10^{-3}}\right\} $

$0.059=\frac{0.059}{2} \log \left\{\frac{ M ^{2+}( aq )_{ a }}{10^{-3}}\right\} $

$-2=\log \left\{\frac{M^{2+}( aq )_a}{10^{-3}}\right\} $

$10^{-2} \times 10^{-3}= M ^{2+}( aq )_{ a }=\text { solubility }= s $

$K _{ sp }=4 s ^3=4 \times\left(10^{-5}\right)^3=4 \times 10^{-15} $

$2.$ $\Delta G=-n F E_{\text {cen }}=-2 \times 96500 \times 0.059 \times 10^{-3} kJ / mole $

$=-11.4 kJ / mole $

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