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An oxidation-reduction reaction in which 3 electrons are transferred has a \Delta G^{0} of 17.37\; kJ\; mol^{-1} at 25^{\circ}C. The value of E_{cell}^{0} (in V) is ______\times 10^{-2}. (1\; F=96,500\; C\; mol^{-1})
 

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We know the formula 

\Delta \mathrm{G}^{\circ}=-\mathrm{nF}\textup{E}^{\circ}_{\textup{cell}}

Given 

n = 3 electrons

F =96500 C/mol

\Delta \mathrm{G}^{\circ} = 17.37 \mathrm{~kJ} \mathrm{~mol}^{-1}

Now, 

17.37 \times 1000 =-3 \times 96500 \times \mathrm{E}^{\circ}_{\textup{cell}}

\Rightarrow \mathrm{E}^{\circ}=-6 \times 10^{-2} \mathrm{~V}

Ans = (– 6)

Posted by

Kuldeep Maurya

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