JEE Advanced 2016 · Paper 2 · Chemistry Q19

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ChemistrySingle correct+3 / -1Question 19 of 54

For the following electrochemical cell at 298 K,
$\mathrm{Pt}(s)\mid\ce{H2}(g,1\text{ bar})\mid\ce{H+}(aq,1\text{ M})\parallel\ce{M^{4+}}(aq),\ce{M^{2+}}(aq)\mid\mathrm{Pt}(s)$
$E_{cell}=0.092$ V when $\dfrac{[\ce{M^{2+}}(aq)]}{[\ce{M^{4+}}(aq)]}=10^x$ .

Given: $E^0_{\ce{M^{4+}}/\ce{M^{2+}}}=0.151$ V; $2.303\dfrac{RT}{F}=0.059$ V

The value of $x$ is

−2

−1

Answer and worked solution

Official answer: D

Cell reaction. $\ce{H2 + M^{4+} -> 2H+ + M^{2+}}$ , with $n=2$ . The SHE anode has $E^\circ=0$ , so $E^\circ_{cell}=0.151$ V.

Nernst equation.
$0.092=0.151-\frac{0.059}2\log\frac{[\ce{M^{2+}}]}{[\ce{M^{4+}}]}=0.151-0.0295x.$
So $x=2$ .

(D)

Topics:Nernst equation

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JEE Advanced 2016 · Paper 2, Chemistry question 19