Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2012 · Paper 2 · Chemistry · Electrochemistry

The electrochemical cell shown below is a concentration cell. M M^...

The electrochemical cell shown below is a concentration cell. M∣M2+\mathrm{M} \mid \mathrm{M}^{2+} (saturated solution of a sparingly soluble salt, \left.\mathrm{MX}_{2}\right) \| \mathrm{M}^{2 \perp}\left(0.001 \mathrm{~mol} \mathrm{dm}^{-3}\right) \mid \mathrm{M}.Theemfofthecelldependsonthedifferenceinconcentrationsof. The emf of the cell depends on the difference in concentrations of\mathrm{M}^{2+}ionsatthetwoelectrodes.Theemfofthecellations at the two electrodes. The emf of the cell at298 \mathrm{~K}isis0.059 \mathrm{~V}.<strong>Question:</strong>Thesolubilityproduct. <strong> Question:</strong> The solubility product\left(\mathrm{K}_{s p} ; \mathrm{mol}^{3} \mathrm{dm}^{-9}\right) of \mathrm{MX}_{2}atat298 \mathrm{~K}basedontheinformationavailableforthegivenconcentrationcellis(takebased on the information available for the given concentration cell is (take2.303 \times \mathrm{R} \times 298 / \mathrm{F}=0.059 \mathrm{~V}$ )

Options

  1. A1×10−151 \times 10^{-15}
  2. B4×10−154 \times 10^{-15}
  3. C1×10−121 \times 10^{-12}
  4. D4×10−124 \times 10^{-12}

Correct answer

B. 4×10−154 \times 10^{-15}

Step-by-step solution

M∣M2+M \mid M^{2+} (aq) ∥M2+\| M^{2+} (aq) ∣M\mid M 0.001M0.001 \mathrm{M} Anode : M⟶M2+(aq)+2e−\quad M \longrightarrow M^{2+}(\mathrm{aq})+2 e^{-} Cathode : M2+(aq)+2e−⟶MM^{2+}(\mathrm{aq})+2 e^{-} \longrightarrow M M2+(aq)c⇌M2+(aq)aM^{2+}(\mathrm{aq})_{c} \rightleftharpoons M^{2+}(\mathrm{aq})_{a} Ecell =0−0.0592log⁡{M2+(aq)a10−3}⇒0.059=−0.0592log⁡{M2+(aq)a10−3}−2=log⁡{M2+(aq)a10−3}\begin{array}{c} E_{\text{cell }}=0-\frac{0.059}{2} \log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \\ \Rightarrow 0.059=-\frac{0.059}{2} \log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \\ -2=\log \left\{\frac{M^{2+}(\mathrm{aq})_{a}}{10^{-3}}\right\} \end{array} ⇒10−2×10−3=M2+(aq)a=\Rightarrow \quad 10^{-2} \times 10^{-3}=\mathrm{M}^{2+}(\mathrm{aq})_{a}= solubility =s=s MX2⇌M2++2X−\quad M X_{2} \rightleftharpoons M^{2+}+2 X^{-} Ksp=S.(2S)22SK_{s p}=S .(2 S)^{2} \quad 2 S ⇒Ksp=4s3=4×(10−5)3=4×10−15\Rightarrow \quad K_{s p}=4 s^{3}=4 \times\left(10^{-5}\right)^{3}=4 \times 10^{-15}
Practice Electrochemistry on Quantrex Academy →

More from Electrochemistry

At 300 K, the molar conductivities of the aqueous solutions of three salts at two… JEE Advanced 2026Given at 298 K: E^ _ Fe^ 2+ /Fe = X Volt; E^ _ Fe^ 3+ /Fe = Y Volt. The E^ _ Fe^ 3+ /Fe^… JEE Main 2026For a general redox reaction Anode : Red_1 arrow Ox_1^ n_1^+ + n_1 e^- Cathode : Ox_2 +… JEE Main 2026Consider the following data. Electrolyte Λ^ ∘ _m (S cm ^2 mol ^ -1 ) BaCl_2 x_1 H_2SO_4… JEE Main 2026One half cell in a voltaic cell is constructed by dipping silver rod in AgNO_3 solution… JEE Main 2026At 298 K , the molar conductivity of x\% (w/w) MX solution (aqueous) is 123.5 S cm^2 mol^… JEE Main 2026An electrochemical cell, consist of the following two redox couples, M^ x+ (aq)/M(s) [E_… JEE Main 2026Consider the following two half-cell reactions along with the standard reduction… JEE Main 2026 Full Electrochemistry list All JEE Advanced PYQs