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AIEEE 2009 Solved Paper

Section: Chemistry

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Question : 15 of 89

Marks: +1, -0

Given :

E^{\circ}_{\mathrm{Fe}^{3+}/\mathrm{Fe}} = -0.036\;\mathrm{V} ; E^{\circ}_{\mathrm{Fe}^{2+}/\mathrm{Fe}} = -0.439\;\mathrm{V}

The value of standard electrode potential for the change,

\mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq})

$-0.072\;\mathrm{V}$
$0.385\;\mathrm{V}$
$0.770\;\mathrm{V}$
$0.270$

Solution:

Given

\mathrm{F e}^{3+}\;+3\mathrm{e}^{-} \rightarrow \mathrm{F e}, E^{\circ} \mathrm{Fe}^{3+}/\mathrm{Fe}

= -0.036\;\mathrm{V} \dots (i)

\mathrm{F e}^{2+}\;+2\mathrm{e}^{-} \rightarrow \mathrm{F e}, E^{\circ} \mathrm{Fe}^{2+}/\mathrm{Fe}

= -0.439\;\mathrm{V} \dots (i i)

we have to calculate

\mathrm{F e}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{F e}^{2+}, \Delta G = ?

To obtain this equation subtract equ

\text{(i i)}

from

\text{(i)}

we get

\mathrm{F e}^{3+} + \mathrm{e}^{-} \rightarrow \mathrm{F e}^{2+} \dots (i i i)

As we know that

\Delta G = -n F E

Thus for reaction

\text{(i i i)}

\Delta G = \Delta G_1 - \Delta G

-n F E^{\circ} = -n F E_1 - (-n F E_2)

-n F E^{\circ} = n F E_2 - n F E_1

-1 F E^{\circ} = 2 \times 0.439 F - 3 \times 0.036 \mathrm{F}

-1 F E^{\circ} = 0.770 F

\therefore E^{\circ} = -0.770\;\mathrm{V}

\mathrm{O}^{--} > \mathrm{F}^{-} > \mathrm{N a}^{+} > \mathrm{M g}^{++} > \mathrm{A l}^{3+}

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