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NCERT Class XII Chemistry
Chapter - Electrochemistry
Questions with Solutions

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Question : 14 of 32
Marks: +1, -0
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O→O21 \text{ mol of } \mathrm{H_2O} \to \mathrm{O_2}?
(ii) 1 mol of FeO→Fe2O31 \text{ mol of } \mathrm{FeO} \to \mathrm{Fe_2O_3}?
Solution:  
(i) The electrode reaction for 1 mol of H2O\mathrm{H_2O} is given as
H2O→12O2\mathrm{H}_2 \mathrm{O} \rightarrow \frac{1}{2} \mathrm{O}_2
i.e., O2−→12O2+2e−\mathrm{O}^{2-} \rightarrow \frac{1}{2} \mathrm{O}_2 + 2 \mathrm{e}^-
∴ Quantity of electricity required =2F= 2 \mathrm{F}
=2×96500C=193000C= 2 \times 96500 \mathrm{C} = 193000 \mathrm{C}
(ii) The electrode reaction for 1 mol of FeO\mathrm{FeO} is given as
FeO→12Fe2O3\mathrm{FeO} \rightarrow \frac{1}{2} \mathrm{Fe}_2 \mathrm{O}_3
i.e., Fe2+→Fe3++e−\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^-
∴ Quantity of electricity required =1F=96500C= 1 \mathrm{F} = 96500 \mathrm{C}
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