Chemistry - Electrochemistry Question with Solution | TestHub

Reduction : ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2+6{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2+2{\mathrm{H}}_2\mathrm{O}$

Moles of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2=\frac{18.45}{123}=0.15\mathrm{mol}$

$6{\mathrm{e}}^{-}=6\mathrm{F}$ for $1\mathrm{mol}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2$

For $0.15\mathrm{mol}{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NO}}_2=6\times 0.15=0.9\mathrm{F}$

Current efficiency $=75\%$

$\begin{array}{rlrlrlr}& \text{ Number of Faraday required }=\frac{0.9\mathrm{F}\times 100}{75}=1.2F& & & & & \end{array}$

$=1.2\times 96500C$

$$\begin{gathered} \text{displaylines}Energyconsumed=Charge\times potentialdifference \\ =1.2\times 96500C\times 3V=347400J=347.4kJ \end{gathered}$$