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TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Chemistry

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Question : 144 of 160

Marks: +1, -0

The

E^{\circ}

\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+}=1.6\,\mathrm{V}

\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}=0.76\,\mathrm{V}

E^{\circ}

\mathrm{Fe}^{3+}

\mathrm{Ce}^{3+}

is

$+0.84\,\mathrm{V}$
$-0.84\,\mathrm{V}$
$-2.32\,\mathrm{V}$
$+1.5\,\mathrm{V}$

Solution:

\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+}\rightleftharpoons\mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}

\;E_{\mathrm{Ce}^{4+}/\mathrm{Ce}^{3+}}^{\circ}=1.6\,\mathrm{V}

\;E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\circ}=0.76\,\mathrm{V}

E_{\;\text{cell}\;}^{\circ}=E_{\;\text{cathode}\;}^{\circ}-E_{\;\text{anode}\;}^{\circ}

\;\mathrm{Fe}^{3+}+e^{-}\rightarrow\mathrm{Fe}^{2+}\;(E_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}^{\circ}=0.76\,\mathrm{V})

\;\mathrm{Ce}^{4+}+e^{-}\rightarrow\mathrm{Ce}^{3+}\;(E_{\mathrm{Ce}^{4+}}^{\circ}/\mathrm{Ce}^{3+}=1.6\,\mathrm{V})

\mathrm{Fe}^{3+}

is oxidising

\mathrm{Ce}^{3+}

, i.e. it is itself getting reduced.

Alsso,

E_{\;\text{cell}\;}^{\circ}=E_{\;\text{red}\;}^{\circ}-E_{\;\text{oxidation}\;}^{\circ}

=0.76-1.6=-0.84\,\mathrm{V}

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