Question

The energy consumed if the potential drop across the cell is $3.0V$ and $12.3$ g of nitrobenzene is reduced to aniline, with current efficiency for the process be 50%, will be:

• A. $494.8kJ$
• B. $347.4kJ$
• C. $394.8kJ$
• D. $698.8kJ$

Answer 1

The correct option is B $347.4kJ$

$C_6{H}_{5}N{O}_2+6H^{\oplus }+6e^{-}\to C_6{H}_{5}N{H}_2+2H_{2}O{N}^3+6e^{-}\to N^{3-}$

$\therefore$ Equivalent weight (Ew) of $C_6{H}_{5}N{O}_2=\frac{Mw}{6}=\frac{123}{6}$

Now, W = $\frac{Z.I.t}{96500}$

Since, current efficiency is 50 %

$\therefore$ I = $\frac{50I_0}{100}(I_0=Initialcurrent)$

$\therefore$ 12.3 = $\frac{123\times I\times t\times 50}{6\times 100\times 96500}$

Now, energy used = $8\times V$

= $115800\times 3=347.4kJ$