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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 17 of 32
Marks: +1, -0
Using the standard electrode potentials predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I−(aq)\mathrm{Fe^{3+}}_{(aq)} \text{ and } \mathrm{I^{-}}_{(aq)}
(ii) Ag+(aq) and Cu(s)\mathrm{Ag^{+}}_{(aq)} \text{ and } \mathrm{Cu}_{(s)}
(iii) Fe3+(aq) and Br−(aq)\mathrm{Fe^{3+}}_{(aq)} \text{ and } \mathrm{Br^{-}}_{(aq)}
(iv) Ag(s) and Fe3+(aq)\mathrm{Ag}_{(s)} \text{ and } \mathrm{Fe^{3+}}_{(aq)}
(v) Br2(aq) and Fe2+(aq)\mathrm{Br}_{2(aq)} \text{ and } \mathrm{Fe^{2+}}_{(aq)}
Given : E12I2/I−∘=0.54 V,E^\circ_{\frac{1}{2}\mathrm{I}_2/\mathrm{I}^{-}} = 0.54\,\text{V},
ECu2+/Cu∘=0.34 V,E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}} = 0.34\,\text{V},
E12Br2/Br−∘=1.09 V,E^\circ_{\frac{1}{2}\mathrm{Br}_2/\mathrm{Br}^{-}} = 1.09\,\text{V},
EAg+/Ag∘=0.80 V,E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} = 0.80\,\text{V},
EFe3+/Fe2+∘=0.77 VE^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} = 0.77\,\text{V}
Solution:  
A reaction is feasible if EMF of the cell reaction is positive.
(i) Fe(aq)3++I(aq)−→Fe(aq)2++12I2\mathrm{Fe}^{3+}_{(aq)} + \mathrm{I}^{-}_{(aq)} \rightarrow \mathrm{Fe}^{2+}_{(aq)} + \frac{1}{2} \mathrm{I}_2,
i.e., Pt∣I2∣I(aq)−∣∣Fe(aq)3+∣Fe(aq)2+∣Pt\text{i.e., } \mathrm{Pt}|\mathrm{I}_2|\mathrm{I}^{-}_{(aq)}||\mathrm{Fe}^{3+}_{(aq)}|\mathrm{Fe}^{2+}_{(aq)}|\mathrm{Pt}
∴Ecell∘=EFe3+/Fe2+∘−E12I2/I−∘\therefore E^\circ_{\text{cell}} = E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} - E^\circ_{\frac{1}{2}\mathrm{I}_2/\mathrm{I}^{-}}
=0.77−0.54=0.23 V(= 0.77 - 0.54 = 0.23\,\text{V}(Feasible))
(ii) Ag(aq)++Cu(s)→Ag(s)+Cu(aq)2+,\mathrm{Ag}^{+}_{(aq)} + \mathrm{Cu}_{(s)} \rightarrow \mathrm{Ag}_{(s)} + \mathrm{Cu}^{2+}_{(aq)},
i.e., Cu∣Cu(aq)2+∣∣Ag(aq)+∣Ag\text{i.e., } \mathrm{Cu}|\mathrm{Cu}^{2+}_{(aq)}||\mathrm{Ag}^{+}_{(aq)}|\mathrm{Ag}
Ecell∘=EAg+/Ag∘−ECu2+/Cu∘E^\circ_{\text{cell}} = E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}} - E^\circ_{\mathrm{Cu}^{2+}/\mathrm{Cu}}
=0.80−0.34=0.46 V(= 0.80 - 0.34 = 0.46\,\text{V}(Feasible))
(iii) Fe3+(aq)+Br−(aq)→Fe2+(aq)+12Br2,\mathrm{Fe}^{3+}(aq) + \mathrm{Br}^{-}(aq) \rightarrow \mathrm{Fe}^{2+}(aq) + \frac{1}{2}\mathrm{Br}_2,
i.e., Br2∣Br(aq)−∣∣Fe(aq)3+∣Fe(aq)2+\text{i.e., } \mathrm{Br}_2|\mathrm{Br}^{-}_{(aq)}||\mathrm{Fe}^{3+}_{(aq)}|\mathrm{Fe}^{2+}_{(aq)}
Ecell∘=EFe3+/Fe2+∘−E12Br2/Br−∘E^\circ_{\text{cell}} = E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} - E^\circ_{\frac{1}{2}\mathrm{Br}_2/\mathrm{Br}^{-}}
=0.77−1.09=−0.32 V= 0.77 - 1.09 = -0.32\,\text{V}(Not feasible)
(iv)Ag(s)+Fe(aq)3+→Ag(aq)++Fe(aq)2+\mathrm{Ag}_{(s)} + \mathrm{Fe}^{3+}_{(aq)} \rightarrow \mathrm{Ag}^{+}_{(aq)} + \mathrm{Fe}^{2+}_{(aq)}
i.e., Ag(s)∣Ag(aq)+∣∣Ag(aq)3+∣Ag(aq)2+\text{i.e., } \mathrm{Ag}_{(s)}|\mathrm{Ag}^{+}_{(aq)}||\mathrm{Ag}^{3+}_{(aq)}|\mathrm{Ag}^{2+}_{(aq)}
Ecell∘=EFe3+/Fe2+∘−EAg+/Ag∘E^\circ_{\text{cell}} = E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}} - E^\circ_{\mathrm{Ag}^{+}/\mathrm{Ag}}
=0.77−0.80=−0.03 V= 0.77 - 0.80 = -0.03\,\text{V} (Not feasible)
(v) 12Br2(aq)+Fe(aq)2+→Br(aq)−+Fe(aq)2+,\frac{1}{2}\mathrm{Br}_{2(aq)} + \mathrm{Fe}^{2+}_{(aq)} \rightarrow \mathrm{Br}^{-}_{(aq)} + \mathrm{Fe}^{2+}_{(aq)},
i.e., Fe(aq)2+∣Fe(aq)3+∣∣Br2∣Br−\text{i.e., } \mathrm{Fe}^{2+}_{(aq)}|\mathrm{Fe}^{3+}_{(aq)}||\mathrm{Br}_2|\mathrm{Br}^{-}
Ecell∘=E12Br2/Br−∘−EFe3+/Fe2+∘E^\circ_{\text{cell}} = E^\circ_{\frac{1}{2}\mathrm{Br}_2/\mathrm{Br}^{-}} - E^\circ_{\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}}
=1.09−0.77=0.32 V= 1.09 - 0.77 = 0.32\,\text{V} (Feasible)
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