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← JEE Advanced 2018 · Paper 1
ChemistryNumerical+3Question 28 of 54

JEE Advanced 2018 · Paper 1, Chemistry question 10

For the electrochemical cell,

Mg(s) ∣ MgX2+(aq,1M) ∣∣ CuX2+(aq,1M) ∣ Cu(s)\ce{Mg(s) | Mg^{2+}(aq, 1 M) || Cu^{2+}(aq, 1 M) | Cu(s)}

the standard emf of the cell is 2.70 V at 300 K. When the concentration of MgX2+\ce{Mg^{2+}} is changed to xx M, the cell potential changes to 2.67 V at 300 K. The value of xx is ______.

(given, FR=11500 K V−1\dfrac FR = 11500\ \mathrm{K\,V^{-1}}, where FF is the Faraday constant and RR is the gas constant, ln⁡(10)=2.30\ln(10) = 2.30)

Answer and worked solution

Official answer: 9.9

Cell reaction. Mg+CuX2+→MgX2++Cu\ce{Mg + Cu^{2+} -> Mg^{2+} + Cu}, with n=2n=2.

Nernst.
E=E∘−RT2Fln⁡[MgX2+][CuX2+]  ⟹  2.67=2.70−3002×11500ln⁡x.E=E^\circ-\frac{RT}{2F}\ln\frac{[\ce{Mg^{2+}}]}{[\ce{Cu^{2+}}]}\implies2.67=2.70-\frac{300}{2\times11500}\ln x.

Solve.
ln⁡x=0.03×23000300=2.30  ⟹  x=10.\ln x=\frac{0.03\times23000}{300}=2.30\implies x=10.

Answer: 10