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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 24 of 32
Marks: +1, -0
The cell in which the following reaction occurs:
2Fe(aq)3++2I(aq)−→2Fe(aq)2++I2(s) has Ecell∘=0.236 V at 298 K.2\mathrm{Fe}^{3+}_{(\mathrm{aq})} + 2\mathrm{I}^{-}_{(\mathrm{aq})} \rightarrow 2\mathrm{Fe}^{2+}_{(\mathrm{aq})} + \mathrm{I}_{2(\mathrm{s})} \text{ has } E^\circ_{\text{cell}} = 0.236\,\text{V} \text{ at } 298\,\text{K}.Calculate the standard Gibb’s energy and the equilibrium constant of the cell reaction.
Solution:  
2Fe3++2e−→2Fe2+2\mathrm{Fe}^{3+} + 2e^{-} \rightarrow 2\mathrm{Fe}^{2+}
2I−→I2+2e−2\mathrm{I}^{-} \rightarrow \mathrm{I}_2 + 2e^{-}
—————————————-\text{----------------------------------------}
2Fe3++2I−→I2+2Fe2+;Ecell∘=0.236 V2\mathrm{Fe}^{3+} + 2\mathrm{I}^{-} \rightarrow \mathrm{I}_2 + 2\mathrm{Fe}^{2+} ; E^\circ_{\text{cell}} = 0.236\,\text{V}
—————————————-\text{----------------------------------------}
ΔG0=−nE0F\Delta G^0 = -n E^0 F
=−2×0.236×96500= -2 \times 0.236 \times 96500
=−45548 J mol−1= -45548\,\mathrm{J\,mol}^{-1}
=−45.54 kJ mol−1= -45.54\,\mathrm{kJ\,mol}^{-1}
ΔG0=−2.303RTlog⁡K\Delta G^{0} = -2.303 R T \log K
log⁡K=−ΔG02.303RT\log K = -\frac{\Delta G^{0}}{2.303 R T}
=−(−45548)2.303×8.314×298= \frac{-(-45548)}{2.303 \times 8.314 \times 298}
=7.98=7.98
⇒K=9.61×107\Rightarrow K = 9.61 \times 10^{7}
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