IIT-JEE 2011 · Paper 2 · Chemistry Q4

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ChemistrySingle correct+3 / -1Question 4 of 60

Consider the following cell reaction:
$\ce{2Fe(s) + O2(g) + 4H+(aq) -> 2Fe^{2+}(aq) + 2H2O(l)}\qquad E^\circ=1.67\ \mathrm V$
At $[\ce{Fe^{2+}}]=10^{-3}$ M, $P(\ce{O2})=0.1$ atm and pH = 3, the cell potential at 25 °C is

1.47 V

1.77 V

1.87 V

1.57 V

Answer and worked solution

Official answer: D

Nernst equation with $n=4$ :
$E=E^\circ-\frac{0.0591}{4}\log\frac{[\ce{Fe^{2+}}]^2}{P_{\ce{O2}}[\ce{H+}]^4}=1.67-\frac{0.0591}{4}\log\frac{(10^{-3})^2}{0.1\times(10^{-3})^4}.$

Quotient. $\dfrac{10^{-6}}{10^{-13}}=10^7$ , so $E=1.67-0.0148\times7=1.67-0.10=1.57$ V.

(D)

Topics:Nernst equation

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IIT-JEE 2011 · Paper 2, Chemistry question 4