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Electrochemistry

Section: Chemistry

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Question : 25 of 30

Marks: +1, -0

Consider the following cell reaction:

\mathrm{O}_2

4\mathrm{H}^{+}

2\mathrm{Fe}^{2+}

2\mathrm{H}_2\mathrm{O}

(1) E° = 1.67 V

[\mathrm{Fe}^{2+}]

10^{-3}\,\mathrm{M}

(\mathrm{O}_2)

= 0.1 atm and pH = 3, the cell potential at 20°C is (1) 1.47 V

[JEE Adv 2011 P2]

1.47
1.77 V
1.87 V
1.57 V

Solution:

For the given electrochemical reaction

2 Fe (s) +

\mathrm{O}_2

(s) +

4\mathrm{H}^{+}

→

2\mathrm{Fe}^{2+}

(aq.) +

2\mathrm{H}_2\mathrm{O}

(l) E° = 1.67 V

From Nernst equation, we have

E_{\text{cell}}

=

E_{\text{cell}}^{\circ}-\frac{0.0591}{n} \log \frac{[\mathrm{Fe}^{2+}]}{[\mathrm{O}_2][\mathrm{H}^{+}]^4}

=

1.67-\frac{0.0591}{4} \log \frac{[10^{-3}]^2}{(10^{-1})(10^{-3})^4}

=

1.67-\frac{0.0591}{4} \log \frac{10^{-6}}{10^{-13}}

=

1.67-\frac{0.0591}{4}\times 7

= 1.57 V

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