An oxidation-reduction reaction in which 3 electrons are transferred has a $∆ G °$ of 17.35 kJ/mol at 25⁰C. Determine the value of $E_{cell}^{\circ} .$
$F = 96500 C / mol$

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answer is 0.06.

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Detailed Solution

$\begin{array}{l} {ΔG}^{0} = - {nFE}_{cell}^{0} \\ E_{cell}^{0} = - \frac{{ΔG}^{0}}{nF} = - \frac{17 .37 \times 10^{3} J}{3 \times 96500} \\ E_{cell}^{0} = - 0.06 V \end{array}$

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