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Electrochemistry — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryElectrochemistry3 Marks
Question
The cell in which the following reaction occurs:
$2\text{Fe}^{3+}(\text{aq})+2\text{I}^-(\text{aq})\rightarrow2\text{Fe}^{2+}(\text{aq})+\text{I}_2(\text{s})\ \text{has}\ \text{E}^\circ_{\text{cell}}$ = 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
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Answer

The cell is
$2\text{Fe}^{3+}(\text{aq})+2\text{I}^-(\text{aq})\rightarrow2\text{Fe}^{2+}(\text{aq})+\text{I}_2(\text{s})2\text{Fe}_2+(\text{aq})+\text{I}_2(\text{s})$
$E^0 cell = 0.236 V^0_a$
$\Delta \mathrm{r} \mathrm{G}^0=-\mathrm{nFE}{ }^0$ cell
$= − 2 \times 0.236 \times 96487 C mol^{−1}$
$= −45541 J$
$\text{Log Kc}=\frac{\text{nFE}^\circ\text{cell}}{2.303\times{\text{RT}}}$
$=\frac{2\times96487\times0.236}{2.303\times8.31\times298}$
$=\frac{45541.864}{5703.1031}$
$= 7.9854$
Or $Kc$ = Antilog $7.9854$
Or $Kc = 9.62 \times 10^7$​​​​​​​

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